Calculate $pH$ of a resultant solution of $25mL$ of $0.1MHCl$, $50mL$ of $0.01MHN{O}_3$ and $25mL$ of $0.1MNaOH$.

The correct option is D $2.3$
$25mL$ of $0.1MHCl$ will completely neutralize $25mL$ of $0.1MNaOH$.

The total volume of the solution will be $25mL+50mL+25mL=100mL$.

Thus, $50mL$ of $0.01MHN{O}_3$ is diluted to $100mL$. This means its concentration will be reduced to one half.

Hence, the hydrogen ion concentration will b $\frac{0.01}{2}=0.005M$ .

The expression for the pH of the solution will be $pH=-log\left[H^{+}\right]=-log0.005=2.3$.

Hence, the $pH$ of the resulting solution is $2.3$