ANSWERS TO AP SOLUBILITY
1. Solid iodine requires an input of energy before it enters either solvent. The I2(s) is more soluble in
ALCOHOL because it takes less energy for the I2 to dissolve in alcohol than in CCl4.
2. Both O2 and N2O enter the aqueous phase with equal ease, but it requires more energy to remove the
N2O from water. Therefore, N2O is more effectively trapped in water than is O 2 , and N2O has a higher
solubility in water.
3. (a) In both cases, a single molecule in the solid phase enters the gas phase so there exists more or
less equal tendencies to maximum randomness in both cases.
(b) The ICN(s) has a greater tendency to remain at minimum energy than does FCN(s).
(c) Combining parts (a) and (b), there is less tendency for ICN(s) to go into the gas phase than FCN(s)
and ICN(s) has a LOWER vapour pressure at 201 K.
4. (a) A mixture of two pure liquids is more random than two pure, separate liquids so that the act of
mixing the solutions increases the randomness.
(b) Since the reaction is exothermic, the tendency to remain at minimum energy favours the product
side.
(c) Since both maximum randomness and minimum energy favour the formation of products, one
would expect to find a great tendency to form products (mixed solutions) so that liquid chloroform
and acetone are MISCIBLE.
Since a water–gasoline mixture is more random than two separate liquids, randomness favours
“products”. If the liquids DO NOT mix, the tendency to remain at minimum energy MUST favour
reactants, that is the reaction is endothermic. However, if the tendencies to minimum energy and
maximum randomness are more or less balanced then an equilibrium exists and the two liquids are
not IMMISCIBLE. Hence, the lack of mixing means the reaction is HIGHLY ENDOTHERMIC.
5.
(a) less
(b) more
6. "Highly soluble” implies ∆HHYD > ∆HCL
“Moderately soluble” implies ∆HHYD  ∆HCL
“Slightly soluble” implies ∆HHYD < ∆HCL
Ag+ + Br– ; KSP = [Ag+][Br–] = 5.4 x 10–13

X
X
7. AgBr
–X
K+ + Br–
0.020
0.020
KBr
–0.020
[Ag+] = X
[Br– ] = 0.020 + X

0.020
(since X is negligible with respect to 0.020)
Now
X • 0.020 = 5.4 x 10–13
and X = 2.7 x 10–11 M = [AgBr]DISSOLVED
so that moles AgBr dissolved = 2.7 x 10–11 mol/L x 1.0 L = 2.7 x 10–11 mol

Ag+ + Cl– ; KSP = [Ag+][Cl–] = 1.8 x 10–10
X
X
8. AgCl
–X
Na+ + Cl–
0.10
0.10
NaCl
–0.10
[Ag+] = X
[Cl–] = X + 0.10
Now

0.10
(since X is negligible with respect to 0.10)
X • 0.10 = 1.8 x 10–10 and
X = 1.8 x 10–9 M = [AgCl]DISSOLVED
so that 
moles AgCl dissolved = 1.8 x 10–9 mol/L x 5.0 L = 9.0 x 10–9 mol
9.
PbSO4
Pb2+
–5.0 x 10–6
5.0 x 10–6

+ SO 24
; KSP = [Pb2+][ SO 24 ] = 1.8 x 10–8
5.0 x 10–6

2
Hebden : Chemistry AP
————————————————————————————————————————————————
2 K+ + SO 24
2X
X
K2SO4
–X
[Pb2+] = 5.0 x 10–6

[ SO 24 ] = 5.0 x 10–6 + X
Solving
 10.
(since 5.0 x 10–6 is negligible with respect to X)
X
(5.0 x 10–6) • X = 1.8 x 10–8

Cu(IO3)2
–1.0 x 10–5
Cu2+
[Cu2+] = 1.0 x 10–5
[ IO 3 ] = X
+ 2.0 x 10–5
and
2 IO 3
+
1.0 x 10–5
Na+ + IO 3
X
X
NaIO3
–X
Solving

2.0 x 10–5
X

(since 2.0 x 10–5 is negligible with respect to X)
(1.0 x 10–5) • X2 = 6.9 x 10–8
 11. (a) AgOH
–X
NaOH
–0.020

and
X = 0.083 M = [NaIO3]
Ag+ + OH– ; KSP = [Ag+][OH–] = 1.4 x 10–8
X
X
Na+ + OH–
0.020
0.020
[Ag+] = X
[OH–] = X + 0.020
Now
; KSP = [Cu2+][ IO 3 ]2 = 6.9 x 10–8


X = 0.0036 M = [K2SO4]

0.020
(since X is negligible with respect to 0.020)
X • 0.020 = 1.4 x 10–8
and
X = 7.0 x 10–7 M = [Ag+]

124.9 g
(b) mass of AgOH = 7.0 x 10–7 x 2.0 L x
= 1.7 x 10–4 g
mol
1mol
4.01 x 10 3 g
x
= 1.454 x 10–5 M
275.8 g
1.00 L

CO 23 
Ag2CO3
2 Ag+
+
–1.454 x 10–5
2.908 x 10–5
1.454 x 10–5


K2CO3
2 K+ + CO 23 

–0.0100
0.0200
0.0100
12. [Ag2CO3] =
[Ag+] = 2.908 x 10–5
[ CO 23  ] = 0.0100 +1.454 x 10–5
and


0.0100
KSP = (2.908 x 10–5)2(0.0100) = 8.46 x 10–12

; KSP = [Ag+]2[ CO 23  ] = ?

AP : SOLUBILITY ANSWERS
3
——————————————————————————————————————————————————
13.
BaSO4
Ba2+
–2.0 x 10–7
2.0 x 10–7
2 K+ +
2X
K2SO4
–X
; KSP = [Ba2+][ SO 24 ] = 1.1 x 10–10
2.0 x 10–7
SO 24

X
[Ba2+] = 2.0 x 10–7
[ SO 24 ] = X 
+ 2.0 x 10–7
Now
SO 24
+


(since 2.0 x 10–7 is negligible with respect to X)
X
(2.0 x 10–7) • X = 1.1 x 10–10
and
X= 5.5 x 10–4 M = [K2SO4]

Mn2+ + S2– ; KSP = [Mn2+][S2–] = 1.4 x 10–15
X
X
 14. MnS
–X
2 NH 4 + S2–
0.040
0.020
(NH4)2S
–0.020
[Mn2+] = X
X + 0.020
[S2–] =

0.020
(since X is negligible with respect to 0.020)
X = 7.0 x 10–14 M = [MnS]
87.0 g
so that mass
= 6.1 x 10–12 g
 MnS = 7.0 x 10–14 x 1.0 L x
mol
Now
15.
1
X • 0.020 = 1.4 x 10–15
2 Ag2O
+
1
2
Ag+ + OH– ; KSP = [Ag+][OH– ] = 2.0 x 10–8
X
X

OH–
1.0 x 10–3
H2O
–X


[Ag+] = X
[OH–] = 1.0 x 10–3 + X
Now
and

1.0 x 10–3
X • (1.0 x 10–3) = 2.0 x 10–8
(since X is negligible with respect to 1.0 x 10–3)
and
X = 2.0 x 10–5 M = [Ag+]

2.0 x 10 5 mol Ag 
0.5 mol Ag2O
so that moles Ag2O =
x
x 2.0 L = 2.0 x 10–5 mol
L
1mol Ag 
50.0 mL
after mixing: [NaOH] = X •
= 0.500 X
100.0 mL

50.0 mL
[Al(OH)3] = 2.0 x 10–6 M x
= 1.0 x 10–6 M
100.0 mL

Al(OH)3
Al3+
+ 3 OH–
; KSP = [Al3+][OH– ]3 = 3.7 x 10–15
–6
–6
–6
–1.0 x 10
1.0 x 10
3.0 x 10
 +
NaOH
Na
+
OH–
–0.500 X
0.500 X
0.500 X
16. Let original [NaOH] = X ,

[Al3+] = 1.0 x 10–6
[OH–] = 3.0 x 10–6 + 0.500 X
Now

0.500 X
(since 3.0 x 10–6 is negligible with respect to 0.500 X)
(1.0 x 10–6) (0.500 X)3 = 3.7 x 10–15 and

X = 3.1 x 10–3 M = [NaOH]
4
Hebden : Chemistry AP
—————————————————————————————————————————————————–
Mg2+ + 2 OH– ; KSP = [Mg2+][OH–]2 = 5.6 x 10–12
X
2X
17. Mg(OH)2
–X
Mg2+ + SO 24
0.050
0.050
MgSO4
–0.050
[Mg2+] = X + 0.050
[OH–] = 2X 

0.050
(since X is negligible with respect to 0.050)
0.050 • (2 X)2 = 5.6 x 10–12 and X = 5.29 x 10–6 M

58.3 g
so that solubility = 5.29 x 10–6 x
= 3.1 x 10–4 g/L
1mol
Now
18. [Cu(OH)2] =
1mol
1.56 x 10 9 g
x
= 1.60 x 10–11 M
1.00 L  97.5 g
Cu(OH)2
–1.60 x 10–11

NaOH
–1.00 x 10–4
Cu2+
+
2 OH–
1.60 x 10–11 3.20 x 10–11

Na+
+
OH–
–4
1.00 x 10
1.00 x 10–4
[Cu2+] = 1.60 x 10–11
[OH–] = 1.00 x 10–4 + 3.20 x 10–11
and

; KSP = [Cu2+][OH– ]2 = ?
1.00 x 10–4
KSP = (1.60 x 10–11) (1.00 x 10–4)2 = 1.60 x 10–19
 ion effect problem!
19. NOTE: This is NOT a common
[Ag2SO4] =
1mol
9.92 x 10 5 g
x
= 3.18 x 10–7 M
311.9 g
1.00 L
3
+3
–23
3 Ag+ + AsO3
4 ; KSP = [Ag ] [ AsO4 ] = 1.00 x 10
Ag3AsO4
+] = 2 x 3.18 x 10–7 = 6.36 x 10–7 M
[Ag

3
3
–23 and
so (6.36 x 10–7
) [ AsO4 ] = 1.00 x 10
and mass of Na3AsO4 = 3.89 x 10–5

20. [Ag3AsO4] =
Ag3AsO4
3 Ag+
[Ag+]
= 4.22 x 10–7
3
[ AsO4 ] = X+ 1.40 x 10–7

mol
207.9 g
–3
L x 1.00 L x 1mol = 8.08 x 10 g

1mol
6.50 x 10 5 g
x
= 1.40 x 10–7 M
462.6 g
1.00 L

+
–1.40 x 10–7
4.22 x 10–7


K3AsO4
3 K+ + AsO3
4

–X
3X
X
Now
–5
[ AsO3
4 ] = 3.89 x 10

X
AsO3
4
–23
; KSP = [Ag+]3[ AsO3
4 ] = 1.00 x 10
1.40 x 10–7

(1.40 x 10–7 is negligible with respect to X)
(4.22 x 10–7)3 • X = 1.00 x 10–23 and
X = 1.33 x 10–4 M = [K3AsO4]
256.2 g

so that mass of K3AsO4 = 1.33 x 10–4 x 1.00 L x
= 3.42 x 10–2 g
1mol
