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* 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{(K_a)/(C)}\\\\\alpha =\sqrt{ (1.78* 10^(-5))/(0.1)}\\\\\alpha = 0.0133* 100 = 1.33 \%`

Hence, this is the required solution.