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, [tex]pK_a = -log\ K_a[/tex]
Taking antilog both side, we get :
[tex]K_a = 1.78\times 10^{-5}[/tex]
Since, acetic acid has 1 hydrogen atom to loose , so it is a monoprotic acid.
Now, percentage dissociation of monoprotic acid is given by :
[tex]\alpha = \sqrt{\dfrac{K_a}{C}}\\\\\alpha =\sqrt{ \dfrac{1.78\times 10^{-5}}{0.1}}\\\\\alpha = 0.0133\times 100 = 1.33 \%[/tex]
Hence, this is the required solution.
