To a 200 ml of 0.1 M weak acid HA solution 90 ml of 0.1 M solution of NaOH be added. Now, what volume of 0.1 M NaOH be added into above solution so that pH of resulting solution be 5.      $\left[{\mathrm{K}}_{\mathrm{a}} \left(\mathrm{HA}\right) = {10}^{-5}\right]$

$$\begin{gathered} \mathrm{HA} + \mathrm{NaOH} \stackrel{ }{\to } \mathrm{NaA} + {\mathrm{H}}_2\mathrm{O} \\ \mathrm{t} = 0 20 9 0 0 \\ \mathrm{t} = \mathrm{t} 11 0 9 0 \\ To \mathrm{have} {\mathrm{p}}^{\mathrm{H}}=5={\mathrm{p}}^{{\mathrm{k}}_{\mathrm{a}}},\frac{\mathrm{salt}}{\mathrm{acid} }=1.\mathrm{This} \mathrm{is} \mathrm{possible} \left[\mathrm{acid}\right]=11-1=10 \mathrm{and} \\ \left[\mathrm{salt}\right]=9+1=10.\mathrm{Hence} 1 \mathrm{millimole} \mathrm{of} \mathrm{NaOH} \mathrm{must} \mathrm{be} \mathrm{added}. \\ \mathrm{Volume} \mathrm{of} \mathrm{NaOH} \mathrm{added}=\frac{\mathrm{number} \mathrm{of} \mathrm{millimoles}}{\mathrm{Molarity}}=\frac{1}{0.1}=10\mathrm{ml} \end{gathered}$$