Answer key to Problem Set on Chapters 16 and 17
1. This is a strong base problem.
Ba(OH)2 (aq) → Ba2+(aq) + 2 OH‾(aq)
0.020 mol Ba(OH) 2
2 mol OH x
 0.040 M OH -  [OH-]
1 L soln
1 mol Ba(OH) 2
pOH = - log[OH‾] = - log(0.040) = 1.40
pH = 14 – pOH = 14 – 1.40 = 12.60
2. This is an acidic buffer problem because a weak acid and a salt of the weak acid are
present in appreciable quantities. From Table 16.3, the Ka of HCN = 4.9 x 10-10.
HCN(aq) + H2O (l) ↔ CN‾(aq) + H3O+(aq)
init
0.30
0.40
0
change -x
x
x
equil
0.30 – x
0.40 + x
x
Ka 
[CN-][H3O  ]
(0.40  x)(x)

 4.9 x 10 - 10
HCN
0.30 - x
Simplifies to:
Ka 
x 
[CN-][H3O  ]
(0.40)(x)

 4.9 x 10 - 10
HCN
0.30
(4.9 x 10 - 10)(0.30)
 3.7 x 10 - 10 M  [H3O]
0.40
[HCN] = 0.30 – (3.7 x 10-10 ) = 0.30 M
[CN‾] = 0.40 + x = 0.40 + (3.7 x 10-10 ) = 0.40 M
[Na+] = 0.40 M
[OH-] 
Kw
1.0 x 10 - 14

 2.7 x 10 - 5

- 10
[H3O ]
3.7 x 10
[H2O] = 55.5 M
1
3. This is a strong acid problem.
HNO3(aq) + H2O(l) ↔ NO3‾(aq) + H3O+(aq)
[HNO3] = 0 M
[NO3‾] = [H3O+] = 0.45 M
[H2O] = 55.5 M
[OH-] 
Kw
1.0 x 10 - 14

 2.2 x 10 - 14 M
[H3O]
0.45
4. This is a weak acid problem. The Ka of HNO2 is 4.5 x 10-4.
HNO2(aq) + H2O(l) → NO2‾(aq) + H3O+(aq)
init
0.45
0
0
change -x
+x
+x
equil. 0.45 – x
x
x
Ka 
[NO2-][H3O  ]
(x)(x)

 4.5 x 10 - 4
[HNO2]
0.45  x
Simplifies to:
Ka 
[NO2-][H3O  ]
(x)(x)

 4.5 x 10 - 4
[HNO2]
0.45
x2 = 4.5 x 10-4(0.45) = 2.0 x 10-4
x 
2.0 x 10 - 4  1.4 x 10 - 2 M  [H3O]  [NO2-]
[HNO2] = 0.45 – x = 0.45 – 1.4 x 10-2 = 0.44 M
[OH-] 
Kw
1.0 x 10 - 14

 7.1 x 10 - 13 M

-2
[H3O ]
1.4 x 10
[H2O] = 55.5 M
2
5. This is a hydrolysis problem. The salt is the salt of a strong acid and strong base, so
neither Na+ (weak conjugate acid) nor NO3‾ (weak conjugate base) will hydrolyze. Both
these ions are spectator ions. Thus, the only [H3O+] or [OH‾] is from the dissociation of
water.
[Na+] = [NO3‾] = 0.45 M
[H3O+] = [OH‾] = 1.0 x 10-7 M
[H2O] = 55.5 M
6. This is also a hydrolysis problem. The K+ is a spectator because it is the weak
conjugate acid of a strong base. Nitrite ion will hydrolyze:
NO2‾(aq) + H2O(l) ↔ HNO2(aq) + OH‾(aq)
0.45
0
0
-x
x
x
0.45 – x
x
x
init
change
equil
Need to find Kb:
Kb 
Kb 
Kw
1.0 x 10 - 14

 2.2 x 10 - 11
Ka
4.5 x 10 - 4
[HNO2][OH-]
(x)(x)

 2.2 x 10 - 11
[NO2 ]
0.45  x
Simplifies to:
Kb 
[HNO2][OH-]
(x)(x)

 2.2 x 10 - 11
[NO2 ]
0.45
x2 = 2.2 x 10-11(0.45) = 1.0 x 10-11
x 
1.0 x 10 - 11  3.2 x 10 - 6 M  [HNO2]  [OH-]
[NO2ˉ] = 0.45 – x = 0.45 – (3.2 x 10-6) = 0.45 M
[K+] = 0.45 M
[H2O] = 55.5 M
3
7. This is a weak base problem. The Kb for CH3NH2 is found on Table 16.5 and has a
value of 4.4 x 10-4.
CH3NH2 (aq) + H2O(l) ↔ CH3NH3+ (aq) + OHˉ(aq)
init.
0.35
0
0
change
-x
x
x
equil.
0.35 – x
x
x
Kb 
[CH3NH3 ][OH-]
(x)(x)

 4.4 x 10 - 4
CH3NH2
0.35  x
Simplifies to:
[CH3NH3 ][OH-]
(x)(x)
Kb 

 4.4 x 10 - 4
CH3NH2
0.35
x2 = 4.4 x 10-4(0.35) = 1.5 x 10-4
x  1.5 x 10
-4
 0.012 M  [CH3NH3]  [OH-]
[CH3NH2] = 0.35 – x = 0.35 – 0.012 = 0.34 M
[H3O] 
- 14
Kw
1 x 10

[OH ]
0.012
 8.1 x 10 - 13 M
[H2O] = 55.5 M
8. This is an alkaline buffer problem.
CH3NH2 (aq) + H2O(l) ↔ CH3NH3+ (aq) + OHˉ(aq)
init.
0.45
0.30
0
change
-x
x
x
equil.
0.45 – x
0.30 + x
x
[CH3NH3 ][OH-]
(0.30  x)(x)
Kb 

 4.4 x 10 - 4
CH3NH2
0.45  x
Simplifies to:
Kb 
[CH3NH3 ][OH-]
(0.30)(x)

 4.4 x 10 - 4
CH3NH2
0.45
4
(4.4 x 10 - 4)(0.45)
x 
 6.6 x 10 - 4 M  [OH-]
0.30
[CH3NH3+] = 0.30 + x = 0.30 - 6.6 x 10-4 = 0.30 M
[CH3NH2] = 0.45 – x = 0.45 - 6.6 x 10-4 = 0.45 M
[H3O] 
- 14
Kw
1 x 10

 1.5 x 10 - 11 M
[OH ]
6.6 x 10 - 4
9. This is a solubility equilibrium problem. The Ksp of Cr(OH)3 is found on Table 17.2
and has a value of 3.0 x 10-29.
equil.
Cr(OH)3 (s) ↔ Cr3+ (aq) + 3 OHˉ(aq)
x
3x
Ksp = [Cr3+][OHˉ]3 = (x)(3x)3 = 3.0 x 10-29
27x4 = 3.0 x 10-29
x4 
3.0 x 10 - 29
 1.1 x 10 - 30
27
molar solubility  x
4
1.1 x 10 - 30  3.2 x 10 - 8
10. This is a solubility problem which introduces a common ion. The Ksp of BaSO4 is
found on Table 17.2 and has a value of 1.1 x 10-10
BaSO4(s) ↔ Ba2+(aq) + SO42- (aq)
init
0
0.0050
change
x
+x
equil.
x
0.0050 + x
Ksp = [Ba2+][SO42-] = (x)(0.0050 + x) = 1.1 x 10-10
Simplifies to:
Ksp = [Ba2+][SO42-] = (x)(0.0050) = 1.1 x 10-10
[Ba2 ] 
Ksp
1.1 x 10 - 10

 2.2 x 10 - 8 M
0.0050
0.0050
5
11. Here you need to find the reaction quotient, Q, and compare it to the Ksp.
Find the new molarity of calcium ions and carbonate ions:
[Ca2  ] 
[CO32-] 
(200 mL)(0.0040 M)
 1.3 x 10 - 3 M Ca2 
600 mL
(400 mL)(0.0003 3 M CO32-)
 2.2 x 10 - 4 M CO32 600 mL
Calculate the reaction quotient, Q:
Q = [Ca2+][ CO32- ] = (1.3 x 10-3 )(2.2 x 10-4) = 2.9 x 10-7
Compare Q and Ksp for calcium carbonate. The Ksp of CaCO3 is 8.7 x 10-9.
2.9 x 10-7 > 8.7 x 10-9.
Q
> Ksp
Since Q > Ksp , the solubility has been exceeded (equilibrium is too far to the right), so a
precipitate will form.
12. This is a complex ion equilibrium problem.
Calculate the moles of Cu2+ and moles of CN ˉ:
mol Cu2+ = (0.100 L )(0.10 mol Cu2+/L) = 0.010 mol Cu2+
mol CN ˉ = (0.100 L)(0.50 mol CN ˉ/L) = 0.050 mol CN ˉ
Cu2+(aq) + 4 CNˉ(aq) ↔ Cu(CN)42-(aq)
init
0.010 mol
0.050 mol
0 mol
change - 0.010 mol - 0.040 mol
0.010 mol
Notice that 0.010 mol of CN ˉ remains in excess. The new concentration of CN ˉ will be:
0.010 mol CN [CN-] 
 0.10 M CN 0.100 L soln
The concentration of complex ion, Cu(CN)42- produced in solution is:
Cu(CN)42 - 
0.010 mol Cu(CN)42  0.10 M Cu(CN)42 0.100 L
6
Using the equilibrium expression:
init
change
equil.
Cu2+(aq) + 4 CNˉ(aq) ↔ Cu(CN)42-(aq)
0
0.10
0.10
+x
+ 4x
-x
x
0.10 + 4x
0.10 – x
[Cu(CN)42-]
Kf 
 1.0 x 1025
2
4
[Cu ][CN ]
Kf 
0.10  x
 1.0 x 1025
4
(x)(0.10  4x)
Kf 
0.10
 1.0 x 1025
4
(x)(0.10)
(from Table 17.4)
Simplifies to:
x 
0.10
 1.0 x 1022 M  [Cu2  ]
25
4
(0.10) (1.0 x 10 )
7