Chem 1622016 Chapter 18 Acidbase & solubility equilibria lecture notes
48
CHEM 1622010 EXAM III
Chapter 16  Equilibria
Solubility product
ET note: Ksp of varied substances only related to solubility if molar ratio of ions are identical.
K
sp
values vs. Relative Solubility
18.
Which of the following compounds would have the greatest molar solubility?
A.
AgCl (K
sp
= 1.8 x 10
10
)
B.
AgBr (K
sp
= 5.0 x 10
13
)
C.
PbCO
3
(K
sp
= 7.4 x 10
14
)
D.
BaSO
4
(K
sp
= 1.1 x 10
10
)
E
.
Ag
2
CrO
4
(K
sp
= 2.4 x 10
12
)
AgCl(s)
Ag
+
(aq) + Cl

(aq)
AgBr(s)
Ag
+
(aq) + Br

(aq)
PbCO
3
(s)
Pb
2+
(aq) + CO
3
2
(aq)
BaSO
4
(s)
Ba
2+
(aq) + SO
4
2
(aq)
Ag
2
CrO
4
(s)
2Ag
+
(aq) + CrO
4
2
(aq)
AgCl(s)
Ag
+
(aq) + Cl

(aq)
AgCl(s)
Ag
+
(aq)
+
Cl

(aq)
Initial Y 0 0
Change X +X +X
Equilibrium YX +X +X
[Ag
+
][Cl

] = K
sp
= 1.8 x 10
10
[X][X] = 1.8 x 10
10
[X] = 1.34 x 10
5
= solubility of AgCl
PbCO
3
(s)
Pb
2+
(aq) + CO
3
2
(aq)
PbCO
3
(s)
Pb
2+
(aq)
+
CO
3
2
(aq)
Initial Y 0 0
Change X +X +X
Equilibrium YX +X +X
[Pb
2+
][CO
3
2
] = K
sp
= 7.4 x 10
14
[X][X] = 7.4 x 10
14
[X] = 2.72 x 10
7
= solubility of PbCO
3
The AgCl, AgBr, PbCO
3
, BaSO
4
all have the same molar ratio of ions, that is, 1:1.
Therefore, they all use the same ICE table
format, and therefore the same formula for finding the solubility, which is
X
2
= K
sp
.
Hence, the most soluble of these would be
the one with the largest K
sp
, which is AgCl.
But Ag
2
CrO
4
has a different ICE table format, and therefore a different formula
for determining its solubility.
So we need to quantitatively compare AgCl to Ag
2
CrO
4
.
Ag
2
CrO
4
(s)
2Ag
+
(aq) + CrO
4
2
(aq)
Ag
2
CrO
4
(s)
2Ag
+
(aq)
+
CrO
4
2
(aq)
Initial Y 0 0
Change X +2X +X
Equilibrium YX +2X +X
[Ag
+
]
2
[CrO
4
2
] = K
sp
= 2.4 x 10
12
[2X]
2
[X] = 2.4 x 10
12
[X] = 8.44 x 10
5
= solubility of Ag
2
CrO
4
= greatest molar solubility, despite not having highest K
sp
.
Chem 1622016 Chapter 18 Acidbase & solubility equilibria lecture notes
49
SKIP (CONCEPT COVERED ON PREVIOUS PAGE)
K
sp
values vs. Relative Solubility
ET note: Relationship of Ksp value and solubility
Given the K
sp’s
of CaF
2
, Mg(OH)
2
and FeCO
3
(1.46 x 10
10
, 2.06 x 10
13
and 3.07 x 10
11
, respectively),
can the relative solubilities be determined?
CaF
2
(s)
Ca
2+
(aq) + 2F

(aq)
CaF
2
(s)
Ca
2+
(aq)
+
2F

(aq)
Initial Y 0 0
Change X X 2X
Equilibrium YX
+X
+2X
[Ca
2+
][F

]
2
= K
sp
[X][2X]
2
= 1.46 x 10
10
4X
3
= 1.46 x 10
10
X = 3.32 x 10
4
M
Mg(OH)
2
(s)
Mg
2+
(aq) + 2OH

(aq)
Mg(OH)
2
(s)
Mg
2+
(aq)
+
2OH

(aq)
Initial Y 0 0
Change X X 2X
Equilibrium YX
+X
+2X
[Mg
2+
][OH

]
2
= K
sp
[X][2X]
2
= 2.06 x 10
13
4X
3
= 2.06 x 10
13
X = 3.72 x 10
5
M
FeCO
3
(s)
Fe
2+
(aq) + CO
3
2
(aq)
FeCO
3
(s)
Fe
2+
(aq)
+
CO
3
2
(aq)
Initial Y 0 0
Change X +X +X
Equilibrium YX
+X
+X
[Fe
2+
][ CO
3
2
] = K
sp
[X][X] = 3.07 x 10
11
X
2
= 3.07 x 10
11
X = 5.54 x 10
6
M
Compound K
sp
Equil. Expr.
Solubility (M)
CaF
2
1.46 x 10
10
[X][2X]
2
= K
sp
3.32 x 10
4
Mg(OH)
2
2.06 x 10
13
[X][2X]
2
= K
sp
3.72 x 10
5
FeCO
3
3.07 x 10
11
[X][X] = K
sp
5.54 x 10
6
Conclusion: K
sp
values are directly related to solubility
only
if the stoichiometric coefficients in the equation, and correspondingly, the forms
of equilibrium expressions, are the same.
Hence, since CaF
2
has a larger K
sp
than Mg(OH)
2
, and has the same form of equilibrium
expression, then CaF
2
is more soluble than Mg(OH)
2
.
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