Question

The pH of a $0.1$ solution of an organic acid is $4.0$. Calculate the dissociation constant of the acid.

Answer 1

$HA\rightleftharpoons H^{+}+A^{-}$

$0.1$

$0.1(1-\alpha )$ $0.1\alpha$ $0.1\alpha$

$K_a=\frac{\left(0.1\alpha \right)\left(0.1\alpha \right)}{0.1(1-\alpha )}$

It is given that $pH=4$ so $-log\left(\right[H^{+}\left]\right)=4$

$\left[H^{+}\right]={10}^{-4}=0.1\alpha$

$\alpha ={10}^{-3}$

$K_a=\frac{\left(0.1\alpha \right)\left(0.1\alpha \right)}{0.1(1-\alpha )}$

As $\alpha <<1$ so $1-\alpha =1$

$K_a=0.1{\alpha }^2={10}^{-7}$