To a beaker containing 1L of 10^–3M CH₃COOH, 40 mg NaOH is added.  The change in pH is recorded is  [p^ka of CH₃COOH = 4.8]

P^H of 1L of 10^-3 M CH₃COOH is given as ${\mathrm{P}}^{\mathrm{H}}=\frac{1}{2}\left[{\mathrm{P}}^{{\mathrm{K}}_{\mathrm{a}}}-\mathrm{logC}\right]=3.9$

On addition of 40 mg NaOH, following reaction occurs.

Resultant solution is a salt of strong base and weak acid. P^H of such a solution is given as

${\mathrm{P}}^{\mathrm{H}}=\frac{1}{2}\left[{\mathrm{P}}^{{\mathrm{K}}_{\mathrm{w}}}+{\mathrm{P}}^{{\mathrm{K}}_{\mathrm{a}}}+\mathrm{logC}\right]$

${\mathrm{P}}^{\mathrm{H}}=\frac{1}{2}[14+4.8-3]$= 7.9

Change in P^H = 7.9 - 3.9 = 4 units