| Metal Has One Oxidation State | Metal Has Two Oxidation States | |
|---|---|---|
(two elements, metal + nonmetal) |
Combine the name of the metal with the root name of the nonmetal and the suffix "ide". |
Combine the name of the metal with the Roman numeral corresponding to its oxidation state1, the root name of the nonmetal, and the suffix "ide". |
(three elements, metal + inorganic anion) |
Combine the name of the metal with the name of the anion. |
Combine the name of the metal with the Roman numeral corresponding to its oxidation state1 and the name of the anion. |
| Formula | Name | |
|---|---|---|
PROBLEM
Name the following compounds or ions:
(a) H3PO4, H2PO4-, HPO42-, PO43-
(b) H2SO4, HSO4-, SO42-
(c) H2CO3, HCO3-, CO32-
SOLUTION
(a) H3PO4 is phosphoric acid, a triprotic acid. H2PO4-, HPO42-, and PO43- are the anions resulting from first, second, and third stage ionization of phosphoric acid.
| Formula | Name |
|---|---|
(b) H2SO4 is sulfuric acid, a diprotic acid. HSO4- and SO42- are the anions resulting from first and second stage ionization of sulfuric acid.
| Formula | Name |
|---|---|
(c) H2CO3 is carbonic acid, a diprotic acid. HCO3- and CO32- are the anions resulting from first and second stage ionization of carbonic acid.
| Formula | Name |
|---|---|
PROBLEM
Name the following acids:
(a) HBr
(b) HCl
(c) HClO
(d) HClO2
(e) HClO3
(f) HClO4
(g) HF
(h) HI
(i) HNO2
(j) HNO3
(k) H2CO3
(l) H2S
(m) H2SO3
(n) H2SO4
(o) H3PO4
SOLUTION
The fifteen acids listed can be broken up into three categories:
| Category | Naming Procedure | Examples | (acids containing only two elements, hydrogen and a nonmetallic element) |
(a) hydrobromic acid
(b) hydrochloric acid (g) hydrofluoric acid (h) hydroiodic acid (l) hydrosulfuric acid2 |
|---|---|---|
(ternary acids containing hydrogen, oxygen, and another element, where oxygen and the other element form a common anion such as ClO3- [chlorate], NO3- [nitrate], CO32- [carbonate], SO42- [sulfate], or PO43- [phosphate]) |
(e) chloric acid
(j) nitric acid (k) carbonic acid (n) sulfuric acid3 (o) phosphoric acid3 |
|
(ternary acids containing hydrogen, oxygen, and another element, where oxygen and the other element form a less common anion such as ClO- [hypochlorite], ClO2- [chlorite], ClO4- [perchlorate], NO2- [nitrite], or SO32- [sulfite]) |
n+1: Add the prefix "per". n-1: Replace the suffix "ic" with "ous". n-2: Add the prefix "hypo" and replace the suffix "ic" with "ous". |
(c) hypochlorous acid
(d) chlorous acid (f) perchloric acid (i) nitrous acid (m) sulfurous acid |
2H2S is an exception. In this case, take the name of the nonmetallic element, sulfur, and add "hydro" as a prefix and "ic" as a suffix.
3H2SO4 and H3PO4 are exceptions. Back to Main Menu
PROBLEM
Identify the ions that exist in (a) sodium acetate, (b) aluminum oxide, and (c) potassium phosphate.
[from Mascetta, Joseph A. 2002, Barron's How to Prepare for the SAT II Chemistry, 7th Edition (Hauppauge, New York: Barron's Educational Series), Diagnostic Test problems 5-7]
SOLUTION
(a) Sodium acetate is NaCH3COO and consists of the sodium (Na+) and acetate (CH3COO-) ions.
(b) Aluminum oxide is Al2O3 and consists of aluminum (Al3+) and oxygen (O2-) ions.
(c) Potassium phosphate is K3PO4 and consists of potassium (K+) and phosphate (PO43-) ions.
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Balancing Chemical Equations
PROBLEM
Write the balanced equation and state the reaction type for each of the following reactions:
(a) zinc + sulfuric acid => zinc sulfate + hydrogen
(b) iron(III) chloride + sodium hydroxide => iron(III) hydroxide + sodium chloride
(c) aluminum hydroxide + sulfuric acid => aluminum sulfate + water
(d) potassium + water =>
(e) magnesium + oxygen =>
(f) silver nitrate + copper => copper(II) nitrate + silver(s)
(g) magnesium bromide + chlorine => magnesium chloride + bromine
[from Mascetta, Joseph A. 2002, Barron's How to Prepare for the SAT II Chemistry, 7th Edition (Hauppauge, New York: Barron's Educational Series), problems 6.1-6.7]
SOLUTION
(a) Zn(s) + H2SO4(aq) => ZnSO4(aq) + H2(g), single replacement, zinc replaces hydrogen
(b) FeCl3 + 3NaOH => Fe(OH)3 + 3NaCl, double replacement, hydroxyl and chlorine are interchanged
(c) Al(OH)3 + H2SO4 => Al2(SO4)3 + H2O
We can rewrite this as
Al(OH)3 + H2SO4 => Al2(SO4)3 + H(OH)
The balanced equation is
2Al(OH)3 + 3H2SO4 => Al2(SO4)3 + 6H(OH)
or
2Al(OH)3 + 3H2SO4 => Al2(SO4)3 + 6H2O, double replacement, hydroxyl and sulfate are interchanged
(d) 2K + 2H2O => 2KOH + H2(g), single replacement, potassium replaces hydrogen
(e) 2Mg + O2 => 2MgO, synthesis
(f) 2AgNO3(aq) + Cu(s) => Cu(NO3)2(aq) + 2Ag(s), single replacement, copper replaces silver
(g) MgBr2 + Cl2 => MgCl2 + Br2, single replacement, chlorine replaces bromine
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PROBLEM
Balance the following equations:
(a) N2O5 => N2O4 + O2
(b) KNO3 => KNO2 + O2
(c) NH4NO3 => N2O + H2O
(d) NH4NO2 => N2 + H2O
(e) NaHCO3 => Na2CO3 + H2O + CO2
(f) P4O10 + H2O => H3PO4
(g) HCl + CaCO3 => CaCl2 + H2O + CO2
(h) Al + H2SO4 => Al2(SO4)3 + H2
(i) CO2 + KOH => K2CO3 + H2O
(j) CH4 + O2 => CO2 + H2O
(k) Be2C + H2O => Be(OH)2 + CH4
(l) Cu + HNO3 => Cu(NO3)2 + NO + H2O
(m) S + HNO3 => H2SO4 + NO2 + H2O
(n) NH3 + CuO => Cu + N2 + H2O
[from Chang, Raymond 2002, Chemistry, Seventh Edition (New York: McGraw-Hill), problem 3.58]
SOLUTION
(a) 2N2O5 => 2N2O4 + O2
(b) 2KNO3 => 2KNO2 + O2
(c) NH4NO3 => N2O + 2H2O
(d) NH4NO2 => N2 + 2H2O
(e) 2NaHCO3 => Na2CO3 + H2O + CO2
(f) P4O10 + 6H2O => 4H3PO4
(g) 2HCl + CaCO3 => CaCl2 + H2O + CO2
(h) 2Al + 3H2SO4 => Al2(SO4)3 + 3H2
(i) CO2 + 2KOH => K2CO3 + H2O
(j) CH4 + 2O2 => CO2 + 2H2O
(k) Be2C + 4H2O => 2Be(OH)2 + CH4
(l) Cu + HNO3 => Cu(NO3)2 + NO + H2O
This equation is more complex than the previous ones because two elements, nitrogen and oxygen, appear in more than two reactants and/or products. We will assume that
A · Cu + B · HNO3 => C · Cu(NO3)2 + D · NO + E · H2O
and solve for A, B, C, D, and E by requiring that the numbers of different atoms are equal on both sides of the equation.
Cu: A = C
H: B = 2E
N: B = 2C + D
O: 3B = 6C + D + E
We have four equations in five unknowns, which is sufficient, since all we have to do is get the relative sizes of the coefficients. Set A = C = 1 and substitute into the last three equations.
B = 2E
B = 2 + D
3B = 6 + D + E
Replace B with 2E in the last two equations.
2E = 2 + D
6E = 6 + D + E => D = 5E - 6 => 2E = 2 + 5E - 6 => E = 4/3 => D = 20/3 - 6 = 2/3 => B = 2 + 2/3 = 8/3
We have A = 1, B = 8/3, C = 1, D = 2/3, and E = 4/3. Since we need integral coefficients, we multiply all of these by three and get A = 3, B = 8, C = 3, D = 2, and E = 4. Thus,
3Cu + 8HNO3 => 3Cu(NO3)2 + 2NO + 4H2O
(m) Assume that
A · S + B · HNO3 => C · H2SO4 + D · NO2 + E · H2O
and solve for A, B, C, D, and E by balancing the numbers of different atoms.
S: A = C
H: B = 2C + 2E
N: B = D
O: 3B = 4C + 2D + E
Set A = C = 1 and substitute into the last three equations.
B = 2 + 2E
B = D
3B = 4 + 2D + E
Replace B with D in the first and third equations.
D = 2 + 2E
3D = 4 + 2D + E => E = D - 4 => D = 2 + 2(D - 4) = 2 + 2D - 8 = 2D - 6 => D = 6 => E = D - 4 = 6 - 4 = 2 => B = 2 + 2E = 2 + (2)(2) = 6
We have A = 1, B = 6, C = 1, D = 6, and E = 2. The balanced equation is
S + 6HNO3 => H2SO4 + 6NO2 + 2H2O
(n) 2NH3 + 3CuO => 3Cu + N2 + 3H2O
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Stoichiometry
PROBLEM
In an experiment 1.056 g of a metal carbonate, containing an unknown metal M, is heated to give the metal oxide and 0.376 g CO2:
MCO3(s) + heat => MO(s) + CO2(g)
What is the identity of the metal M?
[from Kotz, John C., and Treichel, Paul Jr. 1996, Chemistry & Chemical Reactivity, Third Edition (New York: Saunders College Publishing), problem 5.69]
SOLUTION
The molar mass of CO2 is 12.011 g/mole + (2)(15.9994 g/mole) = 44.0098 g/mole, so 0.376 g CO2 = 8.544 x 10-3 mole CO2. For every mole of CO2 produced in the above reaction, one mole of MO is produced. Thus, 8.544 x 10-3 mole MO is produced with a mass of 1.056 g - 0.376 g = 0.68 g. The molar mass of MO is thus (0.68 g) / (8.544 x 10-3 mole) = 79.59 g/mole, and the molar mass of M is 79.59 g/mole - 15.9994 g/mole = 63.59 g/mole, so M is Cu (copper).
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PROBLEM
A compound containing the elements C, H, N, and O is analyzed. When a 1.2359 g sample is burned in excess oxygen, 2.241 g of CO2(g) is formed. The combustion analysis also showed that the sample contained 0.0648 g of H.
(a) Determine the mass, in grams, of C in the 1.2359 g sample of the compound.
(b) When the compound is analyzed for N content only, the mass percent of N is found to be 28.84 percent. Determine the mass, in grams, of N in the original 1.2359 g sample of the compound.
(c) Determine the mass, in grams, of O in the original 1.2359 g sample of the compound.
(d) Determine the empirical formula of the compound.
[from AP Chemistry 2006 Free-Response Questions, College Board, Question 3(a)]
SOLUTION
(a) The molar mass of carbon dioxide (CO2) is 44.011 g/mole. It follows that 2.241 g CO2 = 5.092 x 10-2 mole CO2. Thus, there is 5.092 x 10-2 mole C = 0.6116 g C in the 1.2359 g sample of the compound.
(b) The mass of N in the original sample is (0.2884)(1.2359 g) = 0.3564 g.
(c) The mass of O in the original sample is 1.2359 g - 0.0648 g - 0.6116 g - 0.3564 g = 0.2031 g.
(d) The sample contains
0.0648 g H = 6.429 x 10-2 mole H
5.092 x 10-2 mole C
0.3564 g N = 2.545 x 10-2 mole N
0.2031 g O = 1.269 x 10-2 mole O
For each mole of O, the sample contains
5.065 mole H
4.012 mole C
2.005 mole N
Therefore, the empirical formula of the compound is C4H5N2O.
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Limiting Reactant
PROBLEM
The important industrial chemical carbon tetrachloride, CCl4, is made by the reaction of carbon disulfide with chlorine according to the following unbalanced equation:
CS2(g) + Cl2(g) => S2Cl2(l) + CCl4(l)
What is the theoretical yield of CCl4 if 125 g of CS2 is mixed with 435 g of Cl2? What mass of excess reactant remains when the reaction has gone to completion?
[from Kotz, John C., and Treichel, Paul Jr. 1996, Chemistry & Chemical Reactivity, Third Edition (New York: Saunders College Publishing), problem 5.73]
SOLUTION
The balanced equation is
CS2(g) + 3Cl2(g) => S2Cl2(l) + CCl4(l)
The molar mass of CS2 is 76.143 g/mole. The molar mass of Cl2 is 70.9054 g/mole. The molar mass of CCl4 is 153.8218 g/mole. Initially, there is 125 g CS2 = 1.642 mole CS2 and 435 g Cl2 = 6.135 mole Cl2. To use up all of the CS2 would require (3)(1.642 mole) = 4.925 mole Cl2. Since there is 6.135 mole Cl2, the limiting reactant is CS2 and there will be excess Cl2 remaining. According to the above balanced equation, each mole of CS2 yields one mole of CCl4. Thus, 1.642 mole CCl4 = 252.5 g CCl4 is produced. The mass of Cl2 remaining is (6.135 mole - 4.925 mole)(70.9054 g/mole) = 85.79 g.
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Atomic Structure
PROBLEM
Suppose that a stable element with atomic number 119, symbol Q, has been discovered.
(a) Write the ground-state electron configuration for Q.
(b) Would Q be a metal or a nonmetal? Explain in terms of electron configuration.
(c) On the basis of periodic trends, would Q have the largest atomic radius in its group or would it have the smallest? Explain in terms of electronic structure.
(d) What would be the most likely charge of the Q ion in stable ionic compounds?
(e) Write a balanced equation that would represent the reaction of Q with water.
(f) Assume that Q reacts to form a carbonate compound. (i) Write the formula for the compound formed between Q and the carbonate ion, CO32-. (ii) Predict whether or not the compound would be soluble in water. Explain your reasoning.
[from AP Chemistry 2006 Free-Response Questions, College Board, Question 8]
SOLUTION
(a) Using the following diagram, we fill in the electronic structure along diagonals from upper right to lower left beginning with the 1s, 2s, 2p, 3p, 3d, 4d, 4f, and 5f shells.
1s
2s 2p
3s 3p 3d
4s 4p 4d 4f
5s 5p 5d 5f
6s 6p 6d
7s 7p
8s
For element Q (Z = 119), we get
[1s2][2s22p6][3s23p6][4s23d104p6][5s24d105p6][6s24f145d106p6][7s25f146d107p6][8s1]
Colors identify subshells which belong to the same diagonal. Square brackets indicate periods in the periodic table.
(b) Q has a single valence electron in the 8s subshell and is therefore an alkali metal.
(c) Q would have the largest atomic radius in its group, the alkali metals.
(d) The charge of Q in stable ionic compounds would be +1.
(e) Q would react with water by replacing the hydrogen according to:
2Q + 2H2O => 2QOH + H2
This is a hydrogen displacement redox reaction.
(f) (i) The formula of the compound formed between Q and the carbonate ion would be Q2CO3. (ii) Carbonates containing alkali metal ions are soluble in aqueous solution, so Q2CO3 would be soluble in water.
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Lewis Structures
PROBLEM
Write the Lewis structure(s) or Lewis dot symbol for each of the following molecules and indicate which have resonance structures:
(a) sulfur dioxide (SO2)
(b) ethane (C2H6)
(c) chlorine (Cl2)
(d) hydrogen bromide (HBr)
(e) sodium chloride (NaCl)
[from Mascetta, Joseph A. 2002, Barron's How to Prepare for the SAT II Chemistry, 7th Edition (Hauppauge, New York: Barron's Educational Series), Diagnostic Test problem 64]
SOLUTION
(a) Sulfur dioxide has two Lewis structures, which are resonance structures:
.. .. .. .. .. ..
O===S---O: <-> :O---S===O
.. .. .. ..
(b) ethane:
H H
| |
H--C--C--H
| |
H H
(c) chlorine:
.. ..
:Cl---Cl:
.. ..
(d) hydrogen bromide:
..
H---Br:
..
(e) Sodium chloride is an ionic compound:
..
Na+ :Cl:-
..
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PROBLEM
Write Lewis structures for the following four isoelectronic species: (a) CO, (b) NO+, (c) CN-, (d) N2. Show formal charges.
[from Chang, Raymond 2002, Chemistry, Seventh Edition (New York: McGraw-Hill), problem 9.104]
SOLUTION
(a) CO (10 valence electrons):
-1 +1
:C///O: /// = triple bond
(b) NO+ (10 valence electrons):
0 +1
:N///O: /// = triple bond
(c) CN- (10 valence electrons):
-1 0
:C///N: /// = triple bond
(d) N2 (10 valence electrons)
0 0
:N///N: /// = triple bond
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Valence Shell Electron Pair Repulsion (VSEPR) Model
PROBLEM
Determine if each of the following species is planar or non-planar:
(a) CO32-
(b) NO3-
(c) ClF3
(d) BF3
(e) PCl3
[from Sample Questions for Chemistry Advanced Placement Test, College Board, Question 4]
SOLUTION
(a) CO32- (carbonate) has 24 valence electrons. Its Lewis structure is:
.. .. .. .. .. ..
O===C---O: <=> :O---C===O <=> :O---C---O:
.. | .. .. | .. .. || ..
| | ||
:O: :O: :O:
.. ..
There are three resonance structures. Their shape is approximately trigonal planar.
(b) NO3- (nitrate) has 24 valence electrons. Its Lewis structure is:
.. .. .. .. .. ..
O===N---O: <=> :O---N===O <=> :O---N---O:
.. | .. .. | .. .. || ..
| | ||
:O: :O: :O:
.. ..
There are three resonance structures, which are all planar. Their shape is approximately trigonal planar.
(c) ClF3 (chlorine trifluoride) has 28 valence electrons. Its Lewis structure is:
.. .... ..
:F---Cl---F:
.. | ..
|
:F:
..
There are two lone electron pairs and three bonding pairs surrounding the central chlorine atom. The molecule is T shaped and therefore planar.
(d) BF3 (boron trifluoride) has 24 valence electrons. Its Lewis structure is:
.. .. .. .. .. ..
F===B---F: <=> :F---B===F <=> :F---B---F:
.. | .. .. | .. .. || ..
| | ||
:F: :F: :F:
.. ..
There are three resonance structures, which are all planar. Their shape is approximately trigonal planar.
(e) PCl3 (phosphorus trichloride) has 26 valence electrons. Its Lewis structure is:
.. .. ..
:Cl---P---Cl:
.. | ..
|
:Cl:
..
There is one lone electron pair and three bonding pairs surrounding the central phosphorus atom. The molecule is trigonal pyramidal and therefore non-planar.
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PROBLEM
Write the Lewis structure for each of the following and use it and the valence-shell electron-pair repulsion (VSEPR) model to predict the geometry of the molecule:
(a) carbon dioxide (CO2)
(b) carbon tetrachloride (CCl4)
SOLUTION
(a) The carbon has four valence electrons, while each of the two oxygens has six. Thus, we have a total of sixteen valence electrons to account for. The Lewis structure for carbon dioxide is
.. ..
O===C===O
.. ..
There is a double bond between the carbon and each of the oxygens. Since the central atom, carbon, has two double bonds and no lone pairs, the VSEPR model predicts that the carbon dioxide molecule is linear, which is consistent with experiment.
(b) The carbon has four valence electrons, while each of the four chlorines has seven. Thus, we have a total of 32 valence electrons to account for. The Lewis structure for carbon tetrachloride is
..
:Cl:
|
.. | ..
:Cl---C---Cl:
.. | ..
|
:Cl:
..
There is a single bond between the carbon and each of the chlorines. Since the central atom, carbon, has four single bonds and no lone pairs, the VSEPR model predicts that the carbon tetrachloride molecule is tetrahedral.
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Hybrid Atomic Orbitals
PROBLEM
For each of the following hybrid atomic orbitals, identify (i) which pure atomic orbitals are used, (ii) the shape of the hybrid atomic orbitals, (iii) the angle between bonded atoms, and (iv) an example of a molecule which exhibits this type of hybrid atomic orbital, without any lone pairs, and its Lewis structure:
(a) sp
(b) sp2
(c) sp3
(d) sp3d
(e) sp3d2
SOLUTION
(a) sp: one s and one p orbital, linear, 180°, BeCl2
.. ..
:Cl--Be--Cl:
.. ..
(b) sp2: one s and two p orbitals, trigonal planar, 120°, BF3
..
:F:
.. | ..
:F--B--F:
.. ..
(c) sp3: one s and three p orbitals, tetrahedral, 109.5°, CH4
H
|
H--C--H
|
H
(d) sp3d: one s, three p, and one d orbital, trigonal bipyramidal, 90° and 120°, PCl5
..
:Cl:
|
.. | ..
:Cl---P---Cl:
.. / \ ..
/ \
:Cl: :Cl:
.. ..
(e) sp3d2: one s, three p, and two d orbitals, octahedral, 90°, SF6
.. ..
:F: :F:
\ /
.. \ / ..
:F---S---F:
.. / \ ..
/ \
:F: :F:
.. ..
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Covalent Bonds
PROBLEM
(a) Draw the complete Lewis structure (electron-dot diagram) for CF4, PF5 and SF4.
(b) On the basis of the Lewis structures drawn, answer the following questions about the particular molecule indicated:
(i) What is the F-C-F bond angle in CF4?
(ii) What is the hybridization of the valence orbitals of P in PF5?
(iii) What is the geometric shape formed by the atoms in SF4?
(c) Two Lewis structures can be drawn for the OPF3 molecule, as shown below.
..
:O: :O:
.. || .. .. | ..
:F--P--F: :F--P--F:
.. | .. .. | ..
:F: :F:
.. ..
structure 1 structure 2
(i) How many sigma bonds and how many pi bonds are in structure 1?
(ii) Which of the two structures best represents a molecule of OPF3? Justify your answer in terms of formal charge.
[from 2005 AP Chemistry Free-Response Question 6]
SOLUTION
(a) carbon tetrafluoride (CF4):
..
:F:
|
.. | ..
:F---C---F:
.. | ..
|
:F:
..
phosphorus pentafluoride (PF5):
.. ..
:F: :F:
\ /
.. \ / ..
:F---P---F:
.. | ..
|
:F:
..
sulfur tetrafluoride (SF4)
.. ..
:F: :F:
\ /
.. \ / ..
:F---S---F:
.. .. ..
(b)(i) Carbon tetrafluoride has four electron pairs surrounding the central carbon and no lone pairs. The geometry is tetrahedral. The F-C-F bond angle is 109.5°.
(ii) Phosphorus pentafluoride has five electron pairs surrounding the central phosphorus and no lone pairs. The geometry is trigonal bipyramidal. The hybridization that corresponds to this geometry is sp3d.
(iii) Sulfur tetrafluoride has five electron pairs surrounding the central sulfur. One of these is a lone pair. The corresponding geometry is distorted tetrahedral or seesaw.
(c)(i) Structure 1 has four sigma bonds and one pi bond. The double bond consists of one sigma bond and one pi bond.
(ii) In structure 1, the formal charges are all zero. In structure 2, the formal charges are +1 on the phosphorus, -1 on the oxygen, and zero on the fluorines. When a molecule is neutral, like this one, the Lewis structure in which there are no formal charges is preferred, so structure 1 is the better representation.
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Bond Energies
PROBLEM
(a) From the following data calculate the bond energy of the F2- ion.
F2(g) => 2F(g), DH10 = 156.9 kJ
F -(g) => F(g) + e-, DH20 = 333 kJ
F2-(g) => F2(g) + e-, DH30 = 290 kJ
(b) Explain the difference between the bond energies of F2 and F2-.
[from Chang, Raymond 2002, Chemistry, Seventh Edition (New York: McGraw-Hill), problem 9.108]
SOLUTION
(a) The bond energy of F2-(g) is the energy required to break it up into F(g) and F -(g). F2-(g) can be converted into F2(g) as follows:
DH0 DH20 -DH10
F2-(g) => F(g) + F-(g) => F(g) + F(g) + e- => F2(g) + e-
The total enthalpy change for the above process is
DH30 = DH0 + DH20 - DH10
=> DH0 = DH30 - DH20 + DH10 = 290 kJ - 333 kJ + 156.9 kJ = 113.9 kJ
(b) The bond energy of F2(g) is the energy required to break it up into 2F(g), which is 156.9 kJ. The additional electron in F - increases the repulsion between the electron clouds in F and F -, resulting in a weaker bond in F2- than in F2(g).
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PROBLEM
Experiments show that it takes 1656 kJ/mole to break all the bonds in methane (CH4) and 4006 kJ/mole to break all the bonds in propane (C3H8). Based on these data, calculate the average bond energy of the C-C bond.
[from Chang, Raymond 2002, Chemistry, Seventh Edition (New York: McGraw-Hill), problem 9.118]
SOLUTION
The methane (CH4) molecule has four C-H bonds:
H
|
H--C--H
|
H
Each mole of CH4 contains 4 moles of C-H bonds. Thus, the average bond energy of the C-H bond is (1656 kJ/mole)/4 = 414 kJ/mole. The propane (C3H8) molecule has eight C-H bonds and two C-C bonds:
H H H
| | |
H--C--C--C--H
| | |
H H H
Each mole of C3H8 contains 8 moles of C-H bonds and 2 moles of C-C bonds. Thus, the average bond energy of the C-C bond is [4006 kJ/mole - (8)(414 kJ/mole)] / 2 = 347 kJ/mole
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Gases
PROBLEM
An unknown gas is one of three possible gases: nitrogen, hydrogen, or oxygen. For each of the three possibilities, describe the result expected when the gas is tested using a glowing splint (a wooden stick with one end that has been ignited and extinguished, but still contains hot, glowing, partially burned wood).
[from 2005 AP Chemistry Free-Response Question 5]
SOLUTION
nitrogen: The splint will completely stop burning and cool.
hydrogen: The gas will ignite.
oxygen: The splint will start burning again with a flame.
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PROBLEM
Summarize the following ideal gas laws:
(a) Boyle's law
(b) Charles's law
(c) Gay-Lussac's law
(d) Dalton's law
SOLUTION
(a) Boyle's law, discovered by English scientist Robert Boyle in 1662, states that, for a fixed amount of gas at constant temperature, the product of the pressure P and the volume V is a constant, or PV = constant.
(b) Charles's law, discovered by French chemist Jacques Charles, states that, for a fixed amount of gas at constant pressure, the volume V is proportional to the absolute temperature T or V/T = constant.
(c) Gay-Lussac's law, discovered by French scientist Joseph Gay-Lussac in 1808, states that the ratio between the combining volumes of gases and the volume of the product, if gaseous, can be expressed in small whole numbers.
(d) Dalton's law of partial pressures, discovered by English scientist and school teacher John Dalton in 1801, states that the total gas pressure of a mixture of gases is equal to the sum of the partial pressures of the gases that make up the mixture.
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PROBLEM
A compound which has the empirical formula CH2Br has a vapor density of 6.00 g/L at 375 K and 0.983 atm. Using these data, determine the following:
(a) The molar mass of the compound.
(b) The molecular formula of the compound.
[from AP Chemistry 2006 Free-Response Questions, College Board, Question 3(b)]
SOLUTION
(a) The molar density of an ideal gas is
n/V = p/RT
where p is the pressure, R = 0.0821 L atm / mole K is the gas constant, and T is the temperature. The density of an ideal gas is
r = Mn/V = Mp/RT
where M is the molar mass. Thus, the molar mass of the compound is
M = rRT/p = (6.00 g/L)(0.0821 L atm / mole K)(375 K) / (0.983 atm) = 187.9196 g/mole
(b) The molar mass of CH2Br is 93.9268 g/mole. Therefore, the molecular formula of the compound is C2H4Br2.
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Solubility
PROBLEM
Each of three beakers contains a 0.1 M solution of one of the following solutes: potassium chloride, silver nitrate, or sodium sulfide. The three beakers are labeled randomly as solution 1, solution 2, and solution 3. Mixing solution 1 with solution 2 results in a black precipitate. Mixing solution 2 with solution 3 produces no reaction.
(a) Write the chemical formula of the black precipitate.
(b) Describe the expected results of mixing solution 1 with solution 3.
(c) Identify each of the solutions 1, 2, and 3.
[from 2005 AP Chemistry Free-Response Question 5]
SOLUTION
KCl, AgNO3, and Na2S are all soluble in aqueous solution. The following table shows the solubility of various combinations of the available ions.
| Cl- | NO3- | S2- | |
|---|---|---|---|
soluble |
soluble |
soluble |
|
insoluble |
soluble |
insoluble |
|
soluble |
soluble |
(a) There are only two possible precipitates, AgCl and Ag2S. AgCl is a white precipitate. Ag2S is a black precipitate. It forms on silver when it is exposed to the hydrogen sulfide of the atmosphere. So the black precipitate is Ag2S.
(c) In order for Ag2S to form when solutions 1 and 2 are mixed, one of the solutions must be AgNO3 and the other must be Na2S. In order for no reaction to occur when solutions 2 and 3 are mixed, one must be KCl and the other must be Na2S. Solution 2 is common to both mixtures, so it is Na2S. Solution 1 is AgNO3 and solution 3 is KCl.
(b) The result of mixing solutions 1 and 3, AgNO3 and KCl, is to produce a white precipitate, AgCl. Back to Main Menu
Concentration of Solutions
Molarity is the number of moles of solute per liter of solution.
The gram-equivalent weight or equivalent of a compound is the weight in grams of the compound which yields or reacts with one mole of hydrogen ions or hydroxyl ions.
Normality is the number of equivalents of solute per liter of solution.
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PROBLEM
(a) What is the gram-formula mass of AlCl3?
(b) What is the gram-equivalent mass of AlCl3?
SOLUTION
(a) The gram-formula mass of AlCl3 is the mass, in grams, of one mole of AlCl3, which is 26.98154 g + (3)(35.453 g) = 133.3405 g
(b) The gram-equivalent mass of AlCl3 is the mass, in grams, of AlCl3 which yields or reacts with one mole of hydrogen or hydroxyl ions. Since each of the three Cl- ions in one molecule of AlCl3 can react with one a hydrogen ion, the gram-equivalent mass of AlCl3 is 1/3 of its gram-formula mass or (1/3)(133.3405 g) = 44.44685 g.
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PROBLEM
Determine the equivalent weights (i.e., the weight of one "equivalent") of each of the following:
(a) H2SO4
(b) H3PO4
(c) Ca(OH)2
(d) AlCl3
(e) K2SO4
(f) NaCl
(g) Na3PO4
(h) CuSO4
(i) KOH
(j) Fe2O3
[from Kavanah, Patrick, and Robbins, Jack 1971, Concepts in Modern Chemistry (New York: Cambridge Book Company), problem 14.13]
SOLUTION
(a) One mole of H2SO4 contains two moles of H+ and is thus equal to two equivalents of H2SO4. Therefore, one equivalent of H2SO4 is 0.5 mole of H2SO4 or (0.5 mole)[(2)(1.008 g/mole) + 32.07 g/mole + (4)(16.00 g/mole)] = 49.04 g.
(b) One mole of H3PO4 contains three moles of H+ and is thus equal to three equivalents of H3PO4. Therefore, one equivalent of H3PO4 is 1/3 mole of H3PO4 or (1/3 mole)[(3)(1.008 g/mole) + 30.97 g/mole + (4)(16.00 g/mole)] = 32.66 g.
(c) One mole of Ca(OH)2 contains two moles of OH- and is thus equal to two equivalents of Ca(OH)2. Therefore, one equivalent of Ca(OH)2 is 0.5 mole of Ca(OH)2 or (0.5 mole)[40.08 g/mole + (2)(16.00 g/mole) + (2)(1.008 g/mole)] = 37.048 g.
(d) One mole of AlCl3 contains three moles of Cl-, which is capable of reacting with three moles of H+, and is thus equal to three equivalents of AlCl3. Therefore, one equivalent of AlCl3 is 1/3 mole of AlCl3 or (1/3 mole)[26.98 g/mole + (3)(35.45 g/mole)] = 44.44 g.
(e) One mole of K2SO4 contains one mole of SO42-, which is capable of reacting with two moles of H+, and is thus equal to two equivalents of K2SO4. Therefore, one equivalent of K2SO4 is 0.5 mole of K2SO4 or (0.5 mole)[(2)(39.10 g/mole) + 32.07 g/mole + (4)(16.00 g/mole)] = 87.14 g.
(f) One mole of NaCl contains one mole of Cl-, which is capable of reacting with one mole of H+, and is thus equal to one equivalent of NaCl. Therefore, one equivalent of NaCl is (1 mole)(22.99 g/mole + 35.45 g/mole) = 58.44 g.
(g) One mole of Na3PO4 contains one mole of PO43-, which is capable of reacting with three moles of H+, and is thus equal to three equivalents of Na3PO4. Therefore, one equivalent of Na3PO4 is 1/3 mole of Na3PO4 or (1/3 mole)[(3)(22.99 g/mole) + 30.97 g/mole + (4)(16.00 g/mole)] = 54.65 g.
(h) One mole of CuSO4 contains one mole of SO42-, which is capable of reacting with two moles of H+, and is thus equal to two equivalents of CuSO4. Therefore, one equivalent of CuSO4 is 0.5 mole of CuSO4 or (0.5 mole)[63.55 g/mole + 32.07 g/mole + (4)(16.00 g/mole)] = 79.81 g.
(i) One mole of KOH contains one mole of OH- and is thus equal to one equivalent KOH. Therefore, one equivalent of KOH is (1 mole)(39.10 g/mole + 16.00 g/mole + 1.008 g/mole) = 56.11 g.
(j) One mole of Fe2O3 contains two moles of Fe3+, which is capable of reacting with six moles of OH-, and is thus equal to six equivalents of Fe2O3. Therefore, one equivalent of Fe2O3 is 1/6 mole of Fe2O3 or (1/6 mole)[(2)(55.85 g/mole) + (3)(16.00 g/mole)] = 26.62 g.
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PROBLEM
What is the normality of (a) a 1 M HCl solution, (b) a 0.005 M NaOH solution, (c) a 0.05 M H3PO4 solution, (d) a 0.4 M Mg(OH)2 solution?
[from Kavanah, Patrick, and Robbins, Jack 1971, Concepts in Modern Chemistry (New York: Cambridge Book Company), reasoning exercise 15.6]
SOLUTION
(a) Each mole of HCl contains one mole of H+ ions. Therefore, each mole of HCl is equal to one equivalent of HCl. The normality of a 1 M HCl solution is 1 N.
(b) Each mole of NaOH contains one mole of OH- ions. Therefore, each mole of NaOH is equal to one equivalent of NaOH. The normality of a 0.005 M NaOH solution is 0.005 N.
(c) Each mole of H3PO4 contains three moles of H+ ions. Therefore, each mole of H3PO4 is equal to three equivalents of H3PO4. The normality of a 0.05 M H3PO4 solution is 0.15 N.
(d) Each mole of Mg(OH)2 contains two moles of OH- ions. Therefore, each mole of Mg(OH)2 is equal to two equivalents of Mg(OH)2. The normality of a 0.4 M Mg(OH)2 solution is 0.8 N.
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Boiling Point Elevation
PROBLEM
If 1 mole of each of the following substances was dissolved in 1000 g of water, what would be the boiling point?
(a) sodium chloride (NaCl)
(b) potassium chloride (KCl)
(c) calcium chloride (CaCl2)
(d) cellulose (C6H10O5)
(e) sucrose (C12H22O11)
[from Mascetta, Joseph A. 2002, Barron's How to Prepare for the SAT II Chemistry, 7th Edition (Hauppauge, New York: Barron's Educational Series), Diagnostic Test problem 47]
SOLUTION
(a) The boiling point elevation of a water solution is given by
DTb = Kbm
where Kb = 0.52°C/m and m is the molality of the solution (the number of moles of solute particles per 1000 g of solvent). Sodium chloride is soluble in aqueous solution and yields two solute particles (Na+ and Cl-) per molecule. Thus, the molality of solute particles is (2 moles) / (1 kg) = 2 moles/kg = 2 m and
DTb = (0.52°C/m)(2 m) = 1.04°C
The boiling point is
Tb = 100°C + 1.04°C = 101.04°C
(b) Potassium chloride is soluble in aqueous solution and yields two solute particles (K+ and Cl-) per molecule. Thus, the molality of solute particles is 2 m and the boiling point is Tb = 101.04°C, as in part (a).
(c) Calcium chloride is soluble in aqueous solution and yields three solute particles (Ca2+ and 2Cl-) per molecule. Thus, the molality of solute particles is 3 m and
DTb = (0.52°C/m)(3 m) = 1.56°C => Tb = 100°C + 1.56°C = 101.56°C
(d) Cellulose does not dissociate into more than one particle. Thus, the molality of solute particles is 1 m and
DTb = (0.52°C/m)(1 m) = 0.52°C => Tb = 100°C + 0.52°C = 100.52°C
(e) Sucrose does not dissociate into more than one particle. Thus, the molality of solute particles is 1 m and
DTb = (0.52°C/m)(1 m) = 0.52°C => Tb = 100°C + 0.52°C = 100.52°C
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Freezing Point Depression
PROBLEM
Fishes in the Antarctic Ocean swim in water at about - 2°C. (a) To prevent their blood from freezing, what must be the concentration (in molality) of the blood? Is this a reasonable physiological concentration? (b) In recent years scientists have discovered a special type of protein in these fishes' blood which, although present in quite low concentrations (< 0.001 m), has the ability to prevent the blood from freezing. Suggest a mechanism for its action.
[from Chang, Raymond 2002, Chemistry, Seventh Edition (New York: McGraw-Hill), problem 12.122]
SOLUTION
(a) The freezing point depression of water is
DTf = Kfm
where Kf is the molal freezing-point depression constant, which is 1.86 °C/m for water. Thus,
m > DTf/Kf = (2°C) / (1.86°C/m) = 1.075 m
A list of reference ranges for common human blood tests appears at http://en.wikipedia.org/wiki/Reference_ranges_for_common_blood_tests. If we add up the mean concentrations of all electrolytes, metabolites, ions, and trace metals that are listed, we get 0.154 mole/L = 0.154 M. The corresponding molality is approximately 0.154 m. Thus, the molality needed to depress the freezing point by the required amount is substantially larger than the total molality of the above substances in normal human blood, and appears to be an unreasonable physiological concentration.
(b) The protein could attract water molecules and prevent them from bonding with each other and forming ice crystals.
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Osmotic Pressure
PROBLEM
In the apparatus shown below, what will happen if the membrane is (a) permeable to both water and the Na+ and Cl- ions, (b) permeable to water and Na+ ions but not to Cl- ions, (c) permeable to water but not to Na+ and Cl- ions?
| | | |
|_| |_|
| | | |
+---+ +---+-+---+ +---+
| |/| |
| |/| |
| 0.01 M |/| 0.1 M |
| NaCl |/| NaCl |
| |/| |
| |/| |
+---------+-+---------+
^
|
membrane
[from Chang, Raymond 2002, Chemistry, Seventh Edition (New York: McGraw-Hill), problem 12.104]
SOLUTION
(a) If the membrane is permeable to both water and the Na+ and Cl- ions, the Na+ and Cl- ions will diffuse through the membrane until both sides contain a 0.055 M solution of NaCl. The water level will be the same on both sides.
(b) If the membrane is permeable to both water and Na+ ions, but not to Cl- ions, the Na+ will not diffuse to the left because, if it did, there would be a net positive charge on the left and a net negative charge on the right. The osmotic pressure will be higher on the right side, causing the water level to increase on the right side and decrease on the left side.
(c) If the membrane is permeable to water but not to Na+ or Cl- ions, the osmotic pressure will be higher on the right, and water will diffuse from the left to the right, causing the water level to decrease on the left and increase on the right.
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Chemical Equilibrium
PROBLEM
Answer the following questions that relate to solubility of salts of lead and barium.
(a) A saturated solution is prepared by adding excess PbI2(s) to distilled water to form 1.0 L of solution at 25°C. The concentration of Pb2+(aq) in the saturated solution is found to be 1.3 x 10-3 M. The chemical equation for the dissociation of PbI2(s) in water is shown below.
PbI2 <=> Pb2+(aq) + 2I -(aq)
(i) Write the equilibrium constant expression for the equation.
(ii) Calculate the molar concentration of I -(aq) in the solution.
(iii) Calculate the value of the equilibrium constant Ksp.
(b) A saturated solution is prepared by adding PbI2(s) to distilled water to form 2.0 L of solution at 25°C. What are the molar concentrations of Pb2+ and I - in the solution? Justify your answer.
(c) Solid NaI is added to a saturated solution of PbI2 at 25°C. Assuming that the volume of the solution does not change, does the molar concentration of Pb2+(aq) in the solution increase, decrease, or remain the same? Justify your answer.
(d) The value of Ksp for the salt BaCrO4 is 1.2 x 10-10. When a 500. mL sample of 8.2 x 10-6 M Ba(NO3)2 is added to 500. mL of 8.2 x 10-6 M Na2CrO4, no precipitate is observed.
(i) Assuming that volumes are additive, calculate the molar concentrations of Ba2+(aq) and CrO42-(aq) in the 1.00 L of solution.
(ii) Use the molar concentrations of Ba2+(aq) ions and CrO42-(aq) ions as determined above to show why a precipitate does not form. You must include a calculation as part of your answer.
[from AP Chemistry 2006 Free-Response Questions, College Board, Question 1]
SOLUTION
(a) (i) Ksp = [Pb2+][I -]2
(ii) For each Pb2+ in solution, there are two I -. Thus,
[I -] = 2[Pb2+] = (2)(1.3 x 10-3 M) = 2.6 x 10-3 M
(iii) Ksp = (1.3 x 10-3)(2.6 x 10-3)2 = 8.788 x 10-9
(b) The PbI2 will dissolve into Pb2+ and I - ions in the same proportion as in part (a) until
[Pb2+][I -]2 is equal to Ksp = 8.788 x 10-9. Thus, the final values of [Pb2+] and [I -] will be the same as in part (a).
(c) When solid NaI is added to the saturated solution of PbI2, it breaks up into Na+ and I - ions. The concentration of I - ions increases. In order for the product [Pb2+][I -]2 to remain the same, the concentration of Pb2+ must decrease, and some PbI2 will precipitate.
(d) Nitrates are soluble in aqueous solution. Chromates containing alkali metal ions are also soluble in aqueous solution. Therefore, Ba(NO3)2 and Na2CrO4 are both soluble in aqueous solution.
(i) The molar concentrations of Ba2+(aq) and CrO42-(aq) in the final solution are both 4.1 x 10-6 M.
(ii) In the final solution, [Ba2+][CrO42-] = (4.1 x 10-6)2 = 1.681 x 10-11 < Ksp, so no precipitate forms.
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PROBLEM
Pure nitrosyl chloride (NOCl) gas is heated to 240°C in a 1.00-L container. The NOCl decomposes according to
2NOCl(g) <=> 2NO(g) + Cl2(g)
At equilibrium the total pressure was 1.00 atm and the NOCl pressure was 0.64 atm.
(a) Calculate the partial pressures of NO and Cl2 in the system at equilibrium.
(b) Calculate the equilibrium constant KP.
[from Chang, Raymond 2002, Chemistry, Seventh Edition (New York: McGraw-Hill), problem 14.62]
SOLUTION
Let P0 be the initial pressure of the NOCl and let x be the amount (in atm) of NOCl that decomposes. At equilibrium, the partial pressures of NOCl, NO, and Cl2 are
P(NOCl) = P0 - x = 0.64 atm (1)
P(NO) = x
P(Cl2) = x/2
The total pressure is
P(NOCl) + P(NO) + P(Cl2) = P0 - x + x + x/2 = 1.00 atm (2)
Combining (1) and (2), we get
0.64 atm + 3x/2 = 1.00 atm => x = (2/3)(1.00 atm - 0.64 atm) = 0.24 atm
(a) The partial pressures of NO and Cl2 are P(NO) = 0.24 atm and P(Cl2) = 0.12 atm.
(b) The equilibrium constant is
KP = [P(NO)]2P(Cl2) / [P(NOCl)]2 = (0.24 atm)2(0.12 atm) / (0.64 atm)2 = 0.016875
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PROBLEM
A mixture of 0.47 mole of H2 and 3.59 moles of HCl is heated to 2800°C. Calculate the equilibrium partial pressures of H2, Cl2, and HCl if the total pressure is 2.00 atm. For the reaction
H2(g) + Cl2(g) <=> 2HCl(g)
KP is 193 at 2800°C.
[from Chang, Raymond 2002, Chemistry, Seventh Edition (New York: McGraw-Hill), problem 14.72]
SOLUTION
Let x be the number of moles of H2 that are produced in the reaction. At equilibrium, the number of moles of H2, Cl2, and HCl are
n(H2) = 0.47 + x
n(Cl2) = x
n(HCl) = 3.59 - 2x
The partial pressures are
P(H2) = n(H2)RT/V = (0.47 + x)RT/V
P(Cl2) = n(Cl2)RT/V = xRT/V
P(HCl) = n(HCl)RT/V = (3.59 - 2x)RT/V
The total pressure is
P(H2) + P(Cl2) + P(HCl) = (0.47 + x + x + 3.59 - 2x)RT/V = 4.06RT/V = 2.00 atm => RT/V = (2.00/4.06) atm/mole = (1.00/2.03) atm/mole
The equilibrium constant is
KP = [P(HCl)]2 / [P(H2)P(Cl2)]
= (3.59 - 2x)2(RT/V)2 / [(0.47 + x)(RT/V)(xRT/V)]
= (3.59 - 2x)2 / [(0.47 + x)x] = 193
Cross multiplying, we get
(3.59 - 2x)2 = 193x(0.47 + x) => (3.59)2 - 7.18x + 4x2 = (193)(0.47)x + 193x2
=> 189x2 + [7.18 + (193)(0.47)]x - (3.59)2 = 0 => ax2 + bx + c = 0
where
a = 189
b = 7.18 + (193)(0.47)
c = - (3.59)2
Solve for x using the quadratic formula.
x = [- b ± sqrt(b2 - 4ac)] / 2a => [- b + sqrt(b2 - 4ac)] / 2a = 0.1088 mole
The equilibrium partial pressures are
P(H2) = (0.47 + x)RT/V = (0.47 + 0.1088)(1.00/2.03) atm = 0.2851 atm
P(Cl2) = xRT/V = (0.1088)(1.00/2.03) atm = 0.0536 atm
P(HCl) = (3.59 - 2x)RT/V = [3.59 - (2)(0.1088)](1.00/2.03) atm = 1.661 atm
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Reaction Rate
PROBLEM
The decomposition of gaseous dimethyl ether at ordinary pressures is first order. Its half-life is 25.0 min at 500 °C.
CH3OCH3(g) => CH4(g) + CO(g) + H2(g)
(a) Starting with 8.00 g, what mass remains (in grams) after 125 min and after 145 min?
(b) Calculate the time in minutes required to decrease 7.60 ng (nanograms) to 2.25 ng.
(c) What fraction of the original dimethyl ether remains after 150 min?
[from Kotz, John C., and Treichel, Paul Jr. 1996, Chemistry & Chemical Reactivity, Third Edition (New York: Saunders College Publishing), problem 15.72]
SOLUTION
(a) For a first order process,
q(t)/q0 = (1/2)t/t1/2 (1) => q(t) = q0(1/2)t/t1/2
where q0 is the initial quantity of a reactant, q(t) is the quantity of the reactant that remains at time t, and t1/2 is the half-life.
q(125 min) = (8.00 g)(1/2)(125 min / 25.0 min) = (8.00 g)/32 = 0.25 g
q(145 min) = (8.00 g)(1/2)(145 min / 25.0 min) = (8.00 g)(1.795 x 10-2) = 0.1436 g
(b) From equation (1), we get
ln[q(t)/q0] = (t/t1/2) ln(1/2)
=> t = t1/2 ln[q(t)/q0] / ln(1/2) = (25.0 min) ln(2.25 ng / 7.60 ng) / ln(1/2) = 43.90 min
(c) Using (1), we get
q(150 min)/q0 = (1/2)(150 min / 25.0 min) = (1/2)6 = 1/64
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PROBLEM
Iodide ion, I -, is oxidized to hypoiodite ion, IO-, by hypochlorite, ClO-, in basic solution according to
I -(aq) + ClO-(aq) => IO-(aq) + Cl-(aq)
Three initial-rate experiments were conducted; the results are shown in the following table:
| Experiment | [I -] (mole/L) | [ClO-] (mole/L) | Initial Rate of Formation of IO- (mole/L s) |
|---|---|---|---|
(a) Determine the order of the reaction with respect to each reactant.
(b) For the reaction, write the rate law that is consistent with the result of part (a), calculate the value of the specific rate constant, k, and specify units.
[from 2005 AP Chemistry Free-Response Question 3]
SOLUTION
(a) From experiments 1 and 2, for which the initial concentrations of ClO- are the same, we see that tripling the initial concentration of I - triples the rate, so the reaction is first order with respect to I -.
From experiments 1 and 3, for which the initial concentrations of I - are approximately the same, we see that quadrupling the initial concentration of ClO- quadruples the rate, so the reaction is first order with respect to ClO-.
(b) The rate law is of the form
rate = k[I -][ClO-]
The rate constant is
k = rate / [I -][ClO-]
From the three experiments in the table, we get k = 611.8, 610.3, and 610.7 L/mole s. The average of the three values is k = 610.9 L/mole s. Back to Main Menu
PROBLEM
The catalyzed decomposition of hydrogen peroxide, H2O2(aq), is represented by the following equation:
2H2O2(aq) => 2H2O(l) + O2(g)
The kinetics of the decomposition reaction were studied and the analysis of the results show that it is a first-order reaction. Some of the experimental data are shown in the table below.
| [H2O2] (mole/L) | Time (minutes) |
|---|---|
During the analysis of the data, the graph below was produced.
^
|\
| \
| \
| \
| \
+-----\-------->
Time (minutes)
(a) Label the vertical axis of the graph.
(b) What are the units of the rate constant, k, for the decomposition of H2O2(aq)?
(c) On the graph, draw the line that represents the plot of the uncatalyzed first-order decomposition of 1.00 M H2O2(aq).
[from 2005 AP Chemistry Free-Response Question 3]
SOLUTION
(a) For a first-order reaction, the concentration of H2O2 at time t is
[H2O2](t) = [H2O2](0) e-kt
where [H2O2](0) is the concentration at time t = 0, and k is the rate constant. If we take the natural logarithms of both sides, we get
ln [H2O2](t) = ln [H2O2](t) - kt
A graph of ln [H2O2](t) vs. t is a straight line with a negative slope equal to - k and an intercept of ln [H2O2](0) with the vertical axis. The vertical axis should be labeled "ln [H2O2](t)".
(b) The rate law for the reaction is
rate = - (1/2) d[H2O2] / dt = k[H2O2]
Since the units of [H2O2] are M/s or moles/L s, the units of k are s-1.
(c) For the uncatalyzed decomposition of 1.00 M H2O2, the graph of ln [H2O2](t) vs. time would be a straight line with negative slope starting at the same point at t = 0 but decreasing at a slower rate. Back to Main Menu
Acids and Bases
The solution of a salt derived from a strong acid and a strong base is neutral. The solution of a salt derived from a strong acid and a weak base is acidic. The solution of a salt derived from a weak acid and a strong base is basic. The solution of a salt derived from a weak acid and a weak base is (acidic, basic, approximately neutral) if Ka (>, <, ≈) Kb.
PROBLEM
Which of the following is true of pure water at a temperature of 40 °C?
(a) It is neutral (i.e., neither acidic nor basic).
(b) It has a pH of 7.00.
(c) Both (a) and (b).
(d) Neither (a) nor (b).
SOLUTION
The ionization constant of water is Kw = 10-14 at 25 °C. A solution of pure water at this temperature will have [H+] = [OH -] = 10-7 => pH = - log [H+] = 7.00. At other temperatures, the value of Kw is different and the values of [H+] and [OH -] will be different from their values at 25 °C, but still equal to each other, since each ionization of H2O produces one H+ and one OH -. A neutral solution is a solution in which [H+] = [OH -], so at 40 °C the water will still be neutral, but its pH will not be exactly 7.00. So only (a) is true.
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PROBLEM
Determine whether or not each of the following is a Lewis acid-base reaction:
(a) Al(OH)3(s) + OH -(aq) => Al(OH)4-(aq)
(b) Cl2(g) + H2O(l) => HOCl(aq) + H+(aq) + Cl -(aq)
(c) SnCl4(s) + 2Cl -(aq) => SnCl62-(aq)
(d) NH4+(g) + NH2-(g) => 2NH3(g)
(e) H+(aq) + NH3(aq) => NH4+(aq)
[from Sample Questions for Chemistry Advanced Placement Test, College Board, Question 13]
SOLUTION
A Lewis acid-base reaction is one in which one substance, the Lewis base, donates an electron pair to another substance, the Lewis acid.
(a) Al(OH)3(s) + OH -(aq) => Al(OH)4-(aq)
The Lewis structures for the reactants and product are
H H
| |
:O: :O:
.. | .. .. | ..
H--O--Al + :O--H => H--O--Al--O--H
.. | .. .. | ..
:O: :O:
| |
H H
The hydroxide ion (OH -) donates an electron pair (shown in pink) and is the Lewis base. The aluminum hydroxide (Al(OH)3) accepts an electron pair and is the Lewis acid.
(b) Cl2(g) + H2O(l) => HOCl(aq) + H+(aq) + Cl -(aq)
The Lewis structures for the reactants and products are
.. .. .. .. .. .. .. .. ..
:Cl--Cl: + H--O--H => :Cl + :Cl: + H + :O--H => :Cl--O--H + H + :Cl:
.. .. .. .. .. .. .. .. ..
This reaction is more complex than a simple transfer of an electron pair from one of the reactants to the other, so it is not a Lewis acid-base reaction.
(c) SnCl4(s) + 2Cl -(aq) => SnCl62-(aq)
The Lewis structures for the reactants and product are
.. ..
.. :Cl Cl:
:Cl: ..\ /..
.. | .. .. .. .. \ / ..
:Cl--Sn--Cl: + :Cl: + :Cl: => :Cl---Sn---Cl:
.. | .. .. .. .. / \ ..
:Cl: ../ \..
.. :Cl Cl:
.. ..
The two chlorine ions (Cl -) each donate an electron pair (shown in pink and blue) and are the Lewis base. The tin(IV) chloride or stannic chloride (SnCl4) accepts two electron pairs and is the Lewis acid.
(d) NH4+(g) + NH2-(g) => 2NH3(g)
The Lewis structures for the reactants and product are
H H H H
| | | |
H--N--H + :N: => H--N: + H--N:
| | | |
H H H H
The nitrogen dihyride ion (NH2-) donates an electron pair (shown in blue) and is the Lewis base. The hydrogen ion (H+) (shown in green) from the ammonium ion (NH4+) accepts an electron pair and is the Lewis acid.
(e) H+(aq) + NH3(aq) => NH4+(aq)
The Lewis structures for the reactants and product are
H H
| |
H + :N--H => H--N--H
| |
H H
The ammonia (NH3) donates an electron pair (shown in pink) and is the Lewis base. The hydrogen ion (H+) accepts the electron pair and is the Lewis acid.
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PROBLEM
Decide if each of the following substances should be classified as a Lewis acid or base:
(a) Mn2+
(b) CH3NH2
(c) H2NOH in the reaction H2NOH(aq) + HCl => [H3NOH]Cl(aq)
(d) SO2 in the reaction SO2(g) + BF3(g) <=> O2S-BF3(s)
(e) Zn(OH)2 in the reaction Zn(OH)2(s) + 2OH -(aq) <=> Zn(OH)42-(aq)
[from Kotz, John C., and Treichel, Paul Jr. 1996, Chemistry & Chemical Reactivity, Third Edition (New York: Saunders College Publishing), problem 17.74]
SOLUTION
(a) Mn2+ has two missing valence electrons. It can accept an electron pair, so it is a Lewis acid.
(b) CH3NH2:
H H
| |
H--C--N:
| |
H H
CH3NH2 has a lone electron pair which it can donate, so it is a Lewis base.
(c) H2NOH + HCl => [H3NOH]Cl:
H H
.. | .. .. | ..
H--O--N: + H :Cl: => H--O--N--H :Cl:
.. | .. .. | ..
H H
The H2NOH donates an electron pair, so it is a Lewis base.
(d) SO2(g) + BF3(g) <=> O2S-BF3(s):
.. .. .. ..
:O: :F: :O: :F:
| | .. | | ..
S: + B--F: <=> S---B--F:
|| | .. || | ..
:O: :F: :O: :F:
.. ..
The SO2 donates an electron pair, so it is a Lewis base.
(e) Zn(OH)2(s) + 2OH -(aq) <=> Zn(OH)42-(aq):
H H
| |
:O: :O:
.. | .. .. | ..
H--O: + Zn + :O--H <=> H--O--Zn--O--H
.. | .. .. | ..
:O: :O:
| |
H H
Zn(OH)2 accepts two electron pairs, so it is a Lewis acid.
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PROBLEM
Decide if each of the following substances should be classified as a Lewis acid or base:
(a) BCl3
(b) H2N-NH2, hydrazine
(c) CN - in the reaction Au+(aq) + 2CN -(aq) <=> [Au(CN)2]-(aq)
(d) Fe3+
[from Kotz, John C., and Treichel, Paul Jr. 1996, Chemistry & Chemical Reactivity, Third Edition (New York: Saunders College Publishing), problem 17.75]
SOLUTION
(a) BCl3:
..
:Cl:
| ..
B--Cl:
| ..
:Cl:
..
BCl3 can accept an electron pair, so it is a Lewis acid.
(b) H2N-NH2:
H H
| |
:N--N:
| |
H H
H2N-NH2 can donate two electron pairs, so it is a Lewis base.
(c) Au+(aq) + 2CN -(aq) <=> [Au(CN)2]-(aq):
:N///C: + Au + :C///N: <=> :N///C--Au--C///N: /// = triple bond
The two CN - each donate an electron pair, so the CN - is a Lewis base.
(d) Fe3+ has three missing valence electrons. It can accept an electron pair, so it is a Lewis acid.
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PROBLEM
Equilibrium constants can be measured for the dissociation of Lewis acid-base complexes such as the dimethyl ether complex of BF3, (CH3)2O-BF3. The value of K (here KP) for the reaction
(CH3)2O-BF3(g) <=> BF3(g) + (CH3)2O(g)
is 0.17.
(a) Tell which product is the Lewis acid and which is the Lewis base.
(b) What is the F-B-F angle in BF3? In (CH3)2O-BF3?
(c) What is the hybridization of O in (CH3)2O-BF3? Of the boron atom?
(d) If you place 1.00 g of the complex in a 565 mL flask at 25°C, what is the total pressure in the flask? What are the partial pressures of the Lewis acid, the Lewis base, and the complex?
[from Kotz, John C., and Treichel, Paul Jr. 1996, Chemistry & Chemical Reactivity, Third Edition (New York: Saunders College Publishing), problem 17.110]
SOLUTION
(a) The Lewis structures for the reactant and products are
H H
.. | .. |
:F: H--C--H :F: H--C--H
.. | | .. | |
:F--B------O: <=> :F--B + :O:
.. | | .. | |
:F: H--C--H :F: H--C--H
.. | .. |
H H
Since the oxygen atom in (CH3)2O has two lone electron pairs which can bond with the boron atom in BF3, BF3 is the Lewis acid and (CH3)2O is the Lewis base.
(b) The boron atom in BF3 has three bonding electron pairs and no lone pairs surrounding it. Therefore the shape of BF3 is trigonal planar, and the F-B-F angle is 120°. The boron atom in (CH3)2O-BF3 has three bonding electron pairs and one lone pair surrounding it. Therefore the overall arrangement of the four electron pairs is tetrahedral, and the F-B-F angle is 109.5°.
(c) There are four electron pairs surrounding either the oxygen or the boron atom in (CH3)2O-BF3. Therefore, the hybridization is sp3 in both cases.
(d) The molecular mass of (CH3)2O-BF3 is
M = (2)[12.01 g/mole + (3)(1.008 g/mole)] + 16.00 g/mole + 10.81 g/mole + (3)(19.00 g/mole)
= 113.878 g/mole
Therefore, 1.00 g of (CH3)2O-BF3 equals 8.781 x 10-3 mole of (CH3)2O-BF3. If none of this dissociated, the pressure would be
P0 = nRT/V = (8.781 x 10-3 mole)(0.0821 L atm / mole K)(298 K) / (565 x 10-3 L) = 0.3803 atm
Let x be the number of atm of (CH3)2O-BF3 which dissociate. Then at equilibrium we have
P[(CH3)2O-BF3] = P0 - x
P[(CH3)2O] = P[BF3] = x
The equilibrium constant is
KP = P[(CH3)2O]P[BF3] / P[(CH3)2O-BF3] = x2 / (P0 - x) => KPP0 - KPx = x2
=> x2 + KPx - KPP0 = 0
=> x = [- KP + sqrt(KP2 + 4KPP0)] / 2 = {- 0.17 + sqrt[0.172 + (4)(0.17)(0.3803)]} / 2 = 0.1831 atm
The final partial pressures are
P[(CH3)2O-BF3] = P0 - x = 0.3803 atm - 0.1831 atm = 0.1972 atm
P[(CH3)2O] = P[BF3] = x = 0.1831 atm
The final total pressure is
P = P[(CH3)2O-BF3] + P[(CH3)2O] + P[BF3] = 0.1972 atm + 0.1831 atm + 0.1831 atm
= 0.5633 atm
Alternate Method:
We could solve part (d) using KC instead of KP. The equilibrium constant KC for the above reaction is related to the equilibrium concentrations of the reactant and products according to
KC = [(CH3)2O][BF3] / [(CH3)2O-BF3]
= {P[(CH3)2O] / RT}{P[BF3] / RT} / {P[(CH3)2O-BF3] / RT}
= KP / RT
= (0.17) / [(0.0821 L atm / mole K)(298 K)] = 6.948 x 10-3
This follows from the general fact that, for an ideal gas, the pressure P and concentration n/V are related by
PV = nRT => n/V = P/RT
If none of the initial (CH3)2O-BF3 dissociated, its concentration would be
[(CH3)2O-BF3]0 = (8.781 x 10-3 mole) / (565 x 10-3 L) = 1.554 x 10-2 mole/L = 1.554 x 10-2 M = C0
Let x be the number of moles per liter of (CH3)2O-BF3 which dissociate. Then the equilibrium concentrations of (CH3)2O-BF3, (CH3)2O, and BF3 are
[(CH3)2O-BF3] = C0 - x
[(CH3)2O] = [BF3] = x
We have
KC = [(CH3)2O][BF3] / [(CH3)2O-BF3] = x2 / (C0 - x) => KCC0 - KCx = x2
=> x2 + KCx - KCC0 = 0
=> x = {- KC + sqrt(KC2 + 4KCC0)} / 2
= {- 6.948 x 10-3 + sqrt[(6.948 x 10-3)2 + (4)(6.948 x 10-3)(1.554 x 10-2)]} / 2 = 7.483 x 10-3 M
The equilibrium concentrations of (CH3)2O-BF3, (CH3)2O, and BF3 are
[(CH3)2O-BF3] = C0 - x = 1.554 x 10-2 M - 7.483 x 10-3 M = 8.059 x 10-3 M
[(CH3)2O] = [BF3] = x = 7.483 x 10-3 M
The final partial pressures are
P[(CH3)2O-BF3] = RT[(CH3)2O-BF3] = (0.0821 L atm / mole K)(298 K)(8.059 x 10-3 M)
= 0.1972 atm
P[(CH3)2O] = RT[(CH3)2O] = (0.0821 L atm / mole K)(298 K)(7.483 x 10-3 M) = 0.1831 atm
P[BF3] = RT[BF3] = (0.0821 L atm / mole K)(298 K)(7.483 x 10-3 M) = 0.1831 atm
The final total pressure is
P = P[(CH3)2O-BF3] + P[(CH3)2O] + P[BF3] = 0.1972 atm + 0.1831 atm + 0.1831 atm
= 0.5633 atm
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PROBLEM
Draw electron dot diagrams to show the possible structure of H3O+, H5O2+, H7O3+, and H9O4+.
[from Kavanah, Patrick, and Robbins, Jack 1971, Concepts in Modern Chemistry (New York: Cambridge Book Company), reasoning exercise 15.1]
SOLUTION
H3O+:
..
H--O--H
|
H
H5O2+:
H
.. |
H--O--H-hhhhh-:O: hhhhh = hydrogen bond
| |
H H
H7O3+:
H H
| .. |
:O:-hhhhh-H--O--H-hhhhh-:O: hhhhh = hydrogen bond
| | |
H H H
H9O4+:
H H
| .. |
:O:-hhhhh-H--O--H-hhhhh-:O: hhhhh = hydrogen bond
| | |
H H H
|
h
h
h
|
..
H--O--H
..
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PROBLEM
What is the OH - concentration of a solution which is (a) 0.001 M HCl, (b) 0.001 M H2SO4?
[from Kavanah, Patrick, and Robbins, Jack 1971, Concepts in Modern Chemistry (New York: Cambridge Book Company), reasoning exercise 15.5]
SOLUTION
(a) Hydrochloric acid (HCl) is a strong monoprotic acid which ionizes completely. Therefore, [H+] = 10-3 M and
[OH -] = Kw / [H+] = 10-14 / 10-3 = 10-11 M
(b) Sulfuric acid (H2SO4) is a strong acid which ionizes once according to
H2SO4 => H+ + HSO4-
HSO4- is a weak acid which ionizes according to
HSO4- => H+ + SO42-
with ionization constant Ka2 = 1.3 x 10-2. Let x be the number of moles per liter of H2SO4 which ionize only once to form HSO4- and let y be the number of moles per liter of H2SO4 which ionize twice to form SO42-. At equilibrium, we have
[H+] = x + 2y
[HSO4-] = x
[SO42-] = y
Let f = 0.001 M, the concentration of H2SO4. Since x + y = f, y = f - x, and
[H+] = x + 2(f - x) = 2f - x
[HSO4-] = x
[SO42-] = f - x
The ionization constant
Ka2 = [H+][SO42-]/[HSO4-] = (2f - x)(f - x)/x => Ka2x = 2f2 - 3fx + x2
=> x2 - (3f + Ka2)x + 2f2 = 0
=> x = {(3f + Ka2) ± sqrt[(3f + Ka2)2 - 8f2]} / 2
= (1.587 x 10-2, 1.26 x 10-4)
=> y = f - x = (- 1.487 x 10-2, 8.74 x 10-4)
The "+" solution for x yields a negative value for y, so we reject it and get
x = 1.26 x 10-4
y = 8.74 x 10-4
[H+] = x + 2y = 1.874 x 10-3
[OH -] = Kw/[H+] = (10-14) / (1.874 x 10-3) = 5.336 x 10-12
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PROBLEM
In a 0.050 M solution of hydrofluoric acid, HF, in water at 25°C, only 11% of the molecules dissociate to form ions. Calculate the ionization constant for hydrofluoric acid.
[from Kavanah, Patrick, and Robbins, Jack 1971, Concepts in Modern Chemistry (New York: Cambridge Book Company), reasoning exercise 15.13]
SOLUTION
The ionization constant is
Ka = [H+][F -]/[HF]
Since 11% of the HF dissociate, we have
[H+] = [F -] = (0.11)(0.050 M)
[HF] = (0.050 M)(1 - 0.11)
and
Ka = (0.11)2(0.050)2/[(0.050)(1 - 0.11)] = 6.8 x 10-4
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PROBLEM
Identify each of the following as an acid, base, salt, or amphoteric substance:
(a) sodium phosphate (Na3PO4)
(b) hydrogen chloride (HCl)
(c) sodium hydroxide (NaOH)
(d) hydrogen sulfate or bisulfate (HSO4-)
(e) oxalate (C2O42-)
[from Mascetta, Joseph A. 2002, Barron's How to Prepare for the SAT II Chemistry, 7th Edition (Hauppauge, New York: Barron's Educational Series), Diagnostic Test problem 52]
SOLUTION
(a) Sodium phosphate is the salt of a strong base (sodium hydroxide, NaOH) and a weak acid (phosphoric acid, H3PO4).
(b) Hydrochloric acid is a strong acid.
(c) Sodium hydroxide is a strong base.
(d) Bisulfate can either donate a proton or accept a proton and is therefore both a Bronsted acid and a Bronsted base. It is therefore also amphoteric.
(e) Oxalate can accept protons and is therefore a Bronsted base.
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PROBLEM
Determine whether each of the following oxides dissolve in water to form acidic or basic solutions:
(a) sodium oxide (Na2O)
(b) calcium oxide (CaO)
(c) aluminum oxide (Al2O3)
(d) sulfur trioxide (SO3)
[from Mascetta, Joseph A. 2002, Barron's How to Prepare for the SAT II Chemistry, 7th Edition (Hauppauge, New York: Barron's Educational Series), Diagnostic Test problem 68]
SOLUTION
(a) All alkali metal oxides and all alkaline earth metal oxides except beryllium oxide (BeO) dissolve in water to form basic solutions. Sodium oxide dissolves in water to form sodium hydroxide, a base:
Na2O(s) + H2O(l) => 2NaOH(aq)
(b) Calcium oxide dissolves in water to form calcium hydroxide, which is insoluble in aqueous solution:
CaO(s) + H2O(l) => Ca(OH)2(s)
(c) Aluminum oxide is amphoteric.
(d) Most nonmetallic oxides dissolve in water to form acidic solutions. Sulfur trioxide reacts with water to form sulfuric acid:
SO3(g) + H2O(l) => H2SO4(aq)
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PROBLEM
If each of the following salts undergoes hydrolysis, indicate whether the resulting solution will be acidic or basic:
(a) Na2SO4
(b) K2SO4
(c) NaNO3
(d) Cu(NO3)2
[from Mascetta, Joseph A. 2002, Barron's How to Prepare for the SAT II Chemistry, 7th Edition (Hauppauge, New York: Barron's Educational Series), problem 8.5]
SOLUTION
(a) Na2SO4 + 2H2O => 2NaOH + H2SO4
Sodium hydroxide (NaOH) is a strong base and sulfuric acid (H2SO4) is a strong acid, but since there are twice as many moles of NaOH as H2SO4, the solution will be basic.
(b) K2SO4 + 2H2O => 2KOH + H2SO4
Potassium hydroxide (KOH) is a strong base and sulfuric acid (H2SO4) is a strong acid, but since there are twice as many moles of KOH as H2SO4, the solution will be basic.
(c) NaNO3 + H2O => NaOH + HNO3
Sodium hydroxide (NaOH) is a strong base and HNO3 is a strong acid. There will be equal numbers of H+ and OH- ions, so the solution will be neutral.
(d) Cu(NO3)2 + 2H2O => Cu(OH)2 + 2HNO3
Cuprous hydroxide (Cu(OH)2) is a weak base and nitric acid (HNO3) is a strong acid, so the solution will be acidic.
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PROBLEM
The following three mixtures have been prepared: CaO plus water, SiO2 plus water, and CO2 plus water. For each mixture, predict whether the pH is less than 7, equal to 7, or greater than 7. Justify your answer.
[from 2005 AP Chemistry Free-Response Question 5]
SOLUTION
CaO: Calcium oxide is basic. It reacts with water to form calcium hydroxide according to
CaO + H2O => Ca(OH)2
Most metallic oxides are basic. The pH will be greater than 7.
SiO2: SiO2 does not react with water, so the pH will be equal to 7.
CO2: When CO2 is added to water, carbonic acid is formed according to
CO2 + H2O => H2CO3
The pH will be less than 7.
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PROBLEM
Five acids are listed below in order of decreasing acid strength:
HCl > HC2H3O2 > HCN > H2O > NH3
Determine whether the equilibrium constant for each of the following reactions is less than, equal to, or greater than one:
(a) HCl(aq) + CN -(aq) <=> HCN(aq) + Cl -(aq)
(b) HCl(aq) + H2O(l) <=> H3O+(aq) + Cl -
(c) HC2H3O2(aq) + OH -(aq) <=> C2H3O2-(aq) + H2O(l)
(d) H2O(aq) + NH2-(aq) <=> NH3(aq) + OH -(aq)
(e) HCN(aq) + C2H3O2- <=> HC2H3O2(aq) + CN -(aq)
[from Sample Questions for Chemistry Advanced Placement Test, College Board, Question 21]
SOLUTION
The five acids listed are hydrochloric acid (HCl), acetic acid (HC2H3O2), hydrocyanic acid (HCN), water (H2O), and ammonia (NH3). The ionization constants for the above acids are
Ka1 > Ka2 > Ka3 > Ka4 > Ka5
where
Ka1 = [H3O+][Cl -] / [HCl]
Ka2 = [H3O+][C2H3O2-] / [HC2H3O2]
Ka3 = [H3O+][CN -] / [HCN]
Ka4 = [H3O+][OH -]
Ka5 = [H3O+][NH2-] / [NH3]
(a) HCl(aq) + CN -(aq) <=> HCN(aq) + Cl -(aq)
Keq = [HCN][Cl -] / [HCl][CN -] = [HCN][Cl -][H3O+] / [HCl][CN -][H3O+] = Ka1/Ka3 > 1
(b) HCl(aq) + H2O(l) <=> H3O+(aq) + Cl -
Keq = [H3O+][Cl -] / [HCl] > 1
(c) HC2H3O2(aq) + OH -(aq) <=> C2H3O2-(aq) + H2O(l)
Keq = [C2H3O2-] / [HC2H3O2][OH -]
= [C2H3O2-][H3O+] / [HC2H3O2][OH -][H3O+]
= Ka2/Ka4 > 1
(d) H2O(aq) + NH2-(aq) <=> NH3(aq) + OH -(aq)
Keq = [NH3][OH -] / [NH2-] = [NH3][OH -][H3O+] / [NH2-][H3O+] = Ka4/Ka5 > 1
(e) HCN(aq) + C2H3O2- <=> HC2H3O2(aq) + CN -(aq)
Keq = [HC2H3O2][CN -] / [HCN][C2H3O2-] = [HC2H3O2][CN -][H3O+] / [HCN][C2H3O2-][H3O+]
= Ka3/Ka2 < 1
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PROBLEM
Write the molecular and ionic equations for the following neutralizations and predict the pH (< 7, > 7, or ≈ 7) of a solution of the salt formed:
(a) phosphoric acid and calcium hydroxide
(b) acetic acid and magnesium hydroxide
(c) carbonic acid and potassium hydroxide
[from Kavanah, Patrick, and Robbins, Jack 1971, Concepts in Modern Chemistry (New York: Cambridge Book Company), reasoning exercises 14.3, 14.4]
SOLUTION
(a) The molecular equation is
2H3PO4 + 3Ca(OH)2 => 6H2O + Ca3(PO4)2
The ionic equation is
6H+ + 2PO43- + 3Ca2+ + 6OH- => 6H2O + 3Ca2+ + 2PO43-
The net ionic equation is
6H+ + 6OH- => 6H2O
Phosphoric acid is a weak acid with Ka = 7.5 x 10-3. Calcium hydroxide is a weak base with Kb = 5.5 x 10-6. Since Ka > Kb, the solution of the salt Ca3(PO4)2 is acidic with pH < 7.
(b) The molecular equation is
2HCH3COO + Mg(OH)2 => 2H2O + Mg(CH3COO)2
The ionic equation is
2H+ + 2CH3COO- + Mg2+ + 2OH- => 2H2O + Mg2+ + 2CH3COO-
The net ionic equation is
2H+ + 2OH- => 2H2O
Acetic acid is a weak acid with Ka = 1.8 x 10-5. Magnesium hydroxide is a weak base with Kb = 1.8 x 10-11. Since Ka > Kb, the solution of the salt Mg(CH3COO)2 is acidic with pH < 7.
(c) The molecular equation is
H2CO3 + 2KOH => 2H2O + K2CO3
The ionic equation is
2H+ + CO32- + 2K+ + 2OH- => 2H2O + 2K+ + CO32-
The net ionic equation is
2H+ + 2OH- => 2H2O
Since carbonic acid is a weak acid and potassium hydroxide is a strong base, the solution of the salt K2CO3 is basic with pH > 7.
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PROBLEM
In a 0.1 N solution of carbonic acid, H2CO3, in water at 25°C, only 0.29% ionization takes place. Calculate the ionization constant for carbonic acid.
[from Kavanah, Patrick, and Robbins, Jack 1971, Concepts in Modern Chemistry (New York: Cambridge Book Company), reasoning exercise 15.14]
SOLUTION
A 0.1 N solution of carbonic acid contains 0.1 equivalents of H2CO3 per liter of solution. Each mole of H2CO3 is equal to two equivalents of H2CO3. Therefore, a 0.1 N solution of carbonic acid contains 0.05 moles of H2CO3 per liter of solution, so it is a 0.05 M solution of H2CO3. Carbonic acid ionizes according to
H2CO3 + H2O <=> H3O+ + HCO3-
At equilibrium, we have
[H2CO3] = (0.05)(1 - 0.0029) = 0.049855
[HCO3-] = [H3O+] = (0.05)(0.0029) = 1.45 x 10-4
The ionization constant of carbonic acid is
Ka = [H3O+][HCO3-]/[H2CO3] = (1.45 x 10-4)2/(0.049855) = 4.217 x 10-7
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PROBLEM
The pH of a water solution of sodium carbonate may be (A) 1.5, (B) 3.5, (C) 5.5, (D) 8.5.
[from Kavanah, Patrick, and Robbins, Jack 1971, Concepts in Modern Chemistry (New York: Cambridge Book Company), multiple choice question 15.5]
SOLUTION
Sodium carbonate, Na2CO3, is soluble in aqueous solution:
Na2CO3 => 2Na+ + CO32-
The carbonate can hydrolyze with H2O to form the hydrogen carbonate ion HCO3-,
CO32- + H2O <=> HCO3- + OH-
for which the hydrolysis constant is Kh = Kw/Ka2 = (10-14) / (4.8 x 10-11) = 2.08 x 10-4, where Ka2 = 4.8 x 10-11 is the ionization constant of the hydrogen carbonate ion. The carbonate can also react with H3O+ to form the hydrogen carbonate ion,
CO32- + H3O+ <=> HCO3- + H2O
for which the equilibrium constant is 1/Ka2 = 1/(4.8 x 10-11) = 2.083 x 1010. Whichever species, H2O or H3O+, the carbonate reacts with, the result will be more OH - and less H3O+, resulting in a pH which is greater than 7 or basic.
In general, the solution of a salt derived from a strong base and a weak acid, or a strong base and a neutral entity, is basic. The carbonate ion, CO32- is the conjugate base of a weak acid, HCO3-, and is therefore a strong base. The sodium ion, Na+, is neither acidic nor basic. The correct answer is (D).
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PROBLEM
Calculate the hydrolysis constant for the reaction
HCO2- + H2O <=> HCO2H + OH-
and find the concentrations of H3O+, OH-, HCO2-, and HCO2H in a solution of 0.15 M HCO2Na. The dissociation constant of formic acid (HCO2H) is Ka = 1.8 x 10-4.
[from Mahan, Bruce H. 1966, College Chemistry (Reading, Massachusetts: Addison-Wesley), problem 6.16]
SOLUTION
The hydrolysis constant for the hydrolysis of HCO2- is
Kh = Kw/Ka = (10-14) / (1.8 x 10-4) = 5.556 x 10-11
HCO2Na is soluble in aqueous solution. Let x be the number of moles per liter of HCO2- which hydrolyze. At equilibrium,
[HCO2-] = 0.15 - x
[HCO2H] = x
[OH-] = x
The hydrolysis constant is
Kh = [HCO2H][OH-] / [HCO2-] = x2 / (0.15 - x)
Solve for x.
(0.15 - x)Kh = x2
x2 + Khx - 0.15Kh = 0
x = {- Kh + sqrt[Kh2 + 4(0.15)Kh]} / 2
= {- 5.556 x 10-11 + sqrt[(5.556 x 10-11)2 + 4(0.15)(5.556 x 10-11)]} / 2
= 2.887 x 10-6 M
Thus,
[OH-] = [HCO2H] = x = 2.887 x 10-6 M
[HCO2-] = 0.15 - x = 0.15 - 2.887 x 10-6 = 0.15 M
[H3O+] = Kw / [OH-] = (10-14) / (2.887 x 10-6) = 3.464 x 10-9 M
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PROBLEM
A solution of equal concentrations of lactic acid and sodium acetate was found to have pH = 3.08. (a) What are the values of pKa and Ka of lactic acid? (b) What would the pH
be if the acid had twice the concentration of the salt?
SOLUTION
(a) Let L- and Ac- represent the lactate and acetate ions. Lactic and acetic acid ionize according to
HL + H2O <=> L- + H3O+, Ka1 = ?
HAc + H2O <=> Ac- + H3O+, Ka2 = 1.8 x 10-5
Let x be the moles per liter of HL which ionize and let y be the moles per liter of Ac- which combine with H3O+ to form HAc. Assume that the initial concentrations of HL and NaAc are both 1 M. NaAc is soluble in aqueous solution. The equilibrium concentrations are
[HL] = 1 - x = a - x
[L-] = x
[HAc] = y
[Ac-] = 1 - y = a - y
[H3O+] = x - y
where a = 1. Define
z = x - y => x = y + z
Then the equilibrium concentrations can be written as
[HL] = a - y - z
[L-] = y + z
[HAc] = y
[Ac-] = a - y
[H3O+] = z = 10-pH = 10-3.08 = 8.318 x 10-4 M
The equilibrium constants are
Ka1 = [L-][H3O+] / [HL] = (y + z)z / (a - y - z) (1)
Ka2 = [Ac-][H3O+] / [HAc] = (a - y)z / y => Ka2y = az - yz => y = az / (Ka2 + z) (2)
From (2), we get
y = (1)(8.318 x 10-4) / (1.8 x 10-5 + 8.318 x 10-4) = 0.9788
From (1), we get
Ka1 = (0.9788 + 8.318 x 10-4)(8.318 x 10-4) / (1 - 0.9788 - 8.318 x 10-4) = 4.004 x 10-2
Note: This value for Ka1 differs from the accepted value of 1.4 x 10-4 under standard conditions (Kotz and Treichel 1996, p. 820; Nims and Smith 1936, p. 151, Fig. 3), but this is what results from assuming pH = 3.08.
Kotz, John C., and Treichel, Paul Jr. 1996, Chemistry & Chemical Reactivity, Third Edition (New York: Saunders College Publishing).
Nims, Leslie Frederick, and Smith, Paul K. 1936, "The Ionization of Lactic Acid from Zero to Fifty Degrees," J. Biol. Chem., 113, 145.
(b) Use the result from part (a), Ka1 = 4.004 x 10-2.
Let L- and Ac- represent the lactate and acetate ions. Lactic and acetic acid ionize according to
HL + H2O <=> L- + H3O+, Ka1 = 4.004 x 10-2
HAc + H2O <=> Ac- + H3O+, Ka2 = 1.8 x 10-5
Let x be the moles per liter of HL which ionize and let y be the moles per liter of Ac- which combine with H3O+ to form HAc. Assume that the initial concentration of HL is 2 M and the initial concentration of NaAc is 1 M. NaAc is soluble in aqueous solution. The equilibrium concentrations are
[HL] = 2 - x = a - x
[L-] = x
[HAc] = y
[Ac-] = 1 - y = b - y
[H3O+] = x - y
where a = 2 and b = 1. The ionization constants are
Ka1 = [L-][H3O+] / [HL] = x(x - y) / (a - x) => y = x - Ka1(a - x) / x (1)
Ka2 = [Ac-][H3O+] / [HAc] = (b - y)(x - y) / y => x = y + Ka2y / (b - y) (2)
Substitute (2) into (1).
y = y + Ka2y / (b - y) - Ka1{a - [y + Ka2y / (b - y)]} / [y + Ka2y / (b - y)]
Subtract y from both sides.
0 = Ka2y / (b - y) - Ka1{a - [y + Ka2y / (b - y)]} / [y + Ka2y / (b - y)] = f(y)
We can solve this graphically for y. Plot f(y) as a function of y for 0 < y < 1. We see that the value of y that makes f(y) zero is between 0.999 and 0.9999. We can narrow down the solution to y = 0.99951641. Using (2), we get x = 1.036719985 => x - y = 0.037203575 = [H3O+] => pH = - log [H3O+] = 1.429. See LacticAcetic070421.xls.
Note: If we use the correct value for the ionization constant of lactic acid, Ka1 = 1.4 x 10-4, we get y = 0.904076789, x = 0.90424644, x - y = 0.00016965 = [H3O+], and pH = 3.770. See LacticAcetic070421.2.xls.
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PROBLEM
One liter of a 2 M solution of lactic acid (CH3CHOHCO2H) is combined with one liter of a 2 M solution of sodium acetate (NaCH3CO2). (a) What is the pH of the resulting solution? (b) Repeat if the initial concentrations are 4 M instead of 2 M. (c) Repeat if the initial concentrations are 0.1 M instead of 2 M.
SOLUTION
(a) Let "L" and "Ac" represent the lactate (CH3CHOHCO2-) and acetate (CH3CO2-) ions, respectively. Then lactic and acetic acids ionize according to
HL + H2O <=> L- + H3O+, Ka1 = 1.4 x 10-4
HAc + H2O <=> Ac- + H3O+, Ka2 = 1.8 x 10-5
After the two solutions are mixed, we have a solution containing 1 M lactic acid and 1 M sodium acetate. Sodium acetate is soluble in aqueous solution. The lactic acid will ionize to some extent and some of the resulting H3O+ will combine with Ac- to form HAc. Let x be the moles per liter of HL which ionize and let y be the moles per liter of Ac- which combine with H3O+ to form HAc. At equilibrium, we have
[HL] = 1 - x = a - x
[L-] = x
[HAc] = y
[Ac-] = 1 - y = a - y
[H3O+] = x - y
where a = 1. The ionization constants are
Ka1 = [L-][H3O+] / [HL] = x(x - y) / (a - x) => y = x - Ka1(a - x) / x (1)
Ka2 = [Ac-][H3O+] / [HAc] = (a - y)(x - y) / y => x = y + Ka2y / (a - y) (2)
Substitute (2) into (1).
y = y + Ka2y / (a - y) - Ka1{a - [y + Ka2y / (a - y)]} / [y + Ka2y / (a - y)]
Subtract y from both sides.
0 = Ka2y / (a - y) - Ka1{a - [y + Ka2y / (a - y)]} / [y + Ka2y / (a - y)] = f(y)
We can solve this graphically for y. Plot f(y) as a function of y for 0 < y < 1. We see that the value of y that makes f(y) zero is between 0.73 and 0.74. We can narrow down the solution to y = 0.736043742. Using (2), we get x = 0.736093935 => x - y = 5.01931 x 10-5 = [H3O+] => pH = - log [H3O+] = 4.299. See LacticAcetic070407.xls.
(b) If the initial concentrations are 4 M, the solution to the problem is the same except that a = 2. x and y must both be between 0 and 2. Plot f(y) as a function of y for 0 < y < 2. The value of y that makes f(y) zero is between 1.45 and 1.50. We can narrow down the solution to y = 1.47211258. Using (2), we get x = 1.472162777 => x - y = 5.01964 x 10-5 = [H3O+] => pH = - log [H3O+] = 4.299. See LacticAcetic070407.2.xls.
(c) If the initial concentrations are 0.1 M, the solution to the problem is the same except that a = 0.05. x and y must both be between 0 and 0.05. Plot f(y) as a function of y for 0 < y < 0.05. The value of y that makes f(y) zero is between 0.0365 and 0.0370. We can narrow down the solution to y = 0.036778392. Using (2), we get x = 0.036828462 => x - y = 5.00704 x 10-5 = [H3O+] => pH = - log [H3O+] = 4.300. See LacticAcetic070407.3.xls.
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PROBLEM
The dissociation constant of HCN, hydrocyanic acid, is Ka = 4.8 x 10-10. What is the concentration of H3O+, OH-, and HCN in a solution prepared by dissolving 0.160 mole of NaCN in 450 mL of water?
[from Mahan, Bruce H. 1966, College Chemistry (Reading, Massachusetts: Addison-Wesley), problem 6.17]
SOLUTION
NaCN is soluble in aqueous solution. Let x be the number of moles per liter of CN- which hydrolyze to form HCN according to
CN- + H2O <=> HCN + OH-
for which the hydrolysis constant is
Kh = Kw/Ka = (10-14) / (4.8 x 10-10) = 2.083 x 10-5
At equilibrium,
[CN-] = (0.160 mole) / (0.450 L) - x = 0.3556 - x
[OH-] = x
[HCN] = x
We have
Kh = [HCN][OH-] / [CN-] = x2 / (0.3556 - x)
=> Kh(0.3556 - x) = x2
=> x2 + Khx - 0.3556Kh = 0
=> x = {- Kh + sqrt[Kh2 + 4(0.3556)Kh]} / 2
= {- 2.083 x 10-5 + sqrt[(2.083 x 10-5)2 + 4(0.3556)(2.083 x 10-5)]} / 2 = 2.711 x 10-3 M
=> [OH-] = [HCN] = 2.711 x 10-3 M
=> [H3O+] = Kw / [OH-] = (10-14) / (2.711 x 10-3) = 3.688 x 10-12 M
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PROBLEM
Calculate the solubility of lead sulfate in a solution of 0.100 M H3O+ by taking account of the reaction
HSO4- + H2O => H3O+ + SO42-
The solubility product for PbSO4 is Ksp = 1.8 x 10-8. Be sure to justify all approximations.
[from Mahan, Bruce A. 1966, College Chemistry (Reading, Massachusetts: Addison-Wesley), problem 6.19]
SOLUTION
The hydrogen sulfate or bisulfate ion, HSO4-, is a weak acid with ionization constant Ka = 1.2 x 10-2. The addition of PbSO4 to the H3O+ solution causes the H3O+ to combine with SO42-. Let x be the amount of PbSO4 which dissolves per liter of solution and let y be the amount of HSO4- produced per liter. At equilibrium, we have
[Pb2+] = x
[SO42-] = x - y
[H3O+] = 0.1 - y
[HSO4-] = y
The solubility product for PbSO4 is
Ksp = [Pb2+][SO42-] = x(x - y) (1)
The ionization constant of hydrogen sulfate is
Ka = [H3O+][SO42-] / [HSO4-] = (0.1 - y)(x - y) / y (2)
(1) and (2) are two nonlinear equations in two unknowns x and y. We can assume a value for y, solve (1) for x, substitute the result into (2) and solve for a new value of y, and keep repeating this until the values of x and y obtained converge. From (1), we get
Kspy = x2 - xy => x2 - xy - Kspy = 0
=> x = {y ± sqrt[y2 + 4Kspy]} / 2 => {y + sqrt[y2 + 4Kspy]} / 2 (3)
since x cannot be negative. From (2) we get
Kay = 0.1x - 0.1y - xy + y2
=> y2 - (0.1 + x + Ka)y + 0.1x = 0
=> y = {(0.1 + x + Ka) ± sqrt[(0.1 + x + Ka)2 - 4(0.1x)]} / 2 (4)
For a trial value of y, we note that y cannot be greater than 0.1, the initial amount of H3O+ per liter of solution. So start with y = 0.05.
Substituting y = 0.05 into (3), we get x = 5 x 10-2. Substituting this into (4), we get y = (0.121, 4.15 x 10-2) from the (+, -) solution, so we reject the + solution.
We substitute y = 4.15 x 10-2 into (3) and get x = 4.15 x 10-2. We substitute this into (4) and get y = 3.50 x 10-2. If we repeat this process a number of times, the results converge to x = 4.10 x 10-4 and y = 3.66 x 10-4. Thus, the solubility of PbSO4 is 4.10 x 10-4 M.
Note: In order for this method to work, we must choose a way to solve for x and y iteratively in such a way that the solutions converge.
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PROBLEM
A solution is prepared by dissolving 0.200 mole of sodium formate, HCO2Na, and 0.250 moles of formic acid, HCO2H, in approximately 200(±50) mL of water. Calculate the concentrations of H3O+ and OH-. The dissociation constant of formic acid is Ka = 1.8 x 10-4.
[from Mahan, Bruce H. 1966, College Chemistry (Reading, Massachusetts: Addison-Wesley), problem 6.21]
SOLUTION
Sodium formate is soluble in aqueous solution. Let x be the number of moles per liter of HCO2H at equilibrium. Then at equilibrium we have
[HCO2H] = x
[HCO2-] = [(0.250 mole) / (0.200 L)] - x + [(0.200 mole) / (0.200 L)] = 2.25 - x
[H3O+] = [(0.250 mole) / (0.200 L)] - x = 1.25 - x
The dissociation constant of formic acid, which dissociates according to
HCO2H + H2O <=> H3O+ + HCO2-
is
Ka = [H3O+][HCO2-] / [HCO2H] = (1.25 - x)(2.25 - x) / x
=> Kax = 2.8125 - 3.5x + x2
=> x2 - (Ka + 3.5)x + 2.8125 = 0
=> x = {Ka + 3.5 ± sqrt[(Ka + 3.5)2 - 4(2.8125)]} / 2 = (2.2504 M, 1.2498 M)
Reject the + solution because this would yield a negative value for [HCO2-]. Thus, at equilbrium we have
[HCO2H] = x = 1.2498 M
[HCO2-] = 2.25 - x = 1.000225 M
[H3O+] = 1.25 - x = 2.25 x 10-4 M
[OH-] = Kw / [H3O+] = (10-14) / (2.25 x 10-4) = 4.446 x 10-11 M
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PROBLEM
A carbonate buffer solution is prepared by dissolving 30.0 g of Na2CO3 in 350 mL of water and adding 150 mL of 1.00 M HCl. Calculate the pH of the solution.
[from Mahan, Bruce H. 1966, College Chemistry (Reading, Massachusetts: Addison-Wesley), problem 6.27]
SOLUTION
The molecular weight of Na2CO3 is
(2)(22.99 g/mole) + 12.01 g/mole + (3)(16.00 g/mole) = 105.99 g/mole
=> 30.0 g Na2CO3 = [(30.0 g) / (105.99 g/mole)] Na2CO3 = 0.2830 mole Na2CO3
Na2CO3 is soluble in aqueous solution:
Na2CO3 => 2Na+ + CO32-
The initial concentrations of Na+ and CO32- are
[Na+] = [(2)(0.2830 mole)] / (0.500 L) = 1.132 M
[CO32-] = (0.2830 mole) / (0.500 L) = 0.5661 M
The initial concentration of H3O+ is
[H3O+] = (1.00 M)(0.150 L) / (0.500 L) = 0.3 M
Now consider what could happen. The CO32- could hydrolyze according to
CO32- + H2O <=> HCO3- + OH-
The hydrolysis constant for this reaction is
Kh = Kw/Ka2 = (10-14) / (4.8 x 10-11) = 2.08 x 10-4
where Ka2 = 4.8 x 10-11 is the ionization constant for the reaction
HCO3- + H2O <=> H3O+ + CO32-
The CO32- could also be consumed by reacting with H3O+, for which the equilibrium constant is
1/Ka2 = 1 / (4.8 x 10-11) = 2.083 x 1010
Because 1/Ka2 >> 1, all of the excess H3O+ will react with CO32- and be converted into H2O. At equilibrium, we have
[CO32-] = 0.5661 M - 0.3 M = 0.2661 M
[HCO3-] = 0.3 M
The ionization constant
Ka2 = [H3O+][CO32-] / [HCO3-]
=> [H3O+] = Ka2[HCO3-] / [CO32-] = (4.8 x 10-11)(0.3) / (0.2661) = 5.412 x 10-11 M
=> pH = - log [H3O+] = - log(5.412 x 10-11) = 10.27
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PROBLEM
One liter of a 2 M solution of lactic acid (CH3CHOHCO2H) is combined with one liter of a 2 M solution of sodium acetate (NaCH3CO2). (a) What is the pH of the resulting solution? (b) Repeat if the initial concentrations are 4 M instead of 2 M. (c) Repeat if the initial concentrations are 0.1 M instead of 2 M.
SOLUTION
(a) Let "L" and "Ac" represent the lactate (CH3CHOHCO2-) and acetate (CH3CO2-) ions, respectively. Then lactic and acetic acids ionize according to
HL + H2O <=> L- + H3O+, Ka1 = 1.4 x 10-4
HAc + H2O <=> Ac- + H3O+, Ka2 = 1.8 x 10-5
After the two solutions are mixed, we have a solution containing 1 M lactic acid and 1 M sodium acetate. Sodium acetate is soluble in aqueous solution. The lactic acid will ionize to some extent and some of the resulting H3O+ will combine with Ac- to form HAc. Let x be the moles per liter of HL which ionize and let y be the moles per liter of Ac- which combine with H3O+ to form HAc. At equilibrium, we have
[HL] = 1 - x = a - x
[L-] = x
[HAc] = y
[Ac-] = 1 - y = a - y
[H3O+] = x - y
where a = 1. The ionization constants are
Ka1 = [L-][H3O+] / [HL] = x(x - y) / (a - x) => y = x - Ka1(a - x) / x (1)
Ka2 = [Ac-][H3O+] / [HAc] = (a - y)(x - y) / y => x = y + Ka2y / (a - y) (2)
Substitute (2) into (1).
y = y + Ka2y / (a - y) - Ka1{a - [y + Ka2y / (a - y)]} / [y + Ka2y / (a - y)]
Subtract y from both sides.
0 = Ka2y / (a - y) - Ka1{a - [y + Ka2y / (a - y)]} / [y + Ka2y / (a - y)] = f(y)
We can solve this graphically for y. Plot f(y) as a function of y for 0 < y < 1. We see that the value of y that makes f(y) zero is between 0.73 and 0.74. We can narrow down the solution to y = 0.736043742. Using (2), we get x = 0.736093935 => x - y = 5.01931 x 10-5 = [H3O+] => pH = - log [H3O+] = 4.299. See LacticAcetic070407.xls.
(b) If the initial concentrations are 4 M, the solution to the problem is the same except that a = 2. x and y must both be between 0 and 2. Plot f(y) as a function of y for 0 < y < 2. The value of y that makes f(y) zero is between 1.45 and 1.50. We can narrow down the solution to y = 1.47211258. Using (2), we get x = 1.472162777 => x - y = 5.01964 x 10-5 = [H3O+] => pH = - log [H3O+] = 4.299. See LacticAcetic070407.2.xls.
(c) If the initial concentrations are 0.1 M, the solution to the problem is the same except that a = 0.05. x and y must both be between 0 and 0.05. Plot f(y) as a function of y for 0 < y < 0.05. The value of y that makes f(y) zero is between 0.0365 and 0.0370. We can narrow down the solution to y = 0.036778392. Using (2), we get x = 0.036828462 => x - y = 5.00704 x 10-5 = [H3O+] => pH = - log [H3O+] = 4.300. See LacticAcetic070407.3.xls.
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PROBLEM
A solution is prepared by adding 2.05 g of sodium acetate, NaCH3COO, to 100 mL of 0.100 M HCl solution. What is the resulting concentration of H3O+? A subsequent addition of 6.00 mL of 0.100 M HCl is made. What is the new concentration of H3O+?
[from Mahan, Bruce A. 1966, College Chemistry (Reading, Massachusetts: Addison-Wesley), problem 6.29]
SOLUTION
The molecular weight of NaCH3COO is [22.99 + (2)(12.01) + (3)(1.008) + (2)(16.00)] g/mole = 82.034 g/mole.
2.05 g NaCH3COO = (2.05 g)/(82.034 g/mole) NaCH3COO = 2.499 x 10-2 mole NaCH3COO
The initial concentration of NaCH3COO is (2.499 x 10-2 mole) / (0.100 L) = 0.2499 M. Since NaCH3COO is an ionic compound containing an alkali metal ion (Na+), it is soluble in aqueous solution, and the initial concentrations of Na+ and
CH3COO- are both 0.2499 M. Since HCl is a strong acid, the initial concentrations of H3O+ and Cl- are both 0.100 M. Acetic acid, CH3COOH, is a weak acid with ionization constant Ka = 1.85 x 10-5 (Mahan 1966, p. 193), which ionizes in water according to
CH3COOH + H2O <=> H3O+ + CH3COO-
The addition of the sodium acetate causes this reaction to proceed to the left. Let x be the number of moles per liter of H3O+ which combine with CH3COO- to form CH3COOH. At equilibrium,
[CH3COOH] = x
[CH3COO-] = 0.2499 - x
[H3O+] = 0.100 - x
The ionization constant of acetic acid is
Ka = [H3O+][CH3COO-] / [CH3COOH] = (0.100 - x)(0.2499 - x) / x
=> Kax = 0.02499 - 0.100x - 0.2499x + x2
=> x2 - (Ka + 0.3499)x + 0.02499 = 0
=> x = {(Ka + 0.3499) ± sqrt[(Ka + 0.3499)2 - 4(0.02499)]} / 2 = (0.24993, 0.099988) M
Reject the + solution because 0.24993 M is greater than the initial concentration of CH3COO-. We therefore have
x = 0.099988 M
[CH3COOH] = x = 0.099988 M
[CH3COO-] = 0.2499 - x = 0.1499 M
[H3O+] = 0.100 - x = 1.234 x 10-5 M
This is in 100 mL of solution. Now if 6.00 mL of 0.100 M HCl is added, we have a total of 106 mL of solution. The starting concentrations of H3O+, CH3COOH, and CH3COO- are
[H3O+] = [(0.100 L)(1.234 x 10-5 M) + (0.006 L)(0.100 M)] / (0.106 L)
= 5.672 x 10-3 M
[CH3COOH] = (0.100 L)(0.099988 M) / (0.106 L) = 9.433 x 10-2 M
[CH3COO-] = (0.100 L)(0.1499 M) / (0.106 L) =
0.1414 M
The additional HCl causes more of the CH3COO- to combine with H3O+ to form CH3COOH. Let x be the number of moles per liter of CH3COO- which combine with H3O+. At equilibrium, the new concentrations are
[H3O+] = 5.672 x 10-3 - x = A - x
[CH3COOH] = 9.433 x 10-2 + x = B + x
[CH3COO-] = 0.1414 - x = C - x
where A = 5.672 x 10-3, B = 9.433 x 10-2, and C = 0.1414. We have
Ka = [H3O+][CH3COO-] / [CH3COOH]
= (A - x)(C - x) / (B + x)
=> Ka(B + x) = (A - x)(C - x)
=> KaB + Kax = AC - (A + C)x + x2
=> x2 - (Ka + A + C)x + (AC - KaB) = 0
=> x = {(Ka + A + C) ± sqrt[(Ka + A + C)2 - 4(AC - KaB)]} / 2 = (0.1415, 5.658 x 10-3) M
Reject the + solution because 0.1415 M is greater than the initial concentration of H3O+. So x = 5.658 x 10-3 M and the final concentration of H3O+ is
[H3O+] = 5.672 x 10-3 M - 5.658 x 10-3 M = 1.362 x 10-5 M
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Electrochemistry
An electrochemical cell generates electricity through the use of a spontaneous redox reaction.
In an electrolytic cell, electrical energy causes a nonspontaneous chemical reaction to occur. The process is called electrolysis.
An example of an electrochemical cell is the following.
+-+
+---------|V|---------+
| +-+ |
| |
| salt bridge |
+-+ +-------------+ +-+
| | | | +---------+ | | | |
|-| |--| |-| |-| |--| |-|
| | | | | | | | | | | |
| +-+ | | | | | | +-+ |
| Zn | | Cu |
| | | |
+----------+ +----------+
ZnSO4 CuSO4
solution solution
On the (left, right) side we have a (zinc, copper) electrode immersed in a solution of (ZnSO4, CuSO4). Both ZnSO4 and CuSO4 are soluble in aqueous solution. The standard reduction potentials (25°C and 1 M concentrations of Zn2+ and Cu2+) for the reduction of Zn2+ and Cu2+ are
Zn2+ + 2e- => Zn, E0 = - 0.76 V
Cu2+ + 2e- => Cu, E0 = + 0.34 V
Cu2+ is easier to reduce than Zn2+ because its standard reduction potential is more positive, so in the above electrochemical cell the Cu2+ will be reduced and the Zn will be oxidized. The actual half reactions that occur are
Zn => Zn2+ + 2e- => Zn, E0 = + 0.76 V
Cu2+ + 2e- => Cu, E0 = + 0.34 V
A voltmeter would measure an electromotive force (emf) of
Ecell0 = + 0.76 V + 0.34 V = + 1.10 V
The (zinc, copper) electrode is the (anode, cathode). Electrons flow through the wire from the zinc electrode to the copper electrode, from the anode to the cathode. The (anode, cathode) is the (-, +) terminal. A battery is often represented schematically by the following symbol
- |+
-----||-----
|
where the (shorter, longer) line represents the (anode, cathode). In physics, the charge carriers are thought of as positive charges which flow into the - terminal and out of the + terminal, even though physically the charge carriers are electrons which are negatively charged and flow in the opposite direction.
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PROBLEM
Oxalic acid (H2C2O4) is present in many plants and vegetables.
(a) Balance the following equation in acid solution:
MnO4- + C2O42- => Mn2+ + CO2
(b) If a 1.00 g sample of H2C2O4 requires 24.0 mL of 0.0100 M KMnO4 solution to reach the equivalence point, what is the percent by mass of H2C2O4 in the sample?
[from Chang, Raymond 2002, Chemistry, Seventh Edition (New York: McGraw-Hill), problem 19.66]
SOLUTION
(a) Write down the oxidation numbers of each element.
+7-2 +3-2 +2 +4-2
MnO4- + C2O42- => Mn2+ + CO2
Carbon is oxidized from +3 to +4. Manganese is reduced from +7 to +2. The oxidation-reduction half reactions are
C2O42- => CO2 (oxidation)
MnO4- => Mn2+ (reduction)
We balance these two half reactions by first balancing the non-oxygen non-hydrogen elements, then balancing oxygen by adding H2O, then balancing hydrogen by adding H+, then balancing charge by adding e-. The balanced half reactions are
C2O42- => 2CO2 + 2e- (oxidation)
MnO4- + 8H+ + 5e- => Mn2+ + 4H2O (reduction)
Add 5 times the first equation to 2 times the second equation.
5C2O42- +2MnO4- + 16H+ => 10CO2 + 2Mn2+ + 8H2O
(b) The amount of KMnO4 is (0.024 L)(0.0100 M) = 2.4 x 10-4 mole. The amount of C2O42- is (5/2)(2.4 x 10-4 mole) = 6 x 10-4 mole => 6 x 10-4 mole of H2C2O4. The molar mass of H2C2O4 is (2)(1.008 g/mole) + (2)(12.01 g/mole) + (4)(16.00 g/mole) = 90.036 g/mole. The mass of H2C2O4 is (6 x 10-4)(90.036 g/mole) = 5.402 x 10-2 g. The sample is (5.402 x 10-2 g) / (1.00 g) = 5.402% H2C2O4 by mass.
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PROBLEM
Write balanced equations for the following reactions and determine the maximum potential voltage:
(a) Zn + Ag+ =>
(b) Sn + Cu2+ =>
(c) Zn + Fe2+ =>
(d) Pb + MnO4- + H+ =>
(e) Fe + NO3- + H+ =>
[from Kavanah, Patrick, and Robbins, Jack 1971, Concepts in Modern Chemistry (New York: Cambridge Book Company), problem 16.1]
SOLUTION
(a) Referring to the table of half-reactions, the complete unbalanced equation is
Zn + Ag+ => Zn2+ + Ag
The half-reactions are
Zn => Zn2+ + 2e-, E0 = 0.76 V (oxidation)
Ag+ + e- => Ag, E0 = 0.80 V (reduction)
Add the first equation to two times the second equation.
Zn + 2Ag+ => Zn2+ + 2Ag, E0 = 1.56 V
(b) Referring to the table of half-reactions, the complete unbalanced equation is
Sn + Cu2+ => Sn2+ + Cu
The half-reactions are
Sn => Sn2+ + 2e-, E0 = 0.14 V (oxidation)
Cu2+ + 2e- => Cu, E0 = 0.34 V (reduction)
Add the two equations.
Sn + Cu2+ => Sn2+ + Cu, E0 = 0.48 V
(c) Referring to the table of half-reactions, the complete unbalanced equation is
Zn + Fe2+ => Zn2+ + Fe
The half-reactions are
Zn => Zn2+ + 2e-, E0 = 0.76 V (oxidation)
Fe2+ + 2e- => Fe, E0 = - 0.44 V (reduction)
Add the two equations.
Zn + Fe2+ => Zn2+ + Fe, E0 = 0.32 V
(d) Referring to the table of half-reactions, the complete unbalanced equation is
Pb + MnO4- + H+ => Pb2+ + Mn2+
The half-reactions are
Pb => Pb2+ (oxidation)
MnO4- => Mn2+ (reduction)
Balance non-oxygen, non-hydrogen elements, then balance oxygen by adding H2O, then balance hydrogen by adding H+, then balance charge by adding e-. The standard half-cell potentials are as indicated.
Pb => Pb2+ + 2e-, E0 = 0.13 V (oxidation)
MnO4- + 8H+ + 5e- => Mn2+ + 4H2O, E0 = 1.52 V (reduction)
Add five times the first equation to two times the second equation. The maximum potential voltage is obtained by adding the two half-cell potentials.
5Pb + 2MnO4- + 16H+ => 5Pb2+ + 2Mn2+ + 8H2O, E0 = 1.65 V
(e) Referring to the table of half-reactions, the complete unbalanced equation is
Fe + NO3- + H+ => Fe2+ + NO
Iron is oxidized from 0 to +2 while nitrogen is reduced from +5 to +2. The half-reactions are
Fe => Fe2+ (oxidation)
NO3- => NO (reduction)
Balance non-oxygen, non-hydrogen elements, then balance oxygen by adding H2O, then balance hydrogen by adding H+, then balance charge by adding e-. The standard half-cell potentials are as indicated.
Fe => Fe2+ + 2e-, E0 = 0.44 V (oxidation)
NO3- + 4H+ + 3e- => NO + 2H2O, E0 = 0.96 V (reduction)
Add three times the first equation to two times the second equation.
3Fe + 2NO3- + 8H+ => 3Fe2+ + 2NO + 4H2O, E0 = 1.40 V
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PROBLEM
The following are some substances that contain the element nitrogen: KNO3, KNO2, NH4Cl, N2H5, N2H4, NO2, N2, NH3, N2O4, HNO3.
(a) What is the oxidation state of nitrogen in each substance?
(b) In relation to nitrogen, which substances can act as oxidizing agents?
(c) In relation to nitrogen, which substances can act as reducing agents?
(d) In which of the substances is nitrogen completely oxidized?
(e) In which substance does nitrogen have the lowest oxidation number?
(f) Name the substances that can, in relation to nitrogen, act both as an oxidizing agent and as a reducing agent.
[from Kavanah, Patrick, and Robbins, Jack 1971, Concepts in Modern Chemistry (New York: Cambridge Book Company), reasoning exercise 16.6, p. 389]
SOLUTION
(a) KNO3 (potassium nitrate):
+1+5-2
KNO3
KNO2 (potassium nitrite):
+1+3-2
KNO2
NH4Cl (ammonium chloride):
+5-1-1
NH4Cl
N2H5 (dinitrogen pentahydride):
+5/2-1
N2H5
N2H4 (dinitrogen tetrahydride):
+2-1
N2H4
NO2 (nitrogen dioxide):
+4-2
NO2
N2 (nitrogen):
0
N2
NH3 (ammonia):
+3-1
NH3
N2O4 (dinitrogen tetraoxide):
+4-2
N2O4
HNO3 (nitric acid):
+1+5-2
HNO3
(b) Since the electronic structure of nitrogen is 1s22s22p3, it can acquire up to three additional valence shell electrons, resulting in an oxidation state of -3. Thus, all of the above substances can be reduced and all can act as oxidizing agents.
(c) Nitrogen can lose up to five valence shell electrons, resulting in an oxidation state of +5. All of the substances can be oxidized and act as reducing agents except those which have oxidation states of +5: KNO3, NH4Cl, HNO3.
(d) Nitrogen is completely oxidized in KNO3, NH4Cl, and HNO3.
(e) Nitrogen has the lowest oxidation number in N2.
(f) All except KNO3, NH4Cl, and HNO3 can act both as an oxidizing agent and as a reducing agent.
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PROBLEM
Balance the following redox equation:
Cr2O72-(aq) + HNO2(aq) + H+(aq) => Cr3+ + NO3- + H2O(l)
[from Sample Questions for Chemistry Advanced Placement Test, College Board, Question 18]
SOLUTION
Write the oxidation numbers for each element.
+6 -2 +1+3-2 +1 +3 +5-2 +1-2
Cr2O72- + HNO2 + H+ => Cr3+ + NO3- + H2O
Nitrogen is oxidized from +3 to +5. Chromium is reduced from +6 to +3. The half reactions are
HNO2 => NO3- (oxidation)
Cr2O72- => Cr3+ (reduction)
Balance non-oxygen, non-hydrogen elements. Then balance oxygen by adding H2O, balance hydrogen by adding H+, and balance charge by adding e-. The balanced half reactions are
HNO2 + H2O => NO3- + 3H+ + 2e- (oxidation)
Cr2O72- + 14H+ + 6e- => 2Cr3+ + 7H2O (reduction)
Add three times the first equation to the second equation.
3HNO2 + 3H2O + Cr2O72- + 14H+ + 6e- => 3NO3- + 9H+ + 6e- + 2Cr3+ + 7H2O
Subtract 6e-, 9H+, and 3H2O from both sides. The balanced redox equation is
Cr2O72- + 3HNO2 + 5H+ => 2Cr3+ + 3NO3- + 4H2O
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PROBLEM
Balance the equations for each of the following reactions using the half-reaction (ion-electron) method. Also indicate which is the oxidant, which is the reductant, and the changes in oxidation numbers of the main elements involved.
(a) Cu + H+ + NO3- => Cu2+ + NO + H2O
(b) Fe2+ + MnO4- + H+ => Fe3+ + Mn2+ + H2O
(c) PbO2 + Mn2+ + H+ => Pb2+ + MnO4- + H2O
(d) Br2 + SO2 + H2O => H2SO4 + HBr
(e) I2 + S2O32- => I - + S4O62-
(f) Bi2O3 + NaOH + NaOCl => NaBiO3 + NaCl + H2O
[from Kavanah, Patrick, and Robbins, Jack 1971, Concepts in Modern Chemistry (New York: Cambridge Book Company), reasoning exercise 16.10, p. 390]
SOLUTION
(a) Write down the oxidation numbers for each element.
0 +1 +5-2 +2 +2-2 +1-2
Cu + H+ + NO3- => Cu2+ + NO + H2O
Copper is oxidized from 0 to +2 and is the reductant. Nitrogen is reduced from +5 to +2 and is the oxidant. The half-reactions are
Cu => Cu2+ (oxidation)
NO3- => NO (reduction)
Balance non-hydrogen, non-oxygen elements, then balance oxygen by adding H2O, then balance hydrogen by adding H+, then balance charge by adding e-. The balanced half-reactions are
Cu => Cu2+ + 2e- (oxidation)
NO3- + 4H+ + 3e- => NO + 2H2O (reduction)
Add three times the first equation to two times the second equation.
3Cu + 8H+ + 2NO3- => 3Cu2+ + 2NO + 4H2O
(b) Write down the oxidation numbers for each element.
+2 +7-2 +1 +3 +2 +1-2
Fe2+ + MnO4- + H+ => Fe3+ + Mn2+ + H2O
Iron is oxidized from +2 to +3 and is the reductant. Manganese is reduced from +7 to +2 and is the oxidant. The half-reactions are
Fe2+ => Fe3+ (oxidation)
MnO4- => Mn2+ (reduction)
Balance non-hydrogen, non-oxygen elements, then balance oxygen by adding H2O, then balance hydrogen by adding H+, then balance charge by adding e-. The balanced half-reactions are
Fe2+ => Fe3+ + e- (oxidation)
MnO4- + 8H+ + 5e- => Mn2+ + 4H2O (reduction)
Add five times the first equation to the second equation.
5Fe2+ + MnO4- + 8H+ => 5Fe3+ + Mn2+ + 4H2O
(c) Write down the oxidation numbers for each element.
+4-2 +2 +1 +2 +7-2 +1-2
PbO2 + Mn2+ + H+ => Pb2+ + MnO4- + H2O
Manganese is oxidized from +2 to +7 and is the reductant. Lead is reduced from +4 to +2 and is the oxidant. The half-reactions are
Mn2+ => MnO4- (oxidation)
PbO2 => Pb2+ (reduction)
Balance non-hydrogen, non-oxygen elements, then balance oxygen by adding H2O, then balance hydrogen by adding H+, then balance charge by adding e-. The balanced half-reactions are
Mn2+ + 4H2O => MnO4- + 8H+ + 5e- (oxidation)
PbO2 + 4H+ + 2e- => Pb2+ + 2H2O (reduction)
Add two times the first equation to five times the second equation.
2Mn2+ + 8H2O + 5PbO2 + 20H+ => 2MnO4- + 16H+ + 5Pb2+ + 10H2O
Subtract 16H+ and 8H2O from both sides.
5PbO2 + 2Mn2+ + 4H+ => 5Pb2+ + 2MnO4- + 2H2O
(d) Write down the oxidation numbers for each element.
0 +4-2 +1-2 +1+6-2 +1-1
Br2 + SO2 + H2O => H2SO4 + HBr
Sulphur is oxidized from +4 to +6 and is the reductant. Bromine is reduced from 0 to -1 and is the oxidant. The half-reactions are
SO2 => H2SO4 (oxidation)
Br2 => HBr (reduction)
Balance non-hydrogen, non-oxygen elements, then balance oxygen by adding H2O, then balance hydrogen by adding H+, then balance charge by adding e-. The balanced half-reactions are
SO2 + 2H2O => H2SO4 + 2H+ + 2e- (oxidation)
Br2 + 2H+ + 2e- => 2HBr (reduction)
Add the two equations.
Br2 + SO2 + 2H2O => H2SO4 + 2HBr
(e) Write down the oxidation numbers for each element.
0 +2-2 -1 +5/2-2
I2 + S2O32- => I- + S4O62-
Sulfur is oxidized from +2 to +5/2 and is the reductant. Iodine is reduced from 0 to -1 and is the oxidant. The half-reactions are
S2O32- => S4O62- (oxidation)
I2 => I - (reduction)
Balance non-hydrogen, non-oxygen elements, then balance oxygen by adding H2O, then balance hydrogen by adding H+, then balance charge by adding e-. The balanced half-reactions are
2S2O32- => S4O62- + 2e- (oxidation)
I2 + 2e- => 2I - (reduction)
Add the two equations.
I2 + 2S2O32- => 2I - + S4O62-
(f) Write down the oxidation numbers for each element.
+3-2 +1-2+1 +1-2+1 +1+5-2 +1-1 +1-2
Bi2O3 + NaOH + NaOCl => NaBiO3 + NaCl + H2O
Bismuth is oxidized from +3 to +5 and is the reductant. Chlorine is reduced from +1 to -1 and is the oxidant. Write the oxidation and reduction half-reactions.
Bi2O3 => NaBiO3 (oxidation)
NaOCl => NaCl (reduction)
Balance non-hydrogen, non-oxygen elements, then balance oxygen by adding H2O, then balance hydrogen by adding H+, then balance charge by adding e-. The balanced half-reactions are
Bi2O3 + 2Na+ + 3H2O => 2NaBiO3 + 6H+ + 4e- (oxidation)
NaOCl + 2H+ + 2e- => NaCl + H2O (reduction)
Add the oxidation half-reaction to twice the reduction half-reaction.
Bi2O3 + 2Na+ + 3H2O + 2NaOCl + 4H+ => 2NaBiO3 + 6H+ + 2NaCl + 2H2O
Subtract 4H+ from each side.
Bi2O3 + 2Na+ + 3H2O + 2NaOCl => 2NaBiO3 + 2H+ + 2NaCl + 2H2O
Add 2OH- to both sides.
Bi2O3 + 2Na+ + 3H2O + 2NaOCl + 2OH- => 2NaBiO3 + 2H+ + 2NaCl + 2H2O + 2OH-
Combine the Na+ and the OH- on the left side, and combine the H+ and the OH- on the right side.
Bi2O3 + 2NaOH + 3H2O + 2NaOCl => 2NaBiO3 + 2NaCl + 4H2O
Subtract 3H2O from both sides. The balanced net equation is
Bi2O3 + 2NaOH + 2NaOCl => 2NaBiO3 + 2NaCl + H2O
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PROBLEM
Balance the equations for each of the following reactions, indicating what is oxidized and what is reduced. Indicate the oxidizing and reducing agents and the changes in oxidation numbers of the main elements involved.
(a) H2S(g) + H2SO3(aq) => S(s) + H2O
(b) KI(aq) + O2(g) => KI3(aq) + H2O
(c) Fe(NO3)2(aq) + HNO3(aq) => Fe(NO3)3(aq) + NO(g) + H2O
(d) K2Cr2O7 + HCl => KCl + CrCl3 + Cl2 + H2O
(e) CuS + HNO3 => Cu(NO3)2 + SO2 + NO2 +H2O
(f) KMnO4 + HCl => KCl + MnCl2 + Cl2 + H2O
(g) H2S + HNO3 => NO2 + S + H2O
(h) Cl2 + H2O + SO2 => SO42- + Cl- + H+
(i) HI + H2SO4 => H2S + H2O + I2
(j) Cu + HNO3 => Cu(NO3)2 + NO + H2O
[from Kavanah, Patrick, and Robbins, Jack 1971, Concepts in Modern Chemistry (New York: Cambridge Book Company), problem 16.2]
SOLUTION
(a) Write down the oxidation numbers for each element
+1-2 +1+4-2 0 +1-2
H2S(g) + H2SO3(aq) => S(s) + H2O
Sulfur is oxidized from -2 to 0 and reduced from +4 to 0. Sulfur is both an oxidizing and a reducing agent. The oxidation and reduction half-reactions are
H2S => S (oxidation)
H2SO3 => S (reduction)
Balance non-oxygen, non-hydrogen elements, then balance oxygen by adding H2O, then balance hydrogen by adding H+, then balance charge by adding e-.
H2S => S + 2H+ + 2e- (oxidation)
H2SO3 + 4H+ + 4e- => S + 3H2O (reduction)
Add two times the first equation to the second equation.
2H2S + H2SO3 => 3S + 3H2O
(b) Write down the oxidation numbers for each element.
+1-1 0 +1-1/3 +1-2
KI(aq) + O2(g) => KI3(aq) + H2O
Iodine is oxidized from -1 to -1/3 and is the reducing agent. Oxygen is reduced from 0 to -2 and is the oxidizing agent. The oxidation and reduction half-reactions are
KI => KI3 (oxidation)
O2 => H2O (reduction)
Balance non-oxygen, non-hydrogen elements, then balance oxygen by adding H2O, then balance hydrogen by adding H+, then balance charge by adding e-.
3KI => KI3 + 2K+ + 2e- (oxidation)
O2 + 4H+ + 4e- => 2H2O (reduction)
Add two times the first equation to the second equation.
6KI + O2 + 4H+ => 2KI3 + 4K+ + 2H2O
Add 4OH- to both sides.
6KI + O2 + 4H2O => 2KI3 + 4KOH + 2H2O
Subtract 2H2O from both sides.
6KI + O2 + 2H2O => 2KI3 + 4KOH
(c) Write down the oxidation numbers for each element.
+2+5-2 +1+5-2 +3+5-2 +2-2 +1-2
Fe(NO3)2 + HNO3 => Fe(NO3)3 + NO + H2O
Iron is oxidized from +2 to +3 and is the reducing agent. Nitrogen is reduced from +5 to +2 and is the oxidizing agent. Rewrite the above equation with the terms separated into ions.
Fe2+ + 2NO3- + H+ + NO3- => Fe3+ + 3NO3- + NO + H2O
The oxidation and reduction half-reactions are
Fe2+ => Fe3+ (oxidation)
NO3- => NO (reduction)
Balance non-oxygen, non-hydrogen elements, then balance oxygen by adding H2O, then balance hydrogen by adding H+, then balance charge by adding e-.
Fe2+ => Fe3+ + e- (oxidation)
NO3- + 4H+ + 3e- => NO + 2H2O (reduction)
Add three times the first equation to the second equation.
3Fe2+ + NO3- + 4H+ => 3Fe3+ + NO + 2H2O
Add 9NO3- to both sides.
3Fe2+ + 10NO3- + 4H+ => 3Fe3+ + NO + 2H2O + 9NO3-
Combine 3Fe2+ with 6NO3- and 4H+ with 4NO3- on the left side, and 3Fe3+ with 9NO3- on the right side.
3Fe(NO3)2 + 4HNO3 => 3Fe(NO3)3 + NO + 2H2O
(d) Write down the oxidation numbers for each element.
+1+6-2 +1-1 +1-1 +3-1 0 +1-2
K2Cr2O7 + HCl => KCl + CrCl3 + Cl2 + H2O
Chlorine is oxidized from -1 to 0 and is the reducing agent. Chromium is reduced from +6 to +3 and is the oxidizing agent. Rewrite the above equation with the terms written as separate ions.
2K+ + Cr2O72- + H+ + Cl- => K+ + Cl- + Cr3+ + 3Cl- + Cl2 + H2O
The oxidation and reduction half-reactions are
Cl- => Cl2 (oxidation)
Cr2O72- => Cr3+ (reduction)
Balance the non-oxygen, non-hydrogen elements, then balance oxygen by adding H2O, then balance hydrogen by adding H+, then balance charge by adding e-.
2Cl- => Cl2 + 2e- (oxidation)
Cr2O72- + 14H+ + 6e- => 2Cr3+ + 7H2O (reduction)
Add three times the first equation to the second equation.
6Cl- + Cr2O72- + 14H+ => 3Cl2 + 2Cr3+ + 7H2O
Add 8Cl- and 2K+ to both sides.
14Cl- + Cr2O72- + 14H+ + 2K+ => 3Cl2 + 2Cr3+ + 7H2O + 8Cl- + 2K+
Combine 2K+ with Cr2O72- and 14H+ with 14Cl- on the left side and combine 2K+ with 2Cl- and 2Cr3+ with 6Cl- on the right side.
K2Cr2O7 + 14HCl => 2KCl + 2CrCl3 + 3Cl2 + 7H2O
(e) Write down the oxidation numbers for each element.
+2-2 +1+5-2 +2+5-2 +4-2 +4-2 +1-2
CuS + HNO3 => Cu(NO3)2 + SO2 + NO2 + H2O
Sulfur is oxidized from -2 to +4 and is the reducing agent. Nitrogen is reduced from +5 to +4 and is the oxidizing agent. The half-reactions are
CuS => SO2 (oxidation)
HNO3 => NO2 (reduction)
Balance the non-oxygen, non-hydrogen elements, then balance oxygen by adding H2O, then balance hydrogen by adding H+, then balance charge by adding e-.
CuS + 2H2O => SO2 + Cu2+ + 4H+ + 6e- (oxidation)
HNO3 + H+ + e- => NO2 + H2O (reduction)
Add the first equation to six times the second equation.
CuS + 2H2O + 6HNO3 + 6H+ => SO2 + Cu2+ + 4H+ + 6NO2 + 6H2O
Add 2NO3- to both sides.
CuS + 2H2O + 6HNO3 + 6H+ + 2NO3- => SO2 + Cu(NO3)2 + 4H+ + 6NO2 + 6H2O
Combine 2H+ and 2NO3- on the left side and subtract 4H+ and 2H2O from both sides.
CuS + 8HNO3 => Cu(NO3)2 + SO2 + 6NO2 + 4H2O
(f) Write down the oxidation numbers for each element.
+1+7-2 +1-1 +1-1 +2-1 0 +1-2
KMnO4 + HCl => KCl + MnCl2 + Cl2 + H2O
Chlorine is oxidized from -1 to 0 and is the reducing agent. Manganese is reduced from +7 to +2 and is the oxidizing agent. The half-reactions are
HCl => Cl2 (oxidation)
KMnO4 => MnCl2 (reduction)
Balance the non-oxygen, non-hydrogen elements, then balance oxygen by adding H2O, then balance hydrogen by adding H+, then balance charge by adding e-.
2HCl => Cl2 + 2H+ + 2e- (oxidation)
KMnO4 + 2Cl- + 8H+ + 5e- => MnCl2 + K+ + 4H2O (reduction)
Add five times the first equation to two times the second equation.
10HCl + 2KMnO4 + 4Cl- + 16H+ => 5Cl2 + 10H+ + 2MnCl2 + 2K+ + 8H2O
Add 2Cl- to both sides and subtract 10H+ from both sides.
10HCl + 2KMnO4 + 6Cl- + 6H+ => 5Cl2 + 2MnCl2 + 2K+ + 8H2O + 2Cl-
Combine the 6H+ and 6Cl- on the left side and the 2K+ and 2Cl- on the right side.
2KMnO4 + 16HCl => 2KCl + 2MnCl2 + 5Cl2 + 8H2O
(g) Write down the oxidation numbers for each element.
+1-2 +1+5-2 +4-2 0 +1-2
H2S + HNO3 => NO2 + S + H2O
Sulfur is oxidized from -2 to 0 and is the reducing agent. Nitrogen is reduced from +5 to +4 and is the oxidizing agent. The half-reactions are
H2S => S (oxidation)
HNO3 => NO2 (reduction)
Balance the non-oxygen, non-hydrogen elements, then balance oxygen by adding H2O, then balance hydrogen by adding H+, then balance charge by adding e-.
H2S => S + 2H+ + 2e- (oxidation)
HNO3 + H+ + e- => NO2 + H2O (reduction)
Add the first equation to two times the second equation.
H2S + 2HNO3 => 2NO2 + S + 2H2O
(h) Write down the oxidation numbers for each element.
0 +1-2 +4-2 +6-2 -1 +1
Cl2 + H2O + SO2 => SO42- + Cl- + H+
Sulfur is oxidized from +4 to +6 and is the reducing agent. Chlorine is reduced from 0 to -1 and is the oxidizing agent. The half-reactions are
SO2 => SO42- (oxidation)
Cl2 => Cl- (reduction)
Balance non-oxygen, non-hydrogen elements, then balance oxygen by adding H2O, then balance hydrogen by adding H+, then balance charge by adding e-. The balanced half-reactions are
SO2 + 2H2O => SO42- + 4H+ + 2e- (oxidation)
Cl2 + 2e- => 2Cl- (reduction)
Add the two half-reactions. The balanced net equation is
Cl2 + SO2 + 2H2O => SO42- + 2Cl- + 4H+
(i) Write down the oxidation numbers for each element.
+1-1 +1+6-2 +1-2 +1-2 0
HI + H2SO4 => H2S + H2O + I2
Iodine is oxidized from -1 to 0 and is the reducing agent. Sulfer is reduced from +6 to -2 and is the oxidizing agent. The half-reactions are
HI => I2 (oxidation)
H2SO4 => H2S (reduction)
Balance non-oxygen, non-hydrogen elements, then balance oxygen by adding H2O, then balance hydrogen by adding H+, then balance charge by adding e-.
2HI => I2 + 2H+ + 2e- (oxidation)
H2SO4 + 8H+ + 8e- => H2S + 4H2O (reduction)
Add four times the first equation to the second equation.
8HI + H2SO4 => H2S + 4H2O + 4I2
(j) Write down the oxidation numbers for each element.
0 +1+5-2 +2+5-2 +2-2 +1-2
Cu + HNO3 => Cu(NO3)2 + NO + H2O
Copper is oxidized from 0 to +2 and is the reducing agent. Nitrogen is reduced from +5 to +2 and is the oxidizing agent. The half-reactions are
Cu => Cu(NO3)2 (oxidation)
HNO3 => NO (reduction)
Balance non-oxygen, non-hydrogen elements, then balance oxygen by adding H2O, then balance hydrogen by adding H+, then balance charge by adding e-. The balanced half-reactions are
Cu + 2NO3- => Cu(NO3)2 + 2e- (oxidation)
HNO3 + 3H+ + 3e- => NO + 2H2O (reduction)
Add three times the first equation to two times the second equation.
3Cu + 6NO3- + 2HNO3 + 6H+ => 3Cu(NO3)2 + 2NO + 4H2O
Combine the 6NO3- and 6H+ on the left side.
3Cu + 8HNO3 => 3Cu(NO3)2 + 2NO + 4H2O
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PROBLEM
Consider the following oxidation-reduction reaction:
KMnO4 + H2SO3 => K2SO4 + MnSO4 + H2SO4 + H2O
(a) Which species is/are oxidized and reduced?
(b) Write the balanced oxidation and reduction half-reactions.
(c) Write the balanced equation for the full reaction.
[from Mascetta, Joseph A. 2002, Barron's How to Prepare for the SAT II Chemistry, 7th Edition (Hauppauge, New York: Barron's Educational Series), Practice Test 1, problem 51]
SOLUTION
(a) Write the oxidation numbers for each species.
+1+7-2 +1+4-2 +1+6-2 +2+6-2 +1+6-2 +1-2
KMnO4 + H2SO3 => K2SO4 + MnSO4 + H2SO4 + H2O
We see that Mn is reduced from +7 to +2 and S is oxidized from +4 to +6.
(b) Rewrite the unbalanced equation, separating each component, except water, into ions.
K+ + MnO4- + 2H+ + SO32- => 2K+ + SO42- + Mn2+ + SO42- + 2H+ + SO42- + H2O
The unbalanced oxidation and reduction half-reactions are
SO32- => SO42- (oxidation)
MnO4- => Mn2+ (reduction)
Add one H2O to the left side of the oxidation half-reaction and 4H2O to the right side of the reduction half-reaction to balance the oxygens.
SO32- + H2O => SO42- (oxidation)
MnO4- => Mn2+ + 4H2O (reduction)
Add 2H+ to the right side of the oxidation half-reaction and 8H+ to the left side of the reduction half-reaction to balance the hydrogens.
SO32- + H2O => SO42- + 2H+ (oxidation)
MnO4- + 8H+ => Mn2+ + 4H2O (reduction)
Add 2e- to the right side of the oxidation half-reaction and 5e- to the left side of the reduction half-reaction to balance the charge.
SO32- + H2O => SO42- + 2H+ + 2e-(oxidation)
MnO4- + 8H+ + 5e- => Mn2+ + 4H2O (reduction)
These are the balanced oxidation and reduction half-reactions.
(c) Add five times the oxidation half-reaction to two times the reduction half-reaction.
(5)(SO32- + H2O => SO42- + 2H+ + 2e-) (oxidation)
(2)(MnO4- + 8H+ + 5e- => Mn2+ + 4H2O) (reduction)
================================
5SO32- + 5H2O + 2MnO4- + 16H+ + 10e- =>
5SO42- + 10H+ + 10e- + 2Mn2+ + 8H2O
Subtract 10e- from both sides.
5SO32- + 5H2O + 2MnO4- + 16H+ =>
5SO42- + 10H+ + 2Mn2+ + 8H2O
We need to have potassium ions in order to be able to balance the original equation for the reaction, so add 2K+ to both sides.
5SO32- + 5H2O + 2MnO4- + 16H+ + 2K+ =>
5SO42- + 10H+ + 2Mn2+ + 8H2O + 2K+
Combine the 2K+ with 2MnO4- on the left side and the 2K+ with one SO42- on the right side.
5SO32- + 5H2O + 2KMnO4 + 16H+ =>
4SO42- + 10H+ + 2Mn2+ + 8H2O + K2SO4
Combine 10H+ with 5SO32- on the left side and 2Mn2+ with 2SO4- on the right side.
5H2SO3 + 5H2O + 2KMnO4 + 6H+ =>
2SO42- + 10H+ + 2MnSO4 + 8H2O + K2SO4
Combine 4H+ with 2SO42- on the right side.
5H2SO3 + 5H2O + 2KMnO4 + 6H+ =>
2H2SO4 + 6H+ + 2MnSO4 + 8H2O + K2SO4
Subtract 5H2O and 6H+ from both sides.
5H2SO3 + 2KMnO4 =>
2H2SO4 + 2MnSO4 + 3H2O + K2SO4
Alternate Method:
Assume that the coefficients in the balanced equation are A, B, C, D, E, and F and solve for them by requiring that the numbers of each type of atom be equal on both sides of the equation.
A · KMnO4 + B · H2SO3 => C · K2SO4 + D · MnSO4 + E · H2SO4 + F · H2O
K: A = 2C
Mn: A = D
O: 4A + 3B = 4C + 4D + 4E + F (1)
H: 2B = 2E + 2F (2)
S: B = C + D + E (3)
We have five equations in six unknowns. Take A = D = 1. Then C = A/2 = 1/2. Substituting into (1), (2), and (3), we get
4 + 3B = 2 + 4 + 4E + F => 3B = 2 + 4E + F (4)
B = E + F (5)
B = 1/2 + 1 + E = 3/2 + E (6)
Use (5) to write B in terms of E and F in (4) and (6).
3E + 3F = 2 + 4E + F => 2F = 2 + E (7)
E + F = 3/2 + E => F = 3/2 (8)
Substitute F = 3/2 into (7) to solve for E.
E = 2F - 2 = 2(3/2) - 2 = 1
Use (6) to solve for B.
B = 3/2 + 1 = 5/2
Thus, we have A = 1, B = 5/2, C = 1/2, D = 1, E = 1, and F = 3/2 or, multiplying all of these by two, A = 2, B = 5, C = 1, D = 2, E = 2, and F = 3. Thus, the balanced equation is
2KMnO4 + 5H2SO3 => K2SO4 + 2MnSO4 + 2H2SO4 + 3H2O
or
5H2SO3 + 2KMnO4 =>
2H2SO4 + 2MnSO4 + 3H2O + K2SO4
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PROBLEM
If gold foil were placed in a solution containing zinc ions, what would the reaction potential be?
[from Mascetta, Joseph A. 2002, Barron's How to Prepare for the SAT II Chemistry, 7th Edition (Hauppauge, New York: Barron's Educational Series), problem 12.11]
SOLUTION
The half-cell oxidation potentials for zinc and gold are
Zn => Zn2+ + 2e-, E0 = 0.76 V
Au => Au3+ + 3e-, E0 = - 1.50 V
If gold foil were placed in a solution containing zinc ions, the following combination of reactions could occur:
Zn2+ + 2e- => Zn, E0 = - 0.76 V (1)
Au => Au3+ + 3e-, E0 = - 1.50 V (2)
for which the balanced equation, obtained by adding three times (1) to two times (2), is
3Zn2+ + 2Au => 3Zn + 2Au3+
The reaction potential would be E0 = - 0.76 V - 1.50 V = - 2.26 V, which indicates that the reaction would not be spontaneous.
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PROBLEM
Which of the following oxidizing agents become stronger as the concentration of H+ increases? Which are unchanged and which become weaker? (a) Cl2, (b) Cr2O72-, (c) Fe3+, (d) MnO4-.
[from Mahan, Bruce H. 1966, College Chemistry (Reading, Massachusetts: Addison-Wesley), problem 7.4]
SOLUTION
(a) From the table of standard reduction potentials at 25°C, we see that in an electrochemical cell Cl2 is reduced according to
Cl2(g) + 2e- => 2Cl-(aq)
The oxidizing strength of Cl2 is independent of the concentration of H+.
(b) Cr2O72- is reduced according to
Cr2O72-(aq) + 14H+(aq) + 6e- => 2Cr3+(aq) + 7H2O
The oxidizing strength of Cr2O72- is increased if the concentration of H+ is increased.
(c) Fe3+ is reduced according to
Fe3+(aq) + e- => Fe2+(aq)
The oxidizing strength of Fe3+ is independent of the concentration of H+.
(d) MnO4- is reduced according to
MnO4-(aq) + 2H2 + 3e- => MnO2(s) + 4OH-(aq)
The oxidizing strength of MnO4- is increased if the concentration of OH- is decreased or if the concentration of H+ is increased.
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PROBLEM
From the appropriate values of the standard reduction potential E0, calculate DE0 and the equilibrium constant for the reaction
Hg2+ + Hg => Hg22+
[from Mahan, Bruce H. 1966, College Chemistry (Reading, Massachusetts: Addison-Wesley), problem 7.6]
SOLUTION
From the table of standard reduction potentials, we have the following half reactions:
2Hg2+(aq) + 2e- => Hg22+(aq), E0 = +0.92 V (1)
Hg22+(aq) + 2e- => 2Hg(l), E0 = +0.789 V (2)
From (2), we get
2Hg(l) => Hg22+(aq) + 2e-, E0 = -0.789 V (3)
Adding (1) and (3), we get
2Hg2+(aq) + 2Hg(l) => 2Hg22+, DE0 = E0(1) + E0(3) = +0.92 V - 0.789 V = +0.131 V,
2e- transferred
or
Hg2+(aq) + Hg(l) => Hg22+, DE0 = +0.131 V, 1e- transferred
The equilibrium constant K of this reaction is given by
log K = nDE0/0.059 = (1)(0.131 V)/0.059 = 2.220 => K = 102.220 = 166.1
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PROBLEM
A half-cell (A) consisting of a strip of nickel dipping into a 1 M solution of Ni2+, and a half-cell (B) consisting of a strip of zinc dipping into a 1 M solution of Zn2+ were successively connected with a standard hydrogen half-cell. The magnitudes of the individual half-cell potentials were then determined as
(A) Ni2+ + 2e- => Ni, |E0| = 0.25 V
(B) Zn2+ + 2e- => Zn, |E0| = 0.77 V
(a) When both the half-cells (A) and (B) were connected with the hydrogen half-cell, the metallic electrode (Ni or Zn) was found to be negative. What is the correct sign of the electrode potentials?
(b) Of the substances Ni, Ni2+, Zn, Zn2+; which is the strongest oxidant? Which is the strongest reductant?
(c) Will a noticeable reaction occur when metallic nickel is placed in a 1 M solution of Zn2+? Will a noticeable reaction occur when metallic zinc is placed in a 1 M solution of Ni2+?
(d) Zinc forms a complex ion with hydroxide ion, Zn(OH)42-. If hydroxide ion were added to half-cell (B), would its electrode potential become more positive, less positive, or be unaffected?
(e) If half-cells (A) and (B) were connected together, which electrode would be negative? What would the cell voltage be?
[from Mahan, Bruce H. 1966, College Chemistry (Reading, Massachusetts: Addison-Wesley), problem 7.8]
SOLUTION
(a) Since either metallic electrode is found to be negative, and the hydrogen electrode positive, the metallic electrode is being oxidized and is the anode, while hydrogen ions are being reduced at the hydrogen electrode, and the hydrogen electrode is the cathode. Nickel and zinc are both stronger reducing agents than hydrogen. The correct signs of the electrode potentials shown for reactions (A) and (B) are both negative.
(b) Only Ni2+ and Zn2+ are oxidants, with the following electrode potentials:
(A) Ni2+ + 2e- => Ni, E0 = - 0.25 (1) V
(B) Zn2+ + 2e- => Zn, E0 = - 0.77 V
Since E0 is larger for (A) than for (B), Ni2+ is the stronger oxidant.
Only Ni and Zn are reductants, with the following electrode potentials:
(C) Ni => Ni2+ + 2e-, E0 = 0.25 V
(D) Zn => Zn2+ + 2e-, E0 = 0.77 V
Since E0 is larger for (D) than for (C), Zn is the stronger reductant.
(c) Zinc is easier to oxidize than Ni, so if metallic Ni is placed in a 1 M solution of Zn2+, no noticeable reaction will occur, but if metallic Zn is placed in a 1 M solution of Ni2+, a noticeable reaction will occur.
(d) If the hydroxide ion were added to half cell (B), Zn(OH)42- would be formed, consuming Zn2+. The zinc electrode potential would become more negative or less positive.
(e) If half-cells (A) and (B) were connected, Zn would dissolve and Ni would be deposited. Electrons would flow from the Zn electrode to the Ni electrode. The Zn electrode would be negative and the Ni electrode would be positive. The two half-cell reactions would be
Zn => Zn2+ + 2e-, E0 = 0.77 V
Ni2+ + 2e- => Ni, E0 = - 0.25 V
The cell voltage would be
DE0 = 0.77 V - 0.25 V = 0.52 V
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PROBLEM
A galvanic cell consists of a strip of cobalt metal, Co, dipping into 1 M Co2+ solution, and another half-cell in which a piece of platinum dips into a 1 M solution of Cl-. Chlorine gas at 1 atm pressure is bubbled into this solution. The observed cell voltage is 1.63 V, and as the cell operates the cobalt electrode is negative. Given only that the standard potential for the chlorine-chloride ion half-cell is
(1/2)Cl2 + e- => Cl-, E0 = 1.36 V (1)
supply the desired information.
(a) What is the spontaneous cell reaction?
(b) What is the standard potential of the cobalt electrode?
(c) Would the cell voltage increase or decrease if the pressure of chlorine gas increased?
(d) What would the cell voltage be if the concentration of Co2+ were reduced to 0.01 M?
[from Mahan, Bruce H. 1966, College Chemistry (Reading, Massachusetts: Addison-Wesley), problem 7.10]
SOLUTION
(a) Since the cobalt electrode is negative, oxidation is occurring there:
Co => Co2+ + 2e- (2)
The overall cell reaction is obtained by doubling (1) and adding the result to (2):
Cl2 + Co => 2Cl- + Co2+ (3)
(b) The sum of the potentials of the chlorine and cobalt electrodes equals the observed cell voltage. Thus, the standard potential of the cobalt electrode is
E0 = 1.63 V - 1.36 V = 0.27 V
The standard reduction potential for cobalt is - 0.27 V.
(c) If the chlorine gas pressure increased, the potential of the chlorine half-cell would be larger, and the cell voltage would increase.
(d) According to the Nernst equation, the cell voltage DE is given by
DE = DE0 - (0.059 / n) log {[Cl-]2[Co2+]}
where DE0 is the standard cell potential, and n = 2 is the number of moles of transferred electrons represented by the above equation (3). If [Cl-] = 1 M and [Co2+] = 0.01 M, we have
DE = 1.63 V - (0.059 / 2) log(0.01) = 1.63 V + 0.059 V = 1.689 V
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PROBLEM
Consider the following unbalanced redox reaction:
HCl + KMnO4 => H2O + KCl + MnCl2 + Cl2
(a) Determine which species are oxidized and reduced.
(b) Balance this equation.
SOLUTION
(a) Write the oxidation numbers above each species.
+1-1 +1+7-2 +1-2 +1-1 +2-1 0
HCl + KMnO4 => H2O + KCl + MnCl2 + Cl2
From the above, we see that Cl is oxidized and Mn is reduced.
(b) Let the coefficients of the balanced equation be A, B, C, D, E, and F:
A · HCl + B · KMnO4 => C · H2O + D · KCl + E · MnCl2 + F · Cl2
Require that the number of atoms of each type be balanced:
H: A = 2C (1)
Cl: A = D + 2E + 2F (2)
K: B = D (3)
Mn: B = E (4)
O: 4B = C (5)
Assume that A = 1. Then from (1) we get C = 1/2. From (5) we get B = C/4 = 1/8. From (3) we get D = B = 1/8. From (4) we get E = B = 1/8. From (2) we get F = (A - D - 2E)/2 = (1 - 1/8 - 1/4)/2 = 5/16. Thus, the balanced equation is
HCl + (1/8)KMnO4 => (1/2)H2O + (1/8)KCl + (1/8)MnCl2 + (5/16)Cl2
Multiplying all coefficients by 16 we get
16HCl + 2KMnO4 => 8H2O + 2KCl + 2MnCl2 + 5Cl2
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PROBLEM
Propanoic acid, HC3H5O2, ionizes in water according to
HC3H5O2(aq) <=> C3H5O2-(aq) + H+(aq), Ka = 1.34 x 10-5
(a) Write the equilibrium-constant expression for the reaction.
(b) Calculate the pH of a 0.265 M solution of propanoic acid.
(c) A 0.496 g sample of sodium propanoate, NaC3H5O2, is added to a 50.0 mL sample of a 0.265 M solution of propanoic acid. Assuming that no change in the volume of the solution occurs, calculate each of the following: (i) the concentration of the propanoate ion, C3H5O2-(aq), in the solution, (ii) the concentration of the H+(aq) ion in the solution.
(d) The methanoate ion, HCO2-(aq), reacts with water to form methanoic acid and hydroxide ion according to
HCO2-(aq) + H2O(l) <=> HCO2H(aq) + OH -(aq)
Given that [OH -] is 4.18 x 10-6 M in a 0.309 M solution of sodium methanoate, calculate each of the following: (i) the value of Kb for the methanoate ion, HCO2-(aq), (ii) the value of Ka for methanoic acid, HCO2H.
(e) Which acid is stronger, propanoic acid or methanoic acid? Justify your answer.
[from 2005 AP Chemistry Free-Response Question 1]
SOLUTION
(a) Ka = [C3H5O2-][H+] / [HC3H5O2]
(b) Let x be the number of moles per liter of propanoic acid which dissociate. The equilibrium concentrations are
[C3H5O2-] = x = [H+]
[HC3H5O2] = 0.265 - x
We have
Ka = x2 / (0.265 - x) => 0.265Ka - Kax = x2 => x2 + Kax - 0.265Ka = 0
=> x = {- Ka + sqrt[Ka2 + (4)(0.265Ka)]} / 2
= {- 1.34 x 10-5 + sqrt[(1.34 x 10-5)2 + (4)(0.265)(1.34 x 10-5)]} / 2 = 1.878 x 10-3 M
=> [H+] = x = 1.878 x 10-3 M
=> pH = - log [H+] = - log(1.878 x 10-3) = 2.726
We also have
[C3H5O2-] = x = 1.878 x 10-3 M
[HC3H5O2] = 0.265 M - x = 0.265 M - 1.878 x 10-3 M = 0.2631 M
(c) The molar mass of NaC3H5O2 is
M = 22.99 g/mole + (3)(12.011 g/mole) + (5)(1.0079 g/mole) + (2)(16.00 g/mole) = 96.0625 g/mole
The amount of NaC3H5O2 is
0.496 g NaC3H5O2 = (0.496 g) / (96.0625 g/mole) NaC3H5O2 = 5.163 x 10-3 mole NaC3H5O2
The initial concentration of NaC3H5O2 is
[NaC3H5O2] = (5.163 x 10-3 mole) / (0.050 L) = 0.1033 M
NaC3H5O2 is soluble in aqueous solution. The additional C3H5O2- causes some C3H5O2- to recombine with H+ and form HC3H5O2. Let y be the number of moles per liter of C3H5O2- which recombine with H+. The equilibrium concentrations are
[C3H5O2-] = 1.878 x 10-3 + 0.1033 - y = a - y
[H+] = 1.878 x 10-3 - y = b - y
[HC3H5O2] = 0.2631 + y = d + y
where a = 1.878 x 10-3 + 0.1033, b = 1.878 x 10-3, and d = 0.2631. We have
Ka = (a - y)(b - y) / (d + y) => Ka(d + y) = (a - y)(b - y) = ab - (a + b)y + y2
=> y2 - (Ka + a + b)y + (ab - Kad) = 0
=> y = {Ka + a + b ± sqrt[(Ka + a + b)2 - 4(ab - Kad)]} / 2
= (0.1052, 1.843 x 10-3) M => 1.843 x 10-3 M
because y = 0.1052 M would result in a negative value for [H+], which is impossible. Therefore,
[C3H5O2-] = a - y = 0.1033 M
[H+] = b - y = 3.437 x 10-5 M
(d) The equilibrium concentrations are
[OH -] = 4.18 x 10-6 M = [HCO2H]
[HCO2-] = 0.309 M - 4.18 x 10-6 M ≈ 0.309 M
The base ionization constant is
Kb = [HCO2H][OH -] / [HCO2-] = (4.18 x 10-6)2 / 0.309 = 5.655 x 10-11
The acid ionization constant is
Ka = Kw/Kb = 10-14 / (5.655 x 10-11) = 1.769 x 10-4
(e) Ka is larger for methanoic acid, so methanoic acid is stronger than propanoic acid.
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PROBLEM
Answer the following questions about a pure compound that contains only carbon, hydrogen, and oxygen.
(a) A 0.7549 g sample of the compound burns in O2(g) to produce 1.9061 g of CO2(g) and 0.3370 g of H2O(g). (i) Calculate the individual masses of C, H, and O in the 0.7549 g sample. (ii) Determine the empirical formula for the compound.
(b) A 0.5246 g sample of the compound was dissolved in 10.0012 g of lauric acid, and it was determined that the freezing point of the lauric acid was lowered by 1.68°C. The value of Kf of lauric acid is 3.90°C/m. Assume that the compound does not dissociate in lauric acid. (i) Calculate the molality of the compound dissolved in the lauric acid. (ii) Calculate the molar mass of the compound from the information provided.
(c) Without doing any calculations, explain how to determine the molecular formula of the compound based on the answers to parts (a)(ii) and (b)(ii).
(d) Further tests indicate that a 0.10 M aqueous solution of the compound has a pH of 2.6. Identify the organic functional group that accounts for this pH.
(e) Calculate the acid dissociation constant.
[parts (a)-(d) from 2005 AP Chemistry Free-Response Question 2]
SOLUTION
(a) The molar masses of CO2 and H2O are
CO2: M = 12.011 g/mole + (2)(16.00 g/mole) = 44.011 g/mole
H2O: M = (2)(1.0079 g/mole) + 16.00 g/mole = 18.0158 g/mole
(i) The amounts of CO2 and H2O produced are
CO2: 1.9061 g CO2 = (1.9061 g) / (44.011 g/mole) CO2 = 4.331 x 10-2 mole CO2 => 4.331 x 10-2 mole C = 0.5202 g C in sample
H2O: 0.3370 g H2O = (0.3370 g) / (18.0158 g/mole) H2O = 1.871 x 10-2 mole H2O => (2)(1.871 x 10-2 mole H = 3.741 x 10-2 mole H = 3.771 x 10-2 g H in sample
The amount of oxygen in the sample is 0.7549 g - 0.5202 g - 3.771 x 10-2 g = 0.1970 g = 1.231 x 10-2 mole.
(ii) The relative molar amounts of C, H, and O in the sample can be obtained by dividing the number of moles of C and H by the number of moles of O.
C:H:O = 4.331/1.231 : 3.741/1.231 : 1 = 3.518 : 3.038 : 1 ≈ 3.5 : 3 : 1 = 7:6:2
The empirical formula of the compound is C7H6O2.
(b) The freezing point depression DTf is related to the molality m and the constant Kf according to
DTf = iKfm
where i is the van't Hoff factor, which is the number of particles into which the solute molecules dissociate. In this case, i = 1.
(i) The molality of the compound dissolved in the lauric acid is
m = DTf/Kf = (1.68°C) / (3.90°C/m) = 0.4308 m = 0.4308 mole/kg
(ii) The molality of the solution is also given by
m = (moles of compound) / (mass of solvent) = [(mass of compound) / (molar mass of compound)] / (mass of solvent)
The molar mass of the compound is
M = (mass of compound) / [m(mass of solvent)] = (0.5246 g) / [(0.4308 mole/kg)(10.0012 x 10-3 kg)] = 121.8 g/mole
(c) The molecular formula of the compound can be obtained by determining the molar mass corresponding to the empirical formula of the compound and dividing the molar mass of the compound by this mass. The molar mass corresponding to the empirical formula is
M = (7)(12.011 g/mole) + (6)(1.0079 g/mole) + (2)(16.00 g/mole) = 122.1 g/mole
This is approximately the same as the molar mass of the compound, so the molecular formula is also C7H6O2.
(d) The compound is an organic acid. The most common type of organic acid is a carboxylic acid, which contains one or more carboxyl groups (COOH).
(e) The compound contains only one carboxyl group, since there are only two oxygens. The compound dissociates in aqueous solution according to
C6H5COOH + H2O => C6H5COO- + H3O+
The acid dissociation constant is
Ka = [C6H5COO-][H3O+] / [C6H5COOH]
Let x be the number of moles per liter of C6H5COOH which dissociate. Then the equilibrium concentrations are
[C6H5COO-] = x = [H3O+]
[C6H5COOH] = 0.10 - x
and
Ka = x2 / (0.10 - x)
Since the pH of the solution is 2.6,
[H3O+] = 10-pH = 10-2.6 = 2.512 x 10-3 M = x
and the acid dissociation constant is
Ka = (2.512 x 10-3)2 / (0.10 - 2.512 x 10-3) = 6.472 x 10-5
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PROBLEM
A solution of equal concentrations of lactic acid and sodium acetate was found to have pH = 3.08. (a) What are the values of pKa and Ka of lactic acid? (b) What would the pH
be if the acid had twice the concentration of the salt?
SOLUTION
(a) Let L- and Ac- represent the lactate and acetate ions. Lactic and acetic acid ionize according to
HL + H2O <=> L- + H3O+, Ka1 = ?
HAc + H2O <=> Ac- + H3O+, Ka2 = 1.8 x 10-5
Let x be the moles per liter of HL which ionize and let y be the moles per liter of Ac- which combine with H3O+ to form HAc. Assume that the initial concentrations of HL and NaAc are both 1 M. NaAc is soluble in aqueous solution. The equilibrium concentrations are
[HL] = 1 - x = a - x
[L-] = x
[HAc] = y
[Ac-] = 1 - y = a - y
[H3O+] = x - y
where a = 1. Define
z = x - y => x = y + z
Then the equilibrium concentrations can be written as
[HL] = a - y - z
[L-] = y + z
[HAc] = y
[Ac-] = a - y
[H3O+] = z = 10-pH = 10-3.08 = 8.318 x 10-4 M
The equilibrium constants are
Ka1 = [L-][H3O+] / [HL] = (y + z)z / (a - y - z) (1)
Ka2 = [Ac-][H3O+] / [HAc] = (a - y)z / y => Ka2y = az - yz => y = az / (Ka2 + z) (2)
From (2), we get
y = (1)(8.318 x 10-4) / (1.8 x 10-5 + 8.318 x 10-4) = 0.9788
From (1), we get
Ka1 = (0.9788 + 8.318 x 10-4)(8.318 x 10-4) / (1 - 0.9788 - 8.318 x 10-4) = 4.004 x 10-2
Note: This value for Ka1 differs from the accepted value of 1.4 x 10-4 under standard conditions (Kotz and Treichel 1996, p. 820; Nims and Smith 1936, p. 151, Fig. 3), but this is what results from assuming pH = 3.08.
Kotz, John C., and Treichel, Paul Jr. 1996, Chemistry & Chemical Reactivity, Third Edition (New York: Saunders College Publishing).
Nims, Leslie Frederick, and Smith, Paul K. 1936, "The Ionization of Lactic Acid from Zero to Fifty Degrees," J. Biol. Chem., 113, 145.
(b) Use the result from part (a), Ka1 = 4.004 x 10-2.
Let L- and Ac- represent the lactate and acetate ions. Lactic and acetic acid ionize according to
HL + H2O <=> L- + H3O+, Ka1 = 4.004 x 10-2
HAc + H2O <=> Ac- + H3O+, Ka2 = 1.8 x 10-5
Let x be the moles per liter of HL which ionize and let y be the moles per liter of Ac- which combine with H3O+ to form HAc. Assume that the initial concentration of HL is 2 M and the initial concentration of NaAc is 1 M. NaAc is soluble in aqueous solution. The equilibrium concentrations are
[HL] = 2 - x = a - x
[L-] = x
[HAc] = y
[Ac-] = 1 - y = b - y
[H3O+] = x - y
where a = 2 and b = 1. The ionization constants are
Ka1 = [L-][H3O+] / [HL] = x(x - y) / (a - x) => y = x - Ka1(a - x) / x (1)
Ka2 = [Ac-][H3O+] / [HAc] = (b - y)(x - y) / y => x = y + Ka2y / (b - y) (2)
Substitute (2) into (1).
y = y + Ka2y / (b - y) - Ka1{a - [y + Ka2y / (b - y)]} / [y + Ka2y / (b - y)]
Subtract y from both sides.
0 = Ka2y / (b - y) - Ka1{a - [y + Ka2y / (b - y)]} / [y + Ka2y / (b - y)] = f(y)
We can solve this graphically for y. Plot f(y) as a function of y for 0 < y < 1. We see that the value of y that makes f(y) zero is between 0.999 and 0.9999. We can narrow down the solution to y = 0.99951641. Using (2), we get x = 1.036719985 => x - y = 0.037203575 = [H3O+] => pH = - log [H3O+] = 1.429. See LacticAcetic070421.xls.
Note: If we use the correct value for the ionization constant of lactic acid, Ka1 = 1.4 x 10-4, we get y = 0.904076789, x = 0.90424644, x - y = 0.00016965 = [H3O+], and pH = 3.770. See LacticAcetic070421.2.xls.
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PROBLEM
One liter of a 2 M solution of lactic acid (CH3CHOHCO2H) is combined with one liter of a 2 M solution of sodium acetate (NaCH3CO2). (a) What is the pH of the resulting solution? (b) Repeat if the initial concentrations are 4 M instead of 2 M. (c) Repeat if the initial concentrations are 0.1 M instead of 2 M.
SOLUTION
(a) Let "L" and "Ac" represent the lactate (CH3CHOHCO2-) and acetate (CH3CO2-) ions, respectively. Then lactic and acetic acids ionize according to
HL + H2O <=> L- + H3O+, Ka1 = 1.4 x 10-4
HAc + H2O <=> Ac- + H3O+, Ka2 = 1.8 x 10-5
After the two solutions are mixed, we have a solution containing 1 M lactic acid and 1 M sodium acetate. Sodium acetate is soluble in aqueous solution. The lactic acid will ionize to some extent and some of the resulting H3O+ will combine with Ac- to form HAc. Let x be the moles per liter of HL which ionize and let y be the moles per liter of Ac- which combine with H3O+ to form HAc. At equilibrium, we have
[HL] = 1 - x = a - x
[L-] = x
[HAc] = y
[Ac-] = 1 - y = a - y
[H3O+] = x - y
where a = 1. The ionization constants are
Ka1 = [L-][H3O+] / [HL] = x(x - y) / (a - x) => y = x - Ka1(a - x) / x (1)
Ka2 = [Ac-][H3O+] / [HAc] = (a - y)(x - y) / y => x = y + Ka2y / (a - y) (2)
Substitute (2) into (1).
y = y + Ka2y / (a - y) - Ka1{a - [y + Ka2y / (a - y)]} / [y + Ka2y / (a - y)]
Subtract y from both sides.
0 = Ka2y / (a - y) - Ka1{a - [y + Ka2y / (a - y)]} / [y + Ka2y / (a - y)] = f(y)
We can solve this graphically for y. Plot f(y) as a function of y for 0 < y < 1. We see that the value of y that makes f(y) zero is between 0.73 and 0.74. We can narrow down the solution to y = 0.736043742. Using (2), we get x = 0.736093935 => x - y = 5.01931 x 10-5 = [H3O+] => pH = - log [H3O+] = 4.299. See LacticAcetic070407.xls.
(b) If the initial concentrations are 4 M, the solution to the problem is the same except that a = 2. x and y must both be between 0 and 2. Plot f(y) as a function of y for 0 < y < 2. The value of y that makes f(y) zero is between 1.45 and 1.50. We can narrow down the solution to y = 1.47211258. Using (2), we get x = 1.472162777 => x - y = 5.01964 x 10-5 = [H3O+] => pH = - log [H3O+] = 4.299. See LacticAcetic070407.2.xls.
(c) If the initial concentrations are 0.1 M, the solution to the problem is the same except that a = 0.05. x and y must both be between 0 and 0.05. Plot f(y) as a function of y for 0 < y < 0.05. The value of y that makes f(y) zero is between 0.0365 and 0.0370. We can narrow down the solution to y = 0.036778392. Using (2), we get x = 0.036828462 => x - y = 5.00704 x 10-5 = [H3O+] => pH = - log [H3O+] = 4.300. See LacticAcetic070407.3.xls.
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Titrations
PROBLEM
Vitamin C is the simple compound C6H8O6. Besides being an acid, it is also a reducing agent. One method for determining the amount of vitamin C in a sample therefore is to titrate it with a solution of bromine, Br2, a good oxidizing agent:
C6H8O6(aq) + Br2(aq) => 2HBr(aq) + C6H6O6(aq)
Suppose a 1.00 g "chewable" vitamin C tablet requires 27.85 mL of 0.102 M Br2 for titration to the equivalence point. How many grams of vitamin C are in the tablet?
[from Kotz, John C., and Treichel, Paul Jr. 1996, Chemistry & Chemical Reactivity, Third Edition (New York: Saunders College Publishing), problem 5.65]
SOLUTION
The amount of Br2 needed is (27.85 x 10-3 L)(0.102 M) = 2.841 x 10-3 mole Br2. The amount of C6H8O6 consumed is therefore 2.841 x 10-3 mole C6H8O6. The molar mass of C6H8O6 is 176.1256 g/mole. Therefore the amount of vitamin C in the tablet is (2.841 x 10-3 mole)(176.1256 g/mole) = 0.5003 g.
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PROBLEM
The lead content of a sample can be estimated by converting the lead to PbO2 and then dissolving the PbO2 in an acid solution of KI. This liberates I2 according to the equation
PbO2(s) + 4H+(aq) + 2I -(aq) => Pb2+(aq) + I2(aq) + 2H2O(l) (1)
The liberated I2 is then titrated with Na2S2O3 according to
I2(aq) + 2S2O32-(aq) => 2I -(aq) + S4O62-(aq) (2)
The amount of titrated I2 is related to the amount of lead in the sample. If 0.576 g of a lead-containing mineral requires 35.23 mL of 0.0500 M Na2S2O3 for titration of the liberated I2, what is the weight percent of lead in the mineral?
[from Kotz, John C., and Treichel, Paul Jr. 1996, Chemistry & Chemical Reactivity, Third Edition (New York: Saunders College Publishing), problem 5.81]
SOLUTION
The amount of Na2S2O3 used for the titration is (35.23 x 10-3 L)(0.0500 M) = 1.762 x 10-3 mole of Na2S2O3. Since each mole of Na2S2O3 contains one mole of S2O32-, 1.762 x 10-3 mole of S2O32- was used. From (2), this implies that (1/2)(1.762 x 10-3 mole) = 8.808 x 10-4 mole of I2 was consumed, and from (1) this means that 8.808 x 10-4 mole of PbO2 was consumed. Each mole of PbO2 contains one mole of Pb, so 8.808 x 10-4 mole of Pb was consumed. Lead has a molar mass of 207.2 g/mole, so the amount of lead consumed was (8.808 x 10-4 mole)(207.2 g/mole) = 0.1825 g. Thus, the weight percent of lead in the mineral is (0.1825 g) / (0.576 g) = 0.3168 = 31.68%.
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PROBLEM
You have done a laboratory experiment in which you obtained a new iron-containing compound. Its formula could be one of two possibilities: K3[Fe(C2O4)3] or K[Fe(C2O4)2(H2O)2]. In each of these, the C2O42- ion is the oxalate ion. To find which is correct, you dissolve 1.356 g of the compound in acid, which converts the oxalate ion to oxalic acid according to
2H+(aq) + C2O42-(aq) => H2C2O4(aq) (1)
and then titrate the oxalic acid with potassium permanganate. The balanced, net ionic equation for the titration is
5H2C2O4(aq) + 2MnO4-(aq) + 6H+(aq) => 2Mn2+(aq) + 10CO2(g) + 8H2O(l) (2)
The titration requires 34.50 mL of 0.108 M KMnO4. What is the correct formula of the iron-containing compound?
[from Kotz, John C., and Treichel, Paul Jr. 1996, Chemistry & Chemical Reactivity, Third Edition (New York: Saunders College Publishing), problem 5.85]
SOLUTION
The required number of moles of potassium permanganate (KMnO4) is (34.50 x 10-3 L)(0.108 M) = 3.726 x 10-3 mole. Potassium permanganate is soluble in aqueous solution and breaks up into K+ and MnO4- ions, so this yields 3.726 x 10-3 mole of MnO4-. According to equation (2), for every 2 moles of MnO4- consumed, 5 moles of H2C2O4 are consumed, so the number of moles of H2C2O4 used up is
(5/2)(3.726 x 10-3 mole) = 9.315 x 10-3 mole. Thus, 9.315 x 10-3 mole of C2O42- is used up.
If the unknown compound is K3[Fe(C2O4)3], the number of moles present would have been (1/3)(9.315 x 10-3 mole) = 3.105 x 10-3 mole because each mole of K3[Fe(C2O4)3] contains 3 moles of C2O42-. Since the molar mass of K3[Fe(C2O4)3] is (3)(39.0983 g/mole) + 55.847 g/mole + (3)[(2)(12.011 g/mole) + (4)(15.9994 g/mole)] = 437.2 g/mole, the mass of the sample would have been (437.2 g/mole)(3.105 x 10-3 mole) = 1.358 g.
If the unknown compound is K[Fe(C2O4)2(H2O)2], the number of moles present would have been (1/2)(9.315 x 10-3 mole) = 4.658 x 10-3 mole because each mole of K[Fe(C2O4)2(H2O)2] contains 2 moles of C2O42-. Since the molar mass of K[Fe(C2O4)2(H2O)2] is 39.0983 g/mole + 55.847 g/mole + (2)[(2)(12.011 g/mole) + (4)(15.9994 g/mole)] + (2)[(2)(1.0079 g/mole) + 15.9994 g/mole] = 307.0 g/mole, the mass of the sample would have been (307.0 g/mole)(4.658 x 10-3 mole) = 1.430 g.
The mass inferred assuming that the unknown compound is K3[Fe(C2O4)3] matches the mass of the sample. Therefore, the unknown compound is K3[Fe(C2O4)3].
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PROBLEM
A 0.10 M NaOH solution is slowly added to 25 mL of a 0.10 M HCl solution. Calculate the pH of the mixture after (a) 10 mL, (b) 25 mL, and (c) 35 mL have been added.
[from Chang, Raymond 2002, Chemistry, Seventh Edition (New York: McGraw-Hill), pp. 661-664]
SOLUTION
(a) The total amount of hydrochloric acid (HCl) is (0.10 mole/L)(25 x 10-3 L) = 0.0025 mole, which is the amount of H+ since HCl is a strong acid. The total amount of sodium hydroxide (NaOH) is (0.10 mole/L)(10 x 10-3 L) = 0.0010 mole, which is the amount of OH - since NaOH is a strong base. The 0.0010 mole of OH - will neutralize 0.0010 mole of H+, leaving an excess of 0.0025 mole - 0.0010 mole = 0.0015 mole of H+. Since the total volume of the solution is 25 mL + 10 mL = 35 mL = 35 x 10-3 L, the final concentration of H+ is
[H+] = (0.0015 mole) / (35 x 10-3 L) = 4.286 x 10-2 M
and the pH is
pH = - log [H+] = - log(4.286 x 10-2) = 1.368.
(b) The total amount of hydrochloric acid (HCl) is (0.10 mole/L)(25 x 10-3 L) = 0.0025 mole, which is the amount of H+ since HCl is a strong acid. The total amount of sodium hydroxide (NaOH) is (0.10 mole/L)(25 x 10-3 L) = 0.0025 mole, which is the amount of OH - since NaOH is a strong base. The 0.0025 mole of OH - will neutralize 0.0025 mole of H+, resulting in a neutral solution with pH = 7. Since HCl is a strong acid and NaOH is a strong base, the salt formed (sodium chloride, NaCl) does not undergo hydrolysis.
(c) The total amount of hydrochloric acid (HCl) is (0.10 mole/L)(25 x 10-3 L) = 0.0025 mole, which is the amount of H+ since HCl is a strong acid. The total amount of sodium hydroxide (NaOH) is (0.10 mole/L)(35 x 10-3 L) = 0.0035 mole, which is the amount of OH - since NaOH is a strong base. The 0.0025 mole of H+ will neutralize 0.0025 mole of OH -, leaving an excess of 0.0035 mole - 0.0025 mole = 0.0010 mole of OH -. Since the total volume of the solution is 25 mL + 35 mL = 60 mL = 60 x 10-3 L, the final concentration of OH - is
[OH -] = (0.0010 mole) / (60 x 10-3 L) = 1.667 x 10-2 M
The final concentration of H+ is
[H+] = Kw/[OH -] = (10-14)/(1.667 x 10-2) = 6 x 10-13 M
and the pH is
pH = - log [H+] = - log(1.8 x 10-13) = 12.222.
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PROBLEM
A solution is made by mixing exactly 500 mL of 0.167 M NaOH with exactly 500 mL of 0.100 M CH3COOH. Calculate the equilibrium concentrations of H+, CH3COOH, CH3COO-, OH -, and Na+.
[from Chang, Raymond 2002, Chemistry, Seventh Edition (New York: McGraw-Hill), problem 16.28]
SOLUTION
Sodium hydroxide (NaOH) is a strong base which is fully ionized into Na+ and OH - ions. Acetic acid (CH3COOH) is a weak acid with ionization constant Ka = 1.8 x 10-5. After mixing, the solution has a total volume of 1 L and is 0.0835 M NaOH and 0.0500 M CH3COOH with a total of 0.0835 mole NaOH and 0.0500 mole CH3COOH. The 0.0500 mole of CH3COOH is neutralized by 0.0500 mole of NaOH, leaving 0.0335 mole of OH - and 0.0500 mole of CH3COO-. Let x be the number of moles per liter of CH3COO- which hydrolyze according to
CH3COO- + H2O <=> CH3COOH + OH -, Kh = Kw/Ka = (10-14)/(1.8 x 10-5) = 5.556 x 10-10
At equilibrium, we have
[OH -] = 0.0335 + x
[CH3COO-] = 0.0500 - x
[CH3COOH] = x
The hydrolysis constant is
Kh = [CH3COOH][OH -]/[CH3COO-] = x(0.0335 + x) / (0.0500 - x)
=> 0.05Kh - Khx = 0.0335x + x2
=> x2 + (0.0335 + Kh)x - 0.05Kh = 0
=> x = {- (0.0335 + Kh) + sqrt[(0.0335 + Kh)2 + (4)(0.05Kh)]}/2 = 8.292 x 10-10 M
This is negligible compared to 0.0335 M, so
[OH -] = 0.0335 M
[H+] = Kw/[OH -] = (10-14) / (0.0335) = 2.985 x 10-13 M
[Na+] = 0.0835 M
[CH3COO-] = 0.0500 M
[CH3COOH] = 8.292 x 10-10 M
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PROBLEM
Calculate the pH at the equivalence point for the following titration: 0.10 M HCOOH versus 0.10 M NaOH.
[from Chang, Raymond 2002, Chemistry, Seventh Edition (New York: McGraw-Hill), problem 16.28]
SOLUTION
The equation for the neutralization reaction between formic acid (HCOOH), a weak acid, and sodium hydroxide (NaOH), a strong base, is
HCOOH(aq) + NaOH(aq) <=> NaCOOH(aq) + H2O, Ka = 1.7 x 10-4
Suppose we combine equal volumes of 0.1 M HCOOH and 0.1 M NaOH. The molarities of HCOOH and NaOH in the combined volume of solution are both 0.05 M. There are sufficient amounts of H+ from HCOOH and OH- from NaOH to cause neutralization, but since HCOOH is a weak acid, the formate ion (HCOO-) undergoes hydrolysis according to
HCOO-(aq) + H2O <=> HCOOH(aq) + OH-, Kh = Kw/Ka = (10-14) / (1.7 x 10-4) = 5.882 x 10-11
and the final solution will not be exactly neutral. Let x be the moles per liter of HCOO- which hydrolyze. At equilibrium, we have
[HCOO-] = 0.05 - x
[OH -] = x
[HCOOH] = x
The hydrolysis constant
Kh = [HCOOH][OH -]/[HCOO-] = x2/(0.05 - x)
=> 0.05Kh - Khx = x2
=> x2 + Khx - 0.05Kh = 0
=> x = {- Kh ± sqrt[Kh2 + (4)(0.05)Kh)]} / 2
=> {- Kh + sqrt[Kh2 + (4)(0.05)Kh)]} / 2 = 1.715 x 10-6 M
Thus, [OH -] = x = 1.715 x 10-6 M and
[H3O+] = Kw/[OH -] = (10-14) / (1.715 x 10-6) = 5.831 x 10-9 M => pH = - log [H3O+] = 8.234
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PROBLEM
A piece of magnesium metal weighing 1.56 g is placed in 100.0 mL of 0.100 M AgNO3 at 25°C. Calculate [Mg2+] and [Ag+] in solution at equilibrium. What is the mass of the magnesium left? The volume remains constant.
[from Chang, Raymond 2002, Chemistry, Seventh Edition (New York: McGraw-Hill), problem 19.78]
SOLUTION
For the reduction of Mg2+ and Ag+ under standard conditions,
Mg2+(aq) + 2e- => Mg(s), E0 = - 2.37 V
Ag+(aq) + e- => Ag(s), E0 = + 0.80 V
Ag+ is easier to reduce than Mg2+, because the standard reduction potential is more positive. So Ag+ is reduced and Mg is oxidized.
Mg(s) => Mg2+(aq) + 2e-, E0 = + 2.37 V
Ag+(aq) + e- => Ag(s), E0 = + 0.80 V
The balanced equation for the net reaction is obtained by adding the first equation to twice the second equation.
2Ag+(aq) + Mg(s) => 2Ag(s) + Mg2+(aq), Ecell0 = 2.37 V + 0.80 V = 3.17 V
Two moles of electrons are transferred for each mole of Mg oxidized. The standard emf of the cell is related to the equilibrium constant of the redox reaction according to
Ecell0 = (0.0592 V / n) log K
=> K = 10[nEcell0 / (0.0592 V)] = 10[(2)(3.17 V) / (0.0592 V)] = 10107.1 >> 1
From the equation for the net reaction, we see that for every two moles of Ag+ reduced, one mole of Mg is oxidized. The total number of moles of Ag is (0.1 M)(0.1 L) = 0.01 mole. The total number of moles of Mg is (1.56 g) / (24.31 g/mole) = 0.06417 mole. Since K >> 1, the reaction continues until essentially all of the Ag+ is used up. At equilibrium, the amount of Mg2+ in solution is half the initial amount of Ag+ in solution, which is (0.5)(0.01 mole) = 0.005 mole => [Mg2+] = (0.005 mole) / (0.1 L) = 0.05 M.
The equilibrium constant for the net reaction is
K = [Mg2+] / [Ag+]2 => [Ag+] = sqrt([Mg2+] / K) = sqrt(0.05 / 10107.1) = 6.341 x 10-55 M
0.005 mole or (0.005 mole)(24.31 g/mole) = 0.1216 g of the Mg is used up, so the amount which remains is 1.56 g - 0.1216 g = 1.438 g.
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PROBLEM
A 300 mL solution of NaCl is electrolyzed for 6.00 min.
(a) What is the minimum voltage that must be applied?
(b) If the pH of the final solution is 12.24, calculate the average current used.
[from Chang, Raymond 2002, Chemistry, Seventh Edition (New York: McGraw-Hill), problem 19.88]
SOLUTION
(a) In the electrolysis of NaCl solution, the anode and cathode reactions are
2Cl -(aq) => Cl2(g) + 2e-, E0 = - 1.36 V (oxidation at anode)
2H2O(l) + 2e- => H2(g) + 2OH -(aq), E0 = - 0.83 V (reduction at cathode)
The standard potential for the overall reaction is
E0 = - 1.36 V - 0.83 V = - 2.19 V
The battery must supply at least 2.19 V.
(b) After 6.00 min,
pH = 12.24 => pOH = 14 - 12.24 = 1.76 => [OH -] = 10-pOH = 10-1.76 = 1.738 x 10-2 M
The amount of OH - produced is (1.738 x 10-2 M)(0.3 L) = 5.213 x 10-3 mole. Each mole of OH - produced requires the transfer of one mole of electrons, so 5.213 x 10-3 mole of electrons is transferred. The total charge transferred is (5.213 x 10-3 mole)(96,500 C/mole) = 503.1 C. The average current is (503.1 C) / (360 s) = 1.397 C/s = 1.397 A.
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PROBLEM
Zinc is an amphoteric metal; that is, it reacts with both acids and bases. The standard reduction potential is - 1.36 V for the reaction
Zn(OH)42-(aq) + 2e- => Zn(s) + 4OH -(aq)
Calculate the formation constant (Kf) for the reaction
Zn2+ + 4OH -(aq) <=> Zn(OH)42-(aq)
[from Chang, Raymond 2002, Chemistry, Seventh Edition (New York: McGraw-Hill), problem 19.96]
SOLUTION
Consider the oxidation-reduction half reactions
Zn(OH)42-(aq) + 2e- => Zn(s) + 4OH -(aq), E0 = - 1.36 V (oxidation)
Zn(s) => Zn2+(aq) + 2e-, E0 = + 0.76 V (reduction)
If we add the two together, we get
Zn(OH)42-(aq) => Zn2+(aq) + 4OH -(aq), Ecell0 = - 0.60 V
For the reverse reaction, we have
Zn2+(aq) + 4OH - => Zn(OH)42-(aq), Ecell0 = 0.60 V
The equilibrium constant (Kf) for this reaction is related to the standard emf according to
Ecell0 = (RT/nF) ln Kf => Kf = exp[Ecell0nF/RT]
= exp[(0.60 V)(2)(96,500 J/V mole) / (8.314 J/K mole)(298 K)] = 1.989 x 1020
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PROBLEM
When 25.0 mL of a solution containing both Fe2+ and Fe3+ ions is titrated with 23.0 mL of 0.0200 M KMnO4 (in dilute sulfuric acid), all of the Fe2+ ions are oxidized to Fe3+ ions. Next, the solution is treated with Zn metal to convert all of the Fe3+ ions to Fe2+ ions. Finally, 40.0 mL of the same KMnO4 solution are added to the solution in order to oxidize the Fe2+ ions to Fe3+. Calculate the molar concentrations of Fe2+ and Fe3+ in the original solution.
[from Chang, Raymond 2002, Chemistry, Seventh Edition (New York: McGraw-Hill), problem 19.102]
SOLUTION
Let (n1, n2) be the number of moles of (Fe2+, Fe3+) in the original solution. The procedure described can be broken up into three steps:
1. Add KMnO4 solution to convert all Fe2+ to Fe3+, resulting in n1 + n2 moles of Fe3+.
2. Add Zn to convert all Fe3+ to Fe2+, resulting in n1 + n2 moles of Fe2+.
3. Add KMnO4 to convert all Fe2+ to Fe3+, resulting in n1 + n2 moles of Fe3+.
In steps 1 and 3, the oxidation and reduction half reactions that occur are
Fe2+(aq) => Fe3+(aq) + e-, E0 = - 0.77 V (oxidation)
MnO4-(aq) + 8H+ + 5e- => Mn2+(aq) + 4H2O, E0 = + 1.51 V (reduction)
If we add five times the first equation to the second equation, we get
5Fe2+(aq) + MnO4-(aq) + 8H+=> 5Fe3+(aq) + Mn2+(aq) + 4H2O
The amount of MnO4- used in step 1 is (0.02 M)(0.023 L) = 4.6 x 10-4 mole, which is sufficient to oxidize (5)(4.6 x 10-4 mole) = 0.0023 mole of Fe2+. Thus, n1 = 0.0023 mole.
The amount of MnO4- used in step 3 is (0.02 M)(0.040 L) = 8 x 10-4 mole, which is sufficient to oxidize (5)(8 x 10-4 mole) = 0.0040 mole of Fe2+. Thus, n1 + n2 = 0.0040 mole, and n2 = 0.0040 mole - 0.0023 mole = 0.0017 mole.
The molar concentrations of Fe2+ and Fe3+ in the original solution are
[Fe2+] = (0.0023 mole) / (0.025 L) = 0.092 M
[Fe3+] = (0.0017 mole) / (0.025 L) = 0.068 M
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PROBLEM
Calculate the pressure of H2 (in atm) required to maintain equilibrium with respect to the following reaction at 25°C:
Pb(s) + 2H+ <=> Pb2+ + H2(g)
given that [Pb2+] = 0.035 M and the solution is buffered at pH 1.60.
[from Chang, Raymond 2002, Chemistry, Seventh Edition (New York: McGraw-Hill), problem 19.110]
SOLUTION
The two oxidation-reduction half reactions that result in the given reaction are
Pb(s) => Pb2+(aq) + 2e-, E0 = + 0.13 V (oxidation)
2H+(aq) + 2e- => H2(g), E0 = 0.00 V (reduction)
The corresponding voltaic cell has the standard cell potential
Ecell0 = 0.13 V + 0.00 V = 0.13 V
The standard cell potential is related to the equilibrium constant K according to
Ecell0 = (0.0592 V / n) log K => K = 10(Ecelln / 0.0592 V) = 10[(0.13 V)(2) / (0.0592 V)] = 2.465 x 104
The equilibrium constant is also related to the concentrations of H+, Pb2+, and H2 according to
K = [Pb2+]P(H2) / [H+]2 => P(H2) = K[H+]2 / [Pb2+] = (2.465 x 104)(10-1.60)2 / (0.035) = 444.5 atm
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Thermochemistry
PROBLEM
Discuss the Haber process for synthesizing ammonia. In particular:
(a) Who developed the Haber process?
(b) What is the balanced equation for the formation of ammonia from nitrogen and hydrogen?
(c) Is the reaction endothermic or exothermic?
(d) How does the catalyst that is used in the Haber process work?
SOLUTION
(a) Fritz Haber, a German chemist, discovered the process in 1905.
(b) N2(g) + 3H2(g) => 2NH3(g)
(c) The reaction is exothermic, with DH0 = - 92.6 kJ.
(d) Haber discovered that a catalyst consisting of iron plus a few percent of potassium oxides and aluminum oxides catalyze the formation of ammonia from nitrogen and hydrogen. N2 and H2 dissociate into atoms on the surface of the metal catalyst. The nitrogen and hydrogen atoms migrate on the surface and form NH3 molecules which leave the surface.
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Gibbs Free Energy
PROBLEM
Determine which of the following reactions will spontaneously go to completion:
(a) Zn(s) + 2HCl(aq) => ZnCl2(aq) + H2(g)
(b) CaCO3(s) + 2HCl(aq) => CaCl2(s) + H2O(l) + CO2(g)
(c) Ag+(aq) + HCl(aq) => AgCl(s) + H+(aq)
(d) Cu + 2H+ => Cu2+ + H2, E0 = - 0.34 V
(e) H2SO4(aq) + 2NaOH(aq) => 2Na2SO4(aq) + 2H2O(l)
[from Mascetta, Joseph A. 2002, Barron's How to Prepare for the SAT II Chemistry, 7th Edition (Hauppauge, New York: Barron's Educational Series), Diagnostic Test problem 70]
SOLUTION
(a) A reaction is spontaneous in the forward direction if the free energy change DG is negative. The free energy change for the given reaction is
DG = Gf0[ZnCl2(aq)] + Gf0[H2(g)] - Gf0[Zn(s)] - 2Gf0[HCl(aq)]
= - 369.26 kJ/mole + 0 - 0 - (2)(- 131.2 kJ/mole) = - 106.86 kJ/mole < 0 => spontaneous
(b) The free energy change for the reaction is
DG = Gf0[CaCl2(s)] + Gf0[H2O(l)] + Gf0[CO2(g)] - Gf0[CaCO3(s)] - 2Gf0[HCl(aq)]
= - 750.19 kJ/mole - 237.2 kJ/mole - 394.4 kJ/mole - (- 1128.8 kJ/mole) - (2)(- 131.2 kJ/mole)
= 9.41 kJ/mole > 0 => not spontaneous
(c) The free energy change for the reaction is
DG = Gf0[AgCl(s)] + Gf0[H+(aq)] - Gf0[Ag+(aq)] - Gf0[HCl(aq)]
= - 109.7 kJ/mole + 0 - 77.1 kJ/mole - (- 131.2 kJ/mole) = - 55.6 kJ/mole < 0 => spontaneous
(d) E0 = - 34.1 V < 0 => not spontaneous
(e) The free energy change for the reaction is
DG = 2Gf0[Na2SO4(aq)] + 2Gf0[H2O(l)] - Gf0[H2SO4(aq)] - 2Gf0[NaOH(aq)]
= (2)(- 1266.8 kJ/mole) + (2)(- 237.2 kJ/mole) - (- 744.5 kJ/mole) - (2)(- 419.2 kJ/mole)
= - 1425.1 kJ/mole < 0 => spontaneous
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PROBLEM
The combustion of carbon monoxide is represented by the following equation:
CO(g) + (1/2)O2(g) => CO2(g)
(a) Determine the value of the standard enthalpy change, DHrxn0, for the combustion of CO(g) at 298 K using the following information:
C(s) + (1/2)O2(g) => CO(g), DH2980 = - 110.5 kJ/mole
C(s) + O2(g) => CO2(g), DH2980 = - 393.5 kJ/mole
(b) Determine the value of the standard entropy change, DSrxn0, for the combustion of CO(g) at 298 K using the information in the following table:
| Substance | S2980 (J/mole K) |
|---|---|
(c) Determine the standard free energy change, DGrxn0, for the reaction at 298 K.
(d) Is the reaction spontaneous under standard conditions at 298 K? Justify your answer.
(e) Calculate the value of the equilibrium constant, Keq, for the reaction at 298 K.
[from AP Chemistry 2006 Free-Response Questions, College Board, Question 2]
SOLUTION
(a) We have the following standard enthalpy changes:
CO(g) => C(s) + (1/2)O2(g), DH2980 = 110.5 kJ/mole (1)
C(s) + O2(g) => CO2(g), DH2980 = - 393.5 kJ/mole (2)
Adding (1) and (2), we get
CO(g) + C(s) + O2(g) => C(s) + (1/2)O2(g) + CO2(g), DHrxn0 = - 283.0 kJ/mole
Subtracting C(s) and (1/2)O2(g) from both sides,
CO(g) + (1/2)O2(g) => CO2(g), DHrxn0 = - 283.0 kJ/mole
(b) DSrxn0 = S2980[CO2(g)] - S2980[CO(g)] - (1/2)S2980[O2(g)]
= 213.7 J/mole K - 197.7 J/mole K - (1/2)(205.1 J/mole K) = - 86.55 J/mole K
(c) DGrxn0 = DHrxn0 - TDSrxn0 = - 283.0 kJ/mole - (298 K)(- 0.08655 J/mole K) = - 257.2 kJ/mole
(d) The reaction is spontaneous because DGrxn0 < 0.
(e) DGrxn0 = - RT ln Keq => Keq = exp(- DGrxn0/RT)
= exp{- (- 257.2 x 103 J/mole) / [(8.31 J/mole K)(298 K)]} = 1.282 x 1045 Back to Main Menu
Clausius-Clapeyron Equation
PROBLEM
| Temperature (°C) | Water Vapor Pressure (mm Hg) |
|---|---|
(a) From the data in the above table, find the standard enthalpy of vaporization of water.
(b) Find the boiling point temperature of water when the ambient pressure is 1 atm.
(c) What is the standard entropy of vaporization of water?
(d) What is the standard free energy of vaporization of water?
SOLUTION
(a) Use the Clausius-Clapeyron equation,
ln P = - DHvap/RT + C
where P is the vapor pressure, DHvap is the standard enthalpy of vaporization, R = 8.314 J/mole K is the gas constant, T is the absolute temperature, and C is a constant. Write the above equation for any two pressure-temperature pairs, (P1, T1) and (P2, T2).
ln P1 = - DHvap/RT1 + C
ln P2 = - DHvap/RT2 + C
Subtract the second equation from the first equation. The result is
ln P1/P2 = - (DHvap/R)(1/T1 - 1/T2)
or
ln P2/P1 = (DHvap/R)(1/T1 - 1/T2) (1)
=> DHvap = (R ln P2/P1) / (1/T1 - 1/T2)
Using the first and last data points in the above table, we get
DHvap = [(8.314 J/mole K) ln (355.1 mm Hg / 4.55 mm Hg)] / [1/(273.15 K) - 1/(353.15 K)]
= 43680 J/mole = 43.68 kJ/mole
(b) If we solve equation (1) for T2, we get
T2 = 1 / [1/T1 - (R/DHvap) ln P2/P1]
The boiling point temperature is the temperature at which the vapor pressure equals the ambient pressure, which is 1 atm = 760 mm Hg. Using the last data point in the above table for (P1, T1) and P2 = 760 mm Hg, we get
T2 = 1 / {1/(353.15 K) - [(8.314 J/mole K) / (43680 J/mole)] ln[(760 mm Hg) / (355.1 mm Hg)]}
= 372.19 K = 99.04 °C
The standard entropy of vaporization is
DSvap = DHvap/T = (43680 J/mole) / (372.19 K) = 117.4 J/mole K
The standard free energy of vaporization is
DGvap = DHvap - TDSvap = 43680 J/mole - (372.19 K)(117.4 J/mole K) = 0 Back to Main Menu
Nuclear Chemistry
PROBLEM
The following nuclides lie outside the band of stability. Predict whether each is most likely to undergo beta - decay, beta + decay, or alpha decay, and identify the daughter nucleus: (a) copper-68, (b) cadmium-103, (c) berkelium 243, (d) dubnium-260.
SOLUTION
The three types of nuclear decay described are (n is a neutrino, n* is an antineutrino):
beta - decay: n => p + e- + n*, atomic number increases by 1
beta + decay: p => n + e+ + n, atomic number decreases by 1
alpha decay: atomic number decreases by 2, mass number decreases by 4
(a) The average mass number of a Cu atom is 63.55. This indicates that Cu(68, 29) has too many neutrons and will undergo beta - decay.
(b) The average mass number of a Cd atom is 112.4. This indicates that Cd(103, 48) has too few neutrons and will undergo beta + (positron) decay.
(c) Bk (atomic number Z = 97) is a transuranic (Z > 92) element. The mean atomic mass of Bk is 247. The transuranic elements tend to undergo alpha decay except for the more massive isotopes, which undergo beta - decay. Since Bk(243, 97) is one of the lighter isotopes, it undergoes alpha decay.
(d) Db (Z = 105) is also transuranic. The most stable isotope is Db(268, 105). Since Db(260, 105) is one of the lighter isotopes, it undergoes alpha decay.
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Halogens
PROBLEM
Within the halogen group:
(a) Which element is the most chemically active?
(b) Which element has the smallest ionic radius?
(c) Which element has the lowest first ionization potential?
(d) Which element exhibits metallic properties at room temperature?
[from Mascetta, Joseph A. 2002, Barron's How to Prepare for the SAT II Chemistry, 7th Edition (Hauppauge, New York: Barron's Educational Series), Diagnostic Test problems 1-4]
SOLUTION
(a) The halogen group consists of fluorine (F, Z = 9), chlorine (Cl, 17), bromine (Br, 35), iodine (I, 53), and astatine (At, 85). The most chemically active is fluorine.
(b) The element with the smallest ionic radius is fluorine.
(c) The element with the lowest first ionization potential is astatine. [Astatine is radioactive. Its longest lived isotope is At-210, which has a half-life of 8.3 hr.]
(d) Of the common halogens, only iodine is solid at room temperature and exhibits metallic properties.
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Organic Chemistry
PROBLEM
Indicate the asymmetric carbon atoms in each of the following compounds:
CH3 O
| ||
(a) CH3--CH2--CH--CH--C--NH2
|
NH2
H
|
+
/|\
/ | \
/ Br \
(b) H / \ H
|/ \|
+-----------+
| |
H Br
H H H
| | |
(c) H--C--C--C--Cl
| | |
H Cl H
OH CH3
| |
(d) H3C--C--C--CH2OH
| |
H H
CH2OH
|
|
C--O
H /| \ OH
| / | \ |
|/ H \|
(e) C C
|\ OH H /|
| \ | | / |
HO \| |/ H
C--C
| |
| |
H OH
[from Chang, Raymond 2002, Chemistry, Seventh Edition (New York: McGraw-Hill), problems 24.56 and 24.66]
SOLUTION
(a) The complete structural formula is
H
|
H--C--H
|
H H | H O H
| | | | || /
H--C--C--C*-C*-C--N C* = asymmetric carbon atom
| | | | \
H H H N H
/ \
H H
An asymmetric carbon atom is a carbon atom attached to four different atoms or groups of atoms, so only the two indicated are asymmetric.
(b) The complete structural formula is
H
|
C*
/|\
/ | \
/ Br \
H / \ H
|/ \|
C-----------C* C* = asymmetric carbon atom
| |
H Br
(c) The only asymmetric carbon atom is the one in the middle:
H H H
| | |
H--C--C*-C--Cl C* = asymmetric carbon atom
| | |
H Cl H
(d) The complete structural formula is
H H
| |
H O H-C-H H
| | | |
H--C--C*--C*--C--O--H C* = asymmetric carbon atom
| | | |
H H H H
(e) The complete structural formula is
O--H
|
H--C--H
|
C*-O
H /| \ O--H
| / | \ |
|/ H H \|
C* / C* C* = asymmetric carbon atom
|\ O H /|
| \ | | / |
H--O \| |/ H
C*-C*
| |
| |
H O--H
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PROBLEM
Identify the functional group that characterizes each of the following organic compounds and also give the general form of the compound:
(a) alcohol
(b) aldehyde
(c) acid
(d) ester
(e) ether
(f) ketone
SOLUTION
(a) An alcohol contains the hydroxyl (OH) functional group. A hydrocarbon where an end hydrogen is replaced by a hydroxyl group is a primary alcohol. A hydrocarbon where a non-end hydrogen is replaced by a hydroxyl group is a secondary alcohol. Two general forms of alcohols are
R-OH
and
OH
|
R-C-R'
|
H
where R and R' are hydrocarbon radicals.
(b) An aldehyde is formed by dehydrating an oxidized alcohol. It contains a formyl group (CHO),
O
//
-C
\
H
It has the general form
O
//
R-C
\
H
where R is a hydrocarbon radical.
(c) An organic acid or a carboxylic acid can be made by oxidizing an aldehyde. It contains a carboxyl group (COOH),
O
//
-C
\
OH
It has the general form
O
//
R-C
\
OH
where R is a hydrocarbon radical.
(d) An ester is produced by reacting an organic acid and an alcohol. The functional group of an ester is the COOR group,
O
||
--C--O-R
where R is a hydrocarbon radical. An ester has the general form
O
||
R'--C--O-R
where R' is also a hydrocarbon radical.
(e) An ether is produced by dehydrating an alcohol. It has the general form
R-O-R'
where R and R' are hydrocarbon radicals.
(f) A ketone is produced by oxidizing a secondary alcohol. It has the general form
O
||
R-C--R'
where R and R' are hydrocarbon radicals.
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PROBLEM
Draw the structural formulas of butane, butene, butyne, butanoic acid, butanol, and butanal.
[from Kavanah, Patrick, and Robbins, Jack 1971, Concepts in Modern Chemistry (New York: Cambridge Book Company), reasoning exercise 16.2]
SOLUTION
butane (C4H10):
H H H H
| | | |
H--C--C--C--C--H
| | | |
H H H H
butene (C4H8):
H H H H
\ | | |
C==C--C--C--H
/ | |
H H H
butyne (C4H6):
H H
| |
H--C\\\C--C--C--H
| |
H H
butanoic acid (CH3CH2CH2COOH) (obtained from butanal by replacing the H in the formal group with an OH):
H H H
| | |
H--C--C--C--C==O
| | | |
H H H OH
butanol (C4H9OH) (obtained from butane by replacing an end H with OH):
H H H H
| | | |
H--C--C--C--C--OH
| | | |
H H H H
butanal (CH3CH2CH2CHO) (obtained from butanol by removing the H and OH from the end carbon and adding a double-bonded O to the carbon):
H H H
| | |
H--C--C--C--C==O
| | | |
H H H H
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PROBLEM
(a) Write the structures and names of two primary alcohols having the molecular formula C7H16O.
(b) Write the structures and names of two secondary alcohols having the molecular formula C7H16O.
(c) Write the structures and names of two tertiary alcohols having the molecular formula C7H16O.
SOLUTION
(a) A (primary, secondary, tertiary) alcohol is an alcohol in which the carbon bonded to the hydroxide is bonded to (0 or 1, 2, 3) other carbons.
One example of a primary alcohol with the molecular formula C7H16O is
H H H H H H H
| | | | | | |
H--C--C--C--C--C--C--C--O--H 1-heptanol
| | | | | | |
H H H H H H H
It has the name "1-heptanol" because it is formed from heptane by replacing a hydrogen attached to an end carbon (a "1" carbon) with a hydroxyl group. Carbons are numbered beginning from the end closest to the hydroxyl group.
A second example of a primary alcohol with the molecular formula C7H16O is
H
|
H--C--H
|
H H H H | H
| | | | | |
H--C--C--C--C--C--C--O--H 2-methyl-1-hexanol
| | | | | |
H H H H H H
(b) An example of a secondary alcohol with the molecular formula C7H16O is
H H H H H H H
| | | | | | |
H--C--C--C--C--C--C--C--H 2-heptanol
| | | | | | |
H H H H H O H
|
H
A second example is
H
|
H--C--H
|
H H H | H H
| | | | | |
H--C--C--C--C--C--C--H 3-methyl-2-hexanol
| | | | | |
H H H H O H
|
H
(c) An example of a tertiary alcohol with the molecular formula C7H16O is
H
|
H--C--H
|
H H H H | H
| | | | | |
H--C--C--C--C--C--C--H 2-methyl-2-hexanol
| | | | | |
H H H H O H
|
H
A second example is
H
|
H--C--H
|
H H H | H H
| | | | | |
H--C--C--C--C--C--C--H 3-methyl-3-hexanol
| | | | | |
H H H O H H
|
H
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PROBLEM
For each of the following esters, identify the alcohol and organic acid from which the ester is formed, draw the structural formula of the alcohol and organic acid, and draw the structural formula of the ester:
(a) propyl benzoate (balsam)
(b) 3-methylbutyl acetate (bananas)
(c) octyl acetate (oranges)
(d) methyl butyrate or methyl butanoate (apples)
SOLUTION
(a) propanol + benzoic acid => propyl benzoate + water
H H H O ___
| | | || / \
H--C--C--C--O--H HO--C--< >
| | | \___/
H H H
propanol benzoic acid
H H H O ___
| | | || / \
H--C--C--C--O--C--< >
| | | \___/
H H H
propyl benzoate (C10H12O2)
(b) 3-methyl butanol + acetic acid => 3-methylbutyl acetate + water
H H H H O H
| | | | || |
H--C--C--C--C--O--H HO--C--C--H
| | | |
H | H H
|
H--C--H
|
H
3-methyl butanol acetic acid
H H H H O H
| | | | || |
H--C--C--C--C--O--C--C--H
| | | |
H | H H
|
H--C--H
|
H
3-methylbutyl acetate
(c) octanol + acetic acid => octyl acetate + water
H H H H H H H H O H
| | | | | | | | || |
H--C--C--C--C--C--C--C--C--O--H HO--C--C--H
| | | | | | | | |
H H H H H H H H H
octanol acetic acid
H H H H H H H H O H
| | | | | | | | || |
H--C--C--C--C--C--C--C--C--O--C--C--H
| | | | | | | | |
H H H H H H H H H
octyl acetate
(d) methanol + butanoic acid => methyl butanoate + water
H O H H H
| || | | |
H--C--O--H HO--C--C--C--C--H
| | | | |
H O H H H
methanol butanoic acid
H O H H H
| || | | |
H--C--O--C--C--C--C--H
| | | | |
H O H H H
methyl butanoate
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PROBLEM
Acetylsalicylic acid, commonly known as aspirin, has the following structural formula:
O
^ ||
/ \---C--O--H
| | H
| | |
\ /----O---C--C--H
v || |
O H
Show that acetylsalicylic acid is an ester which is formed from two acids.
SOLUTION
Acetylsalicylic acid is an ester of salicylic acid (the alcohol) and acetic acid. Salicylic acid is also known 2-hydroxybenzoic acid.
salicylic acid
O
^ ||
/ \---C--O--H
| | H
| | |
\ /---O--H HO--C--C--H
v || |
O H
acetic acid
salicylic acid + acetic acid => acetylsalicylic acid + water
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Nuclear Magnetic Resonance (NMR)
Certain atomic nuclei have magnetic dipole moments due to the angular momenta of their constituent protons and neutrons. The nuclei act like bar magnets.
In the presence of an external magnetic field, these nuclei can have different alignments relative to the field direction. The different alignments are quantized (i.e., only certain alignments are possible) and associated with different energy levels.
If such a nucleus is subjected to electromagnetic radiation of the right frequency, it can absorb the radiation and make a nuclear spin transition from one of these energy levels to another. The difference between the energy levels depends on the strength of the external field, the nuclear species, and the neighboring atoms if the nucleus is part of a molecule.
By subjecting a sample of molecules to an external magnetic field of constant strength and electromagnetic radiation of varying frequency, and noting the frequencies at which the sample absorbs radiation, we can deduce the presence of certain nuclei and some of the structural characteristics of the molecule. The technique is called nuclear magnetic resonance (NMR) spectroscopy. For the magnetic field strengths typically used, resonance occurs at radio frequencies (n ~ 30 MHz - 3 GHz).
NMR spectrometers work by either keeping the magnetic field strength fixed and varying the radio frequency or by keeping the radio frequency fixed and varying the magnetic field strength.
High-resolution NMR spectroscopy can separate closely spaced resonances that appear to be a single resonance using low-resolution NMR spectrometers.
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PROBLEM
What type of radiation (in wavelength, l) would be minimally required to initiate the radical chlorination of methane?
[from Vollhardt, K. Peter C., and Schore, Neil E. 1999, Organic Chemistry: Structure and Function, Third Edition (New York: W. H. Freeman and Company), exercise 10.2]
SOLUTION
In the radical chlorination of methane, we start with a chlorine molecule and a methane molecule and end up with a hydrogen chloride molecule and a chloromethane molecule:
Cl2 + CH4 => HCl + CH3Cl
This involves
breaking a Cl-Cl bond, DH0 = + 58 kcal/mole
breaking a C-H bond, DH0 = + 105 kcal/mole
forming a H-Cl bond, DH0 = - 103 kcal/mole
forming a C-Cl bond, DH0 = - 85 kcal/mole
The standard enthalpy change for the entire reaction is
DH0 = + 58 kcal/mole + 105 kcal/mole - 103 kcal/mole - 85 kcal/mole = - 25 kcal/mole
The initiation step is the breaking of the Cl-Cl bond, which requires an energy of
E = 58 kcal/mole = (58 kcal/mole)(1 mole / 6.022 x 1023 bonds)(4.186 x 103 J/kcal)
= 4.032 x 10-19 J/bond
The corresponding wavelength is
l = c/n = hc/E = (6.626 x 10-34 J s)(3 x 108 m/s)/(4.032 x 10-19 J) = 4.930 x 10-7 m = 493.0 nm
where n is the frequency, h = 6.626 x 10-34 J s is Planck's constant, and c = 3 x 108 m/s is the speed of light. This wavelength is at the violet end of the visible part of the electromagnetic spectrum.
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PROBLEM
With a 300 MHz NMR instrument, the three signals for 2,2-dimethyl-1-propanol are recorded at 261, 861, and 957 Hz, respectively, downfield from (CH3)4Si. Calculate the chemical shift d for each of the signals and the frequencies that would be recorded at 90 MHz.
H
|
H--C--H
|
H | H
| | |
H--C--C--C--O--H 2,2-dimethyl-1-propanol
| | |
H | H
|
H--C--H
|
H
[from Vollhardt, K. Peter C., and Schore, Neil E. 1999, Organic Chemistry: Structure and Function, Third Edition (New York: W. H. Freeman and Company), exercise 10.3]
SOLUTION
The chemical shift, in parts per million (ppm), is given by
d = [distance of peak from (CH3)4Si in Hz] / [spectrometer frequency in MHz]
We use this formula to calculate the chemical shift for the three signals measured with a 300 MHz instrument. Then we rewrite the formula in the form
[distance of peak from (CH3)4Si in Hz] = d[spectrometer frequency in MHz]
to calculate the position of the peaks that would be obtained with a 90 MHz NMR instrument.
| 300 MHz Peak | d | 90 MHz Peak |
|---|---|---|
The three signals can be identified with three types of equivalent hydrogens: (1) the nine hydrogens which are bonded to the carbons in the three methyl groups, (2) the two hydrogens which are bonded to the carbon which is adjacent to the oxygen, and (3) the single hydrogen which is bonded directly to the oxygen. The type (1, 3) hydrogen has the (least, most) deshielding due to the oxygen and will have its peak at the (highest, lowest) resonant frequency. We therefore identify type (1, 2, 3) with (957, 861, 261) Hz. The relative intensities are (9, 2, 1).
|
|
|
|
| NMR spectrum of 2,2-dimethyl-1-propanol
|
|
| |
| | |
--+--+--+-->
frequency (Hz) Back to Main Menu