20 ml of 0.2 M NaOH is added to 50 ml, of 0.2 M ${\mathrm{CH}}_3\mathrm{COOH}$ to give 70 ml, of the solution. What is the pH of the solution ? The ionization constant of acetic acid is $2 \times {10}^{-5}$

The addition of NaOH converts equivalent amount of acetic acid into sodium acetate. Hence,

Concentration of acetic acid after the addition of

$\mathrm{NaOH} = \frac{30}{70} \times 0.2 \mathrm{M}$

Concentration of ${\mathrm{CH}}_3\mathrm{COONa}$ after the addition of 

$\mathrm{NaOH} = \frac{20}{70} \times 0.2 \mathrm{M}$

Hence, using the expression

$\mathrm{pH}= {\mathrm{pK}}_{\mathrm{a}} + \log \frac{\left[\mathrm{Salt}\right]}{\left[\mathrm{Acid}\right]}$

$$\begin{gathered} =-\log \left(2 \times {10}^{-5}\right) + \log \left(\frac{2}{3}\right) \\ = 4.699 - 0.177 = 4.522 \end{gathered}$$