Chemistry - A Molecular Science

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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