Question

the ionisation constant of acetic acid is 1.74 × 10-⁵. calculate the degree of dissociation of acetic acid in its 0.05M solution. calculate the concentration of acetate ion in the solution and it's pH​

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.

Explanation: