20 mL of a weak monobasic acid (HA) requires 20 mL 0.2 M NaOH for complete titration. If pH of solution upon addition of 10 mL of this alkali to 25 mL of the above solution of HA is 5.8. The ${\mathrm{pK}}_{\mathrm{a}}$ of the weak acid is

meq. of acid = meq of base $\Rightarrow 20 \times M = 20 \times 0.2 = 4$

Molarity of HA = 0.2 M

When 25ml,0.2M acid reacts with 10ml 0.2M NaOH

$$\begin{gathered} \mathrm{HA} + {\mathrm{OH}}^{-} \stackrel{ }{\to } {\mathrm{A}}^{-} + {\mathrm{H}}_2\mathrm{O} \\ 5 (=25\times .2) 2 (10\times .2) 0 \\ 3(=5-2) - 2 \mathrm{mmole} \\ \therefore \mathrm{pH} = {\mathrm{pK}}_{\mathrm{a}} + \log \frac{\left[{\mathrm{A}}^{-}\right]}{\left[\mathrm{HA}\right]} \\ \Rightarrow 5.8 = {\mathrm{pK}}_{\mathrm{a}} + \log \left(\frac{2}{3}\right) \\ \Rightarrow {\mathrm{pK}}_{\mathrm{a}} = 5.98 \end{gathered}$$