Chemistry 122 [Tyvoll]
Spring 2008
Chapter 17 Homework Solutions for Moore
Problems # – 1, 2, 5, 13, 16, 19, 21, 24, 34, 36, 39, 40, 47, 51, 57, 62, 65, 67, 68, 79, 82, 88, 95
1. The buffer capacity of a buffer solution is related to the concentrations of the buffer’s acid and
base. The buffer solution is able to resist changes in pH only until the total amount of added
acid or added base completely neutralizes one component of the buffer solution. The higher the
buffer concentrations are, the higher the buffer capacity is.
2. A weak acid-base conjugate pair must be in the solution. The weak base can neutralize added
acid making more of the weak base, and the weak acid can neutralize added base making more
of the weak acid; hence, the pair together makes a buffer solution capable of resisting pH.
(a) A strong acid, HCl, and a weak acid, CH3COOH, together in a system are able to neutralize
added base, but are not able to neutralize added acid, so this pair does not form a buffer.
(b) The dissolved NaH2PO4 salt contains the weak acid H2PO4–, and the dissolved Na2HPO4 salt
contains its conjugate weak base, HPO42–, so this pair would form a buffer.
(c) The solution contains the weak acid H2CO3 and a dissolved salt, NaHCO3, containing its
conjugate weak base, HCO3–, so this pair would form a buffer.
5. The “end point” is the term used during an acid base titration describing the volume of added
titrant at which an appropriate indicator being used changes color – indicators colors changes
occur only based on the solution’s pH. The “equivalence point” is the point during a titration
describing the volume of added titrant at which the moles of reactant acid are equal to the moles
of reactant base. With the proper choice of indicator, these two points should coincide, though
they usually will not be exactly the same. The pH changes rapidly near the equivalence point, so
the indicator will changes color close to the pH of the equivalence point with almost the same
amount of titrant.
13. The ion product and solubility product expression have the same functional form. Both of them
are the mathematical product of the concentrations of the ions in a solution. However, in the ion
product, the concentrations need not be equilibrium concentrations. In the solubility product
expression, the concentrations must be equilibrium concentrations.
16. To determine the pH of a buffer, look up the pKa (Table 17.1) or look up the Ka
(Table 16.2) and calculate the pKa (pKa = – log Ka). The pKa closest to the desired pH is the best
buffer, since close to equal quantities of the acid and base would be used, giving the solution
approximately equal ability to neutralize added acid or added base.
(a) The CH3COOH/NaCH3COO buffer system has pKa = 4.74.
(b) The acid in HCl/NaCl is HCl. It has Ka = very large. This is not a buffer.
(c) The NH3/NH4Cl buffer system has pKa = 9.25.
The combination that would make the best pH 9 buffer system is (c) NH3/NH4Cl.
19. Compare the [H3O+] to the values of Ka in Table 17.1, since that is equal to the [H3O+] in an 1:1
buffer solution. The Ka closest to the desired [H3O+] is the most suitable.
(a) [H3O+] = 4.5 x 10–3 M, needs a lactic acid/lactate buffer (Ka = 1.4 x 10–4).
(b) [H3O+] = 5.2 x 10–8 M, needs a dihydrogen phosphate/hydrogen phosphate buffer
(Ka = 6.2 x 10–8).
(c) [H3O+] = 8.3 x 10–6 M, needs an acetic acid/acetate buffer (Ka = 1.8 x 10–5).
(d) [H3O+] = 9.7 x 10–11 M, needs a hydrogen carbonate/carbonate buffer (Ka = 4.8 x 10–11).
21. Use pH = pKa + log {[base]/[acid]} @ 4.57 = 4.74 + log {[base]/(0.150)}
− 0.17 = log {[base]/(0.150)} @ 10−0.17 = 0.676 = {[base]/(0.150)}
[base] = (0.676)(0.0150 M) = 0.10 M
g NaCH3COO = (0.500 L)(0.10 mol/L)(82.034 g/mol) = 4.2 g
24. Calculate the initial concentration of NH4+, the conjugate acid, where
C(NH4+) = (5.15 g/0.10 L)(1 mol/80.04 g)(1 mol NH4+/1 mol NH4NO3 = 0.64 M NH4+
Table 17.1 gives pKa of ammonium/ammonia buffer as 9.25, and the initial concentration of
NH3, the conjugate base, is 0.15 M.
so, pH = 9.25 + log {0.15/0.64) = 9.25 + (−0.63) = 8.62
34. At the equivalence point, the solution contains the conjugate base of the weak acid. Weaker
acids have stronger conjugate bases. Stronger bases have higher pH, so the titration of a weaker
acid will have a more basic equivalence point.
36. The color change needs to occur as close as possible to the pH of the equivalence point.
(a) The strong base, NaOH, titrated with a strong acid, HClO4, has a neutral equivalence point.
It would be best to choose bromthymol blue, which is shown changing color in Figure 17.5
at a pH near 7. In practice, any of them would be suitable, because of the extreme change
in the pH of the solution very close to the equivalence point.
(b) The weak acid, CH3COOH, titrated with a strong base, KOH, has a basic equivalence
point, due to the presence of the weak base CH3COO– in the solution. It would be best to
choose phenolphthalein, which is shown changing color in Figure 17.5 at a pH near 9.
(c) The weak base, NH3, titrated with a strong acid, HBr, has an acidic equivalence point, due
to the presence of the weak acid NH4+ in the solution. It would be best to choose methyl
red, which is shown changing color in Figure 17.8 at a pH near 5.
36. (d) The strong base, KOH, titrated with a strong acid, HNO3, has a neutral equivalence point. It
would be best to choose bromthymol blue, which is shown changing color in Figure 17.5 at
a pH near 7. In practice, any of them would be suitable, because of the extreme change in
the pH of the solution very close to the color in Figure 17.8 at a pH near 5.
39. M NaOH = {(12.4 mL H2SO4/20.0 mL NaOH)(1 L H2SO4/1000 mL H2SO4)(
x (1000 mL NaOH/1 L NaOH)(0.205 mol H2SO4/1 L H2SO4)(2 mol H+/1 mol H2SO4)
x (1 mol NaOH/1 mol H+)} = 0.254 mol NaOH/L NaOH = 0.254 M
40. g C6H8O6 = {(24.4 mL NaOH)(1 L/1000 mL)(0.110 mol NaOH/1 L NaOH)
x (1 mol C6H8O6/1 mol NaOH)(176.1238 g C6H8O6/1 mol C6H8O6) = 0.473 g C6H8O6
47. The pH in rain always contains some dissolved CO2 from the air. As described in
Section 17.3, atmospheric carbon dioxide dissolves and reacts with rainwater to form a solution
of weak carbonic acid. Rainwater in equilibrium with dissolved CO2 has a pH of 5.6. Any water
less acidic than this, even if it is acidic by the chemical definition, is still not called “acid rain.”
51. Adapt the method developed in Problem-Solving Example 17.8.
(a) BaCrO4 (s) ' Ba2+ (aq) + CrO42– (aq)
(b) Mn(OH)2(s) ' Mn2+ (aq) + 2 OH– (aq)
(c) Ag2SO4 (s) ' 2 Ag+ (aq) + SO42– (aq)
Ksp = [Ba2+][CrO42–]
Ksp = [Mn2+][OH–]2
Ksp = [Ag+]2[SO42–]
57. Write the chemical equation and the equilibrium expression for the dissociation of the solute:
SrF2 (s) ' Sr2+ (aq) + 2 F– (aq)
Ksp = [Sr2+][F– ]2
At equilibrium, the moles of solid that dissolve per liter is
M = {(0.011 g SrF2/100. mL)(1 mol SrF2/125.62 g SrF2)(1000 mL/1 L) = 8.8 x 10–4 M
The stoichiometry of the equation shows that the concentration of strontium ion is the same as
the moles of solid that dissolve per liter, 8.8 x 10–4 M, and the fluoride ion concentration is two
times that value, or, 2(8.8 x 10–4 M) = 1.8 x 10–4 M.
Ksp = (8.8 x 10–4)(1.8 x 10–4)2 = 2.7 x 10–9
62. Use the equilibrium expression for Ksp and the known values to determine the unknown value.
Get the value of Ksp from Appendix H. Na2CO3 produces carbonate ions in solution, so we use
the base hydrolysis reaction where CO32– (aq) + H2O (l) ' HCO3– (aq) + OH– (aq)
ICE Table
I
C
E
[CO32–] (aq)]
0.25
−x
0.25 − x
[HCO3–] (aq)]
0
+x
x
[OH–] (aq)]
0
+x
x
62. Kb = (x)(x)/(0.25 – x) = 2.1 x 10–4
x = [OH–] = 0.0072 M and
[CO32–] = 0.25 – x = 0.25 – 0.0072 = 0.24 M
Finally, Ksp = [Zn2+][CO32–] = [Zn2+](0.24) = 1.5 x 10–11, [Zn2+] = 6.2 × 10–11 mol/L
65. (a) Write the chemical equation and the equilibrium expression for the dissociation of the solute:
Fe (OH)2 (s) ' Fe2+ (aq) + 2 OH– (aq)
Ksp = [Fe2+][OH–]2
At equilibrium, the moles of solid per liter are:
M = (6.0 x 10−1 mg Fe(OH)2/L)(1 g/1000 mg)(1 mol Fe(OH)2/89.860 g Fe(OH)2)
M = 6.7 x 10–6 M
The stoichiometry of the equation shows that the iron(II) ion concentration is the identical to
the number of moles of solid that dissolve per liter, 6.7 x 10–6 M, and the hydroxide ion
concentration is two times that value, 2(6.7 x 10–6 M) = 1.3 × 10–5 M, so
Ksp = (6.7 x 10–6)(1.3 x 10–5)2 = 1.1 x 10–15
(b) First we determine the concentration of the iron (II) ion in the Fe2+ ion solution:
[Fe2+] = (1.0 μg Fe2+/L)(10−6 g/1 μg)(1 mol Fe2+/55.845 g Fe2+) = 1.8 x 10–8 M
Next, use the calculated Ksp from (a) to calculate the hydroxide ion concentration in this
solution:
Ksp = 1.1 x 10–15 = (1.8 x 10–8)[OH–]2 @ [OH–] = 2.6 x 10–4 M
An equilibrium [OH–] of 2.6 x 10–4 M or higher will keep the [Fe2+] at or below 1.0 μg/L
67. Write the chemical equation and the equilibrium expression for the dissociation of the solute:
AgCl (s) ' Ag+ (aq) + Cl– (aq)
Ksp = [Ag+][Cl–]
Get the value of Ksp = 1.8 x 10–10 from Table 17.2 or Appendix H. The soluble AgNO3 salt
produces silver ions in the solution. [Ag+] = 0.05 mol/L.
Ksp = 1.8 x 10–10 = (0.05)[Cl–] @ [Cl– ] = 4 x 10–9 M
68. Given the pH, we can determine the concentration of the aqueous ion, OH–. The precipitation of
sparingly soluble Zn(OH)2 will occur above a certain concentration of Zn2+.
[OH– ] = 10–pOH = 10pH – 14.00 = 1010.00 – 14.00 = 1.0 x 10–4 M
Ksp = [Zn2+][OH–]2 = 4.5 x 10–17 (from Appendix H)
Ksp = 4.5 x 10–17 = [Zn2+](1.0 x 10–4)2 @ [Zn2+] = 4.5 x 10–9 M
An equilibrium [Zn2+] of 4.5 x 10–9 M or lower can exist in a solution with pH 10.00. Above
that concentration, Zn(OH)2 will precipitate.
79. Use pH = pKa + log {[base]/[acid]} @ 9.00 = 9.25 + log {[0.93]/ [NH4+]}
Table 17.1 gives pKa of ammonium/ammonia buffer as 9.25. The pH is 9.00. The initial
concentration of NH3, the conjugate base, is 0.93 M. The conjugate acid, NH4+, is being added,
but its initial concentration is unknown, so use [NH4+]to identify that quantity.
9.00 − 9.25 = −0.25 = log {(0.93 M)/ [NH4+]} @ 10−0.25 = {(0.93 M)/ [NH4+]}
[NH4+] = (0.93 M)(10–0.25 = 1.7 M
82. [NO2–] = (7.50 g KNO2/L)(1 mol KNO2/85.11 g KNO2)(1 mol NO2–/1 mol KNO2)
[NO2–] = 0.0881 M
In Table 16.2, the Ka of dihygydrogen phosphate is given: 4.5 x 10–4. As described in Section
17.1, calculate the pKa: pKa = –logKa = –log (4.5 x 10–4) = 3.35
The pH is 4.00. The initial concentration of NO2–, the conjugate base, is 0.0881 M. The
conjugate acid, HNO2, is being added, but its initial concentration is unknown, so use [HNO2]
to identify it. Use pH = pKa + log {[base]/[acid]}
4.00 = 3.35 + = 0.65 = log {[0.0881 M]/[HNO2]}
100.65 = {[0.0881 M]/[HNO2]} = 4.47 @ [HNO2] = {[0.0881 M]/ 4.47} = 0.020 M
88. There is no effect on the equilibrium state if more solid is added, since the quantity
of that solid present does not affect the equilibrium state.
95. The pH 10 buffer is made with a hydrogen carbonate/carbonate buffer system (according to
Table 17.1). The lab technician wrote down the wrong acid/base conjugate pair: carbonic
acid/hydrogen carbonate, H2CO3/HCO3–.
pH = pKa + log {[base]/[acid]} @ 10.00 = 10.32 + log {[CO32– ]/[HCO3–]
−0.32 = log {[CO32– ]/[HCO3–]
95. If the technician put equimolar quantities (a 1:1 ratio) of HCO3– and CO32– into the same
solution, the concentration ratio of [CO32–]/[HCO3–] would be 1.00.
pH = 10.32 + log {[CO32−][HCO3−]} = 10.32 + log (1.00) = 10.32