4 mole of  ${\mathrm{NH}}_4\mathrm{OH}$ and 1 mole of  ${\mathrm{H}}_2{\mathrm{SO}}_4$ are mixed and diluted to 1 lit solution. The ${\mathrm{pK}}_{\mathrm{b}}$ of  ${\mathrm{NH}}_4\mathrm{OH}$ is 4.8. The $\mathrm{pH}$of solution is

Ammonium sulphate is created when 1 mole of $H_{2}S{O}_4$ reacts with 2 moles of ammonium hydroxide. Ammonium hydroxide, 2 moles, is still unreacted.

As a result, the resultant solution is a simple buffer solution.

$\left[\mathrm{salt}\right]=2 \mathrm{M} \mathrm{and} \mathrm{base} = 2 \mathrm{M}$

The expression for the $\mathrm{pOH}$ of the basic buffer solution is as given below.

$\mathrm{pOH}={\mathrm{pK}}_{\mathrm{b}}+\log \frac{\left[\text{ salt }\right]}{\left[\text{ base }\right]}$

But ${\mathrm{pK}}_{\mathrm{b}}=4.8$

Substitute values in the above equation.

$\mathrm{pOH}=4.8+\log \left(\frac{2}{2}\right)=4.8$

$\mathrm{pH}=14-\mathrm{pOH}=14-4.8=9.2.$

Therefore, the correct option is (B).