Equal volumes of $0.1\mathrm{M}$ potassium hydroxide and $0.1\mathrm{M}$ sulphuric acid are mixed. The $\mathrm{pH}$ of resulting solution is

It is given to find the $\mathrm{pH}$ of the resulting solution.

$0.1\times 2=0.2 \mathrm{N}$

Since the above case normality of sulphuric acid is more than the normality of the potassium hydroxide then the normality of the resulting solution after mixing is given as:

$\frac{N_1{V}_1-N_2{V}_2}{V_1+V_2}=N_3$

$\Rightarrow \frac{(0.2-0.1)\times V}{2V}=N_3$

$\Rightarrow \frac{0.1\times V}{2V}=N_3$

$\Rightarrow 0.05=N_3$

Since the normality of the resulting solution is $0.05 \mathrm{N}$ thus it behaves as an acid so the $\mathrm{pH}$ value of the resulting solution will be less than 7 .

Hence the correct answer (D) <7