To a $200 mL$ of $0.1 M$ 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?

$\left[\left(K_{\alpha }\left(\mathrm{HA}\right)={10}^{-5}\right)\right]$

$\mathrm{HA}+ {\mathrm{OH}}^{-}⟶ {\mathrm{A}}^{-}+ {\mathrm{H}}_2\mathrm{O}$

 20                 9                   -

 11                 -                   9

 11             $0.1\times V$         9

$11-0.1\times V$ -             $9+0.1\times V$

$\mathrm{pH}=p{K}_a=5$

$\left[{\mathrm{A}}^{-}\right]=\left[\mathrm{HA}\right]\text{ or }V=10\mathrm{ml}$

Hence, option C is correct.