Equal volumes of two solutions with $\mathrm{pH}$=3 and $\mathrm{pH}$=11 are mixed. Then the $\mathrm{pH}$ of resulting solution is

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

$\mathrm{pH}$ of $1^{\text{st }}solution=3$

$\left[{\mathrm{H}}^{+}\right]={10}^{-3}$

pH of $2^{\text{nd }}$ solution =11

$\mathrm{pOH}$ is (14-11)=03

$\left[{\mathrm{OH}}^{-}\right]={10}^{-3}$

now, they are mixed

$\left[{\mathrm{H}}^{+}\right]={10}^{-3}\text{ and }\left[{\mathrm{OH}}^{-}\right]={10}^{-3}$

equal concentrations mix together and neutralise each other.

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

Hence the correct answer (B) 7