Question

113. 0.02 molar solution of acetic acid is 3 % ionised. Its ionisation constant is a) 0.9 x 10-3 b) 1.8 x 10-5c) 18 x 10-5 d) 1.8 x 10-4​

Answer 1

Given:

concentration of acetic acid = 0.02M

Solution of acetic acid is 3% ionized

To find:

Ionisation constant

Solution:

The required chemical reaction is-

      CH₃COOH  +  H₂O →  H₃O⁺  +  CH₃COO⁻

at t=0,    C                           0              0

at t=t'     C-Cα                      Cα            Cα

as we known,

               $K_a = \frac{[H_3O^+] [ CH_3COO^-]}{[CH_3COOH]}$

                        $K_a = \frac{C\alpha C\alpha }{C-C\alpha }$

                        $K_a = \frac{C^2\alpha^2}{C(1- \alpha)}$

                        $K_a = \frac{c \alpha ^2}{(1-\alpha )}$

here,  $K_a$ = Ionisation constant

          C = concentration

         α =  ionized

It is given, C = 0.02M

and α = 3% = 0.03

so, putting all the values, we get

   $K_a = \frac{(0.02) \times (0.03)^2}{(1-0.03)}$

   $K_a = \frac{0.000018}{0.97}$

   $K_a = 1.86 \times 10^-^5$

Therefore, Ionisation constant of a solution of acetic acid is found to be 1.86 × 10⁻⁵.