Question

What is the equilibrium constant for the hydrolysis of acetate ion at 250C? Ka = 1.74x10-5

Answer 1

Answer:

The dissociation equilibrium is CH

3

COOH⇌CH

3

COO

−

+H

+

.

Let α be the degree of dissociation.

The equilibrium concentrations of CH

3

COOH,CH

3

COO

−

and H

+

are c(1−α),c(α) and c(α) respectively.

The equilibrium constant expression is K

c

=

[CH

3

COOH]

[CH

3

COO

−

][H

+

]

.

K

c

=

c(1−α)

(cα)(cα)

≈cα

2

α=

c

K

a

=

0.05

1.74×10

−5

=1.865×10

−2

[CH

3

CO

−

]=[H

+

]=cα=0.05×1.865×10

−2

=9.33×10

−4

M

pH=−log[H

+

]=−log(9.33×10

−4

)=3.03

The concentration of acetate ion and its pH are 9.33×10

−4

and 3.03 respectively.