A solution of weak acid $HA$ was titrated with base $NaOH$. The equivalent point was reached when $40mL$ of $0.1MNaOH$ has been added. Now $20mL$ of $0.1MHCl$ were added to titrated solution, the $pH$ was found to be $5.0$. What will be the $pH$ of the solution obtained by mixing $20mL$ of $0.2MNaOH$ and $20mL$ of $0.2MHA$?

The correct option is B $9$
$HA+NaOH\to NaA+H_{2}O$

The number of millimoles of salt $NaA$ is $40\times 0.1=4$
$A^{-}+H^{+}\to HA$
Initial millimoles 4 2
Final millimoles 2 2

For this acidic buffer solution, $\left[A^{-}\right]=\left[HA\right]$

$pH=p{K}_a+log\frac{\left[A^{-}\right]}{\left[HA\right]}$

Hence, $p{K}_a=5$

Now,

$HA+NaOH\to NaA+H_{2}O$

For the hydrolysis of $A^{-}$

$\left[Na{A}^{-}\right]=\frac{\text{millimoles of acid}}{\text{total volume}}=\frac{20\times 0.2}{20+20}=0.1$

$pH=\frac{1}{2}(p{K}_w+p{K}_a+logC)=\frac{1}{2}(14+5-1)=9$.