Chapter 10 Solutions
constant expression as their molar concentrati
ons raised to an exponent equal to their
coefficient in the balanced equation, while
solids enter the expression as 1. Substances on
the right (products) are in the numerator, and
substances on the left (reactants) are in the
denominator. Thus, the equilibrium consta
nt expression for the dissolution of CaCO
is 3
2-
2+
2-
-9
3
sp
3
[Ca ][CO ]
K
= [Ca ][CO ] = 4.8 10
1
2 +
=×
Eq. 10.3
The equilibrium constant, which involves only
a product of concentrations (there is no
denominator), is called the
solubility product constant
and given the symbol
Ksp
†. A
small value for K
means that the product of the concentrations of the ions present at sp
equilibrium is very small, which indicates that
the substance is not very soluble. The K
of sp
CaCO
is 4.8x10 3
-9 at 298K, K
<< 1, so it is an insoluble compound. Table 10.5 shows the sp
solubility product constants of a few other inso
luble salts. In a saturated solution of AgCl
(K
= 1.8x10sp
-10
), [Ag
1+] ~ 10
-5 M, while in a saturated solution of Ag
S (K 2
= 8x10sp
-48
),
[Ag
1+] ~ 10
-17
M. Thus, silver chloride may be fairly insoluble, but Ag
S is even more 2
insoluble.
† Do not think that this equilibrium constant is somehow different
because it has its own name. The ‘K’ in K
signifies an sp
equilibrium constant, and the ‘sp’ subscript simply indicates that the equilibrium process is the dissolution of an ionic solid. Table 10.5
Some solubility products (K
) in water at 25 sp
oC
Compound K
Compound sp
Ksp
AgCl 1.8x10
-10
Ag
S 6.3x10 2
-50
Ag
CrO 2
1.1x10 4
-12
CaSO
2.4x10 4
-5^
AgI 8.3x10
-17
BaSO
1.1x10 4
-10
Answers to student exercises in Example 10.12
d) Pb
2+ + CrO
2-^4
→
PbCrO
4
e) Ba
2+ + SO
2- 4
→
BaSO
4
f) 3Co
2+ + 2PO
3-^4
→
Co
(PO 3
) 42
g) No reaction
Example 10.13 The K
of zinc carbonate is 1.0x10sp
-10
. Write the process and the expression that
correspond to that value.
The process is the dissolution of ZnCO
in water. In this process, the ionic bonds between 3
the Zn
2+ ions and the CO
2- 3
ions are broken, but the covalent C-O bonds within the
carbonate ion remain intact. The chemical process and K
expression are sp
]
2+
2-
2+
2-
-10
33
sp3
ZnCO (s)
Zn + CO
K = [Zn ][CO
= 1.0 10
×
U
Example 10.14
The K
of iron(III) hydroxide is 1.6x10sp
-39. Write the process and the expression that
correspond to that value. Three OH
1- ions are required for each Fe
3+ ion, so the chemical formula is Fe(OH)
. In 3
addition, three hydroxide ions must be produced in the dissolution, so the hydroxide ion concentration must be cubed in the K
expression. sp
]
3+
1-
3+
1- 3
-39
3s
p
Fe(OH) (s)
Fe + 3OH K = [Fe ][OH
= 1.6 10
×
U
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