CHM 2046
Quiz 2A
Spring, 2015
Answer all questions. Be careful to give your final answer to the correct sig. figs.
Assume all temperatures are 25 ºC.
1.
a)
(3 points) HPO42- is a weak base. Write down the base dissociation (base ionization)
equation for HPO42- in water.
HPO42- (aq) + H2O (l) ⇌ H2PO4- (aq) + OH- (aq)
b)
(2 points) Complete the expression for the base-dissociation constant (Kb) for HPO42-.
Kb =
c)

[H2PO4 ][OH- ]
2[HPO4 ]
(8 points) Calculate the pH of a 0.50 M solution of HPO42- (Kb = 1.6 x 10-7).
[init]
[change]
[equil]
HPO42- (aq) + H2O (l) ⇌
0.50
-x
(0.50-x)
H2PO4- (aq) + OH- (aq)
0
0
+x
+x
x
x
x2
x2
Kb =

 1.6x10 -7  x  [OH- ]  2.8 x 10- 4
(0.50  x) 0.50
 pOH = 3.55
2.
a)

pH = 10.45
(3 points) The Ka of benzoic acid (C6H5COOH) is 6.3 x 10-5. What is the Kb of its
conjugate base, benzoate (C6H5COO-)?
Kw 1.0 x 10 -14
Kb 

 1.6 x 10 -10
-5
K a 6.3 x 10
b)
Which one of the following salts will give a basic solution when dissolved in
water? Circle your choice.
(2 points)
CaCl2 , NH2Et2Cl , KBr , NaClO3 , SrI2 , none of these
Write an equation for the reaction that occurs when the salt dissolves in
water and makes the solution basic, or explain why none do.
(2 points)
ClO3- (aq) + H2O (l) ⇌ HClO3 (aq) + OH- (aq)
(2 points)
Which of the following is a weak acid? Circle your choice.
HCl, KClO4, NaHSO4, HI, CaCl2, NaNO3, NH3, NaF, none of these
Write an equation for the reaction that occurs when your choice dissolves in
water and makes the solution acidic, or briefly explain why none of them are weak acids.
(2 points)
HSO4- (aq) + H2O (l) ⇌ SO42- (aq) + H3O+ (aq)
3.
a)
What is the pH of a buffer solution prepared from adding 60.0 mL of 0.22 M
nitrous acid (HNO2) solution to 50.0 mL of 0.24 M sodium nitrite (NaNO2) solution? (Ka
for HNO2 is 7.1 x 10-4).
(60.0mL)(0.22M)
[HNO2 ] 
 0.120M (extra sig. fig.)
110.0mL
(50.0mL)(0.24M)
[NO2 ] 
 0.109M
110.0mL
(10 points)
HNO2 (aq) ⇌ H+ (aq) + NO2- (aq)
[init]
0.120
0
0.109
[change] -x
+x
+x
[equil]
(0.120-x) x
(0.109+x)
Ka 
x(0.109  x 0.109x

 7.1 x 10 -4
(0.120 - x)
0.120
 x = 7.82 x 10-4
 [H+] = x = 7.82 x 10-4 M = 7.8 x 10-4 M (sig. figs.)
 pH = -log (7.8 x 10-4)
pH = pKa + log [base]
[acid]
pH = 3.11
pH = 3.15 + (-0.04)
pH = 3.11
b) (10 points) What is the new pH if 10.0 mL of a 0.20 M solution of NaOH is now added?
n(HNO2) = MV = (0.120 M)(110.0 x 10-3 L) = 0.0132 mol
n(NO2-) = MV = (0.109 M)(110.0 x 10-3 L) = 0.0120 mol
n(OH-) added = MV = (0.20 M)(10.0 x 10-3 L) = 0.0020 mol
HNO2 +
OH- 
NO2- + H2O
n(init)
0.0132
0.0120
addition
0.0020
n(new)
0.0112
new [HNO2] =
new [NO2-] =
0
0.0112mol
120x10
0.0140
120x10
3
3
0.0140
= 0.0933M
OR
OR METHOD (2): pH = pKa + log
[base]
[acid]
OR METHOD (3): use number of moles
= 0.1167M
pH = pKa + log
n(base)
n(acid)
METHOD (1):
HNO2 ⇌ H+ + NO2[init]
0.0933
0
0.1167
[change] -x
+x +x
[equil]
(0.0933)-x x
(0.1167)+x
+
-4
[H ] = x = (7.1 x 10 )(0.0933)/(0.1167)
[H+] = 5.7 x 10-5  pH = 3.25
4.
 pH = 3.15 + 0.097
 pH = 3.25
(4 points) Calculate the solubility of the salt MX2 (containing M2+ and X- ions) in water (Ksp =
4.4 x 10-8).
MX2 (s) ⇌ M3+ (aq) + 2 X- (aq)
I (solid)
0
0
C -x
+x
+2x
E (solid)
x
2x
5.
Either way,
Ksp = 4.4 x 10-8 = [M2+][X-]2
 Ksp = x(2x)2
 4.4 x 10-8 = 4x3
x3 = 1.1 x 10-8
 x = 2.2 x 10-3 M
 solubility of MX2 = 2.2 x 10-3 M
(4 points) Now calculate the solubility of the same salt MX2 in an aqueous solution containing
0.10 M NaX, where NaX is a very soluble salt.
MX2 (s) ⇌ M3+ (aq) + 2 X- (aq)
I (solid)
0
0.10
C -x
+x
+2x
E (solid)
x
(0.10 + 2x)
Ksp = 4.4 x 10-8 = [M2+][X-]2
 Ksp = x(0.10 + 2x)2
 4.4 x 10-8 ≈ x (0.10)2 = x(1.0 x 10-2)
x = 4.4 x 10-8 /1.0 x 10-2 = 4.4 x 10-6
 solubility of MX2 = 4.4 x 10-6 M