Question

A certain weak acid has $K_a=1.0\times {10}^{-4}$. Calculate the equilibrium constant for its reaction with a strong base.

Answer 1

Equilibrium reaction is

$HA+BOH\rightleftharpoons BA+H_{2}O$

weak strong

$HA+B^{+}+{OH}^{-}\rightleftharpoons B^{+}+A^{-}+H_{2}O$

or $HA+{OH}^{-}\rightleftharpoons A^{-}+H_{2}O$

$K=\frac{\left[A^{-}\right]}{\left[HA\right]\left[{OH}^{-}\right]}$

In the above reaction

$K_h=\frac{\left[HA\right]\left[{OH}^{-}\right]}{\left[A^{-}\right]}$

so, $K=\frac{1}{K_h}$

We know that

$K_h=\frac{K_w}{K_a}=\frac{{10}^{-14}}{1.0\times {10}^{-4}}={10}^{-10}$