When $100\mathrm{ml}$ of $\mathrm{N}/10\mathrm{NaOH}$ are added to $50\mathrm{ml}$ of $\mathrm{N}/5\mathrm{HCl}$, the $\mathrm{pH}$ of the resulting solution is

It is given to find the $\mathrm{pH}$ of the resulting solution When $100\mathrm{ml}$ of $\mathrm{N}/10\mathrm{NaOH}$ are added to 50 $\mathrm{ml}$ of $\mathrm{N}/5\mathrm{HCl}$

Equivalents of $\mathrm{NaOH}$= Normality of $\mathrm{NaOH}\times \mathrm{Volume}$ of $\mathrm{NaOH}$

$=\frac{1}{10}\times 100$

$=10\text{ milliequivalents. }$

Equivalents of $\mathrm{HCl}$= Normality of $\mathrm{HCl}\times$Volume of $\mathrm{HCl}$

$=\frac{1}{5}\times 50$

$=10\text{ milliquivalents. }$

Here, equivalents of $\mathrm{NaOH}$= equivalents of $\mathrm{HCl}$

Therefore, the resulting solution will be neutral.

$\therefore \mathrm{pH}$ of resulting solution is 7 .

Hence the correct answer (A) 7