1
Ch. 18 Practice Test Key
CHEM 185
1) What is the solubility (in g L–1) of Tl(OH)3 [thallium(III) hydroxide] in water.
Tl(OH)3 (s)  Tl3+ (aq) + 3 OH– (aq)
----0
0
----+x
+3x
----x
3x
I
C
E
Ksp = [Tl3+][OH–]3 = 6.3 x 10–46
Ksp = [Tl3+][OH–]3 = (x)(3x)3 = (x)(27x3) = 27x4
x = Ksp
27
¼
= 6.3 x 10–46
27
¼
= 2.1¦97 x 10–12 mol Tl3+
L
1 mol Tl(OH)3 255.4053¦20 g Tl(OH)3 = 5.6¦11x 10–10 g
1 mol Tl3+
1 mol Tl(OH)3
L
----->
5.6 x 10–10 g L–1
2) Determine the Equilibrium constant for the reaction between CaF2 (s) and nitric acid:
(I) CaF2 (s)  Ca2+ + 2 F– (aq)
(II) [H3O+ (aq) + F– (aq)  H2O (l) + HF (aq)] x 2
KI = Ksp
KII = (1/Ka)2
CaF2 (s) + 2 H3O+ (aq)  Ca2+ (aq) + 2 H2O (l) + 2 HF (aq)
Koverall = KIKII = (Ksp)(1/Ka)2 =
5.3 x 10–9
1
6.6 x 10–4
2
= 0.012¦16 -----> 0.012 = K
3) Determine the Equilibrium constant for the reaction between Sn(OH)2 (s) and HClO2 (aq):
(I) Sn(OH)2 (s)  Sn2+ + 2 OH– (aq)
(II) [HClO2 (aq) + H2O (l)  ClO2– (aq) + H3O+ (aq)] x 2
(III) [H3O+ (aq) + OH– (aq)  2 H2O (l)] x 2
K1 = Ksp
KII = (Ka)2
KIII = (1/Kw)2
Sn(OH)2 (s) + 2 HClO2 (aq)  Sn2+ (aq) + 2 ClO2– (aq) + 2 H2O (l)
Koverall = K1KIIKIII = (Ksp)(Ka)2(1/Kw)2 = 1.4 x 10–28
1.1 x 10–2
2
1
1.00 x 10–14
2
= 1.6¦94 x 10–4
----->
1.7 x 10–4 = K
4) What is the solubility (in g L–1) of Ag2CO3 in a solution of 0.280 M AgClO3.
AgClO3 (aq) -----> Ag+ (aq) + ClO3– (aq)
I
C
E
(common ion effect problem)
Ag2CO3 (s)  2 Ag+ (aq) + CO32– (aq)
----0.280
0
----0.280 + 2x
+x
----0.280 + 2x
x
Ksp = [Ag+]2[CO32–] = 8.5 x 10–12
Ksp = [Ag ] [CO3 ] = (0.280 + 2x) (x) ~ (0.280) (x)
+ 2
x=
2–
2
2
M = 0.280
= 3.2¦94 x 1010 >> 100
–12
Ksp
8.5 x 10
(assumption valid)
Ksp = 8.5 x 10–12 = 1.0¦84 x 10–10 mol CO32– 1 mol Ag2CO3 275.745¦60 g Ag2CO3 = 2.9¦89 x 10–8 g
(0.280)2 0.0784¦00
L
1 mol CO32–
1 mol Ag2CO3
L
----->
3.0 x 10–8 g L–1
2
5) Determine the Equilibrium constant for the reaction between Fe(OH)3 (s) and HNO2 (aq):
(I) Fe(OH)3 (s)  Fe3+ + 3 OH– (aq)
(II) [HNO2 (aq) + H2O (l)  NO2– (aq) + H3O+ (aq)] x 3
(III) [H3O+ (aq) + OH– (aq)  2 H2O (l)] x 3
K1 = Ksp
KII = (Ka)3
KIII = (1/Kw)3
Fe(OH)3 (s) + 3 HNO2 (aq)  Fe3+ (aq) + 3 NO2– (aq) + 3 H2O (l)
Koverall = K1KIIKIII = (Ksp)(Ka)3(1/Kw)3 = 4 x 10–38
7.2 x 10–4
3
1
1.00 x 10–14
3
= 1.¦49 x 10–5
----->
1 x 10–5 = K
6) What is the Solubility (in g L–1) of PbCl2 in a solution of 0.114 M FeCl3.
FeCl3 (aq) -----> Fe3+ (aq) + 3 Cl– (aq)
0.114 M
0.114 M
3 (0.114 M) = 0.342 M
(common ion effect problem)
PbCl2 (s)  Pb2+ (aq) + 2 Cl– (aq)
----0
0.342
----+x
0.342 + 2x
----x
0.342 + 2x
I
C
E
Ksp = [Pb ][Cl ] = (x)(0.342 + 2x) ~ (x)(0.342)
2+
x =
– 2
2
Ksp = [Pb2+][Cl–]2 = 1.6 x 10–5
M =
0.342 = 2.1¦37 x 104 > 100
Ksp
1.6 x 10–5
(assumption valid)
2
Ksp
= 1.6 x 10–5 = 1.3¦67 x 10–4 mol Pb2+ 1 mol PbCl2 278.1¦05 g PbCl2
2
(0.342)
0.116¦96
L
1 mol Pb2+
1 mol PbCl2
= 0.038¦01 g
L
-----> 0.038 g L–1
7) What is the solubility (in g L–1) of Li3PO4 in water.
I
C
E
Li3PO4 (s) 
-------------
3 Li+ (aq) +
0
+ 3x
3x
PO43– (aq)
0
+x
x
Ksp = [Li+]3[PO43–] = 3.2 x 10–9
Ksp = [Li+]3[PO43–] = (3x) 3(x) = (27x3)(x) = 27x4
x = Ksp
27
¼
= 3.2 x 10–9
27
¼
= 3.2¦99 x 10–3 mol PO43– 1 mol Li3PO4 115.794¦36 g Li3PO4 = 0.38¦20 g
L
1 mol PO43–
1 mol Li3PO4
L
----->
0.38 g L–1
8) Determine the Equilibrium constant for the reaction between ScF3 (s) and perchloric acid:
(I) ScF3 (s)  Sc3+ + 3 F– (aq)
(II) [H3O+ (aq) + F– (aq)  H2O (l) + HF (aq)] x 3
KI = Ksp
KII = (1/Ka)3
H3O+ (aq) + ClO4– (aq)
ScF3 (s) + 3 H3O+ (aq)  Sc3+ (aq) + 3 H2O (l) + 3 HF (aq)
Koverall = KIKII = (Ksp)(1/Ka)3 = 4.2 x 10–18
1
6.6 x 10–4
3
= 1.4¦60 x 10–8 -----> 1.5 x 10–8 = K