What is the $pH$ of a solution in which $25\text{mL}$ of $0.1\text{M}NaOH$ is added to $25\text{mL}$ of $0.08\text{M}HCl$ and final solution is diluted to $500\text{mL}?$

The correct option is B $11$
Milli-equivalents of $HCl=25\times 0.08=2$
Milli-equivalents of $NaOH=25\times 0.1=2.5$
As milli-equivalents of $NaOH$ is more than milli-equivalents of $HCl$, then the remaining concentration of $\left[O{H}^{-}\right]=\frac{2.5-2}{500}={10}^{-3}$
$\left[H^{+}\right]={10}^{-11}$
$pH=-\log \left({10}^{-11}\right)=11$