20ml of 0.1M ${\mathrm{H}}_2{\mathrm{SO}}_4$solution is added to 30ml of 0.2M ${\mathrm{NH}}_4\mathrm{OH}$ solution. The pH of resultant mixture is ( $p^{K_b}$ of ${\mathrm{NH}}_4\mathrm{OH}$= 4.7 )

                                                    ${\mathrm{H}}_2{\mathrm{SO}}_4 + 2{\mathrm{NH}}_4\mathrm{OH} \to {\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4+2{\mathrm{H}}_2\mathrm{O}$

                 Given :                20ml of 0.2N +  30ml of 0.2N

no.of milliequivalents =            4                        6

In an equation the  no.of milliequivalents should be same

i.e    no.of milliequivalents of  acid =no.of milliequivalents of base = no.of milliequivalents of salt formed (VN)=4

no.of milliequivalents of base remains = 6-4 =2

Here , ${\mathrm{V}}_{\mathrm{A}}{\mathrm{N}}_{\mathrm{A}} < {\mathrm{V}}_{\mathrm{B}}{\mathrm{N}}_{\mathrm{B}}$

Hence   ${\mathrm{P}}^{\mathrm{OH}}={\mathrm{P}}^{{\mathrm{k}}_{\mathrm{b}}}+\log \frac{\left[\mathrm{salt}\right]}{\left[\mathrm{base}\right]}$ 

                    $=4.7+\log \frac{4}{2}=4.7+0.3=5$   

  ${\mathrm{P}}^{\mathrm{H}} = 14-5 =9$