Question

Calculate the % dissociation of 0.10 M solution of acetic acid (pKa = 4.75)

Answer 1

Given :

Molarity of acetic acid solution, M = 0.10 M.

pKa of acetic acid, pKa = 4.75 .

To Find :

Percentage dissociation of 0.10 M solution of acetic acid.

Solution :

We know, $pK_a = -log\ K_a$

Taking antilog both side, we get :

$K_a = 1.78\times 10^{-5}$

Since, acetic acid has 1 hydrogen atom to loose , so it is a monoprotic acid.

Now, percentage dissociation of monoprotic acid is given by :

$\alpha = \sqrt{\dfrac{K_a}{C}}\\\\\alpha =\sqrt{ \dfrac{1.78\times 10^{-5}}{0.1}}\\\\\alpha = 0.0133\times 100 = 1.33 \%$

Hence, this is the required solution.