Question

The solution of weak monoprotic acid which is $0.10M$ has $pH=3$. Calculate $K_a$ of weak acid.

Answer 1

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

$ka=\frac{\left[H^{+}\right]\left[A^{-}\right]}{\left[HA\right]}$

For weak acids, $\left[H^{+}\right]=\left[A^{-}\right]$

$\therefore Ka=\frac{{\left[H^{+}\right]}^2}{\left[HA\right]}\approx \frac{{\left[{10}^{{-P}^H}\right]}^2}{0.1}\approx \frac{{\left({10}^{-3}\right)}^2}{\left({10}^{-1}\right)}\approx {10}^{-5}$