For a reaction,

${\mathrm{CH}}_3\mathrm{COOH}\left(\mathrm{aq}\right)\rightleftharpoons {\mathrm{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{CH}}_3{\mathrm{COO}}^{-}\left(\mathrm{aq}\right)$

or $\mathrm{HAc}\left(\mathrm{aq}\right)\rightleftharpoons {\mathrm{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{Ac}}^{-}\left(\mathrm{aq}\right)$

Evaluate the pH of the solution resulting on addition of 0.05 M acetate ion to 0.05 M acetic acid solution $\left({\mathrm{K}}_{\mathrm{a}}=1.8\times {10}^{-5}\right)$

 

                        $\mathrm{HAc}\left(\mathrm{aq}\right)\rightleftharpoons {\mathrm{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{Ac}}^{-}\left(\mathrm{aq}\right)$

$\begin{array}{llll}\text{ Initial conc. }& \left(\mathrm{M}\right)0.05& 0 & 0.05\end{array}$

$\begin{array}{lccc}\text{ Equil. }& 0.05-\mathrm{x} & \mathrm{x}& 0.05+\mathrm{x}\end{array}$

conc. (M)

${\mathrm{K}}_{\mathrm{a}}=\frac{\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{Ac}}^{-}\right]}{\left[\mathrm{HAc}\right]}=\frac{\left\{\right(\mathrm{x}\left)\right(0.05+\mathrm{x}\left)\right\}}{(0.05-\mathrm{x})}$

As K, is small for a very weak acid, .x << 0.05

Hence, (0.05 + x) $\approx$ (0.05 - x) $\approx$ 0.05

Thus, $1.8\times {10}^{-5}=\frac{\left(\mathrm{x}\right)(0.05+\mathrm{x})}{(0.05-\mathrm{x})}$

$=\frac{\mathrm{x}(0.05)}{(0.05)}=\mathrm{x}=\left[{\mathrm{H}}^{+}\right]$

$=1.8\times {10}^{-5}\mathrm{M}$

$\mathrm{pH}=-\log \left(1.8\times {10}^{-5}\right)=4.74$