20 mL of 0.1 M ${\text{H}}_2{\text{SO}}_4$  solution is added to 30 mL of 0.2 M  ${\text{NH}}_4\text{O}H$ solution. The pH of the resultant mixture is ${\text{[pK}}_b{\text{of\hspace{0.17em} NH}}_4\text{OH }=\text{ 4}.\text{7}]$ 

$N{H}_{4}OH+H_{2}S{O}_4\to N{H}_4^{(+)}+H_{2}O$

Intial            6          4       0      0
At Eq           2          0       4      4

${n_H}^{(+)}=0.1\times 20\times 2=0.4=4mm$
$n_{N{H}_{4}OH}OH=30\times 0.2=0.6=6mm$

Mm= millimoles

$P^{OH}=Pkb+\log \frac{\left(salt\right)}{\left(Base\right)}\frac{\left(N{H}_4^{+}\right)}{\left(N{H}_{4}OH\right)}$

$P^{OH}=4.7+\log \frac{4}{2}({\log }_2=0.30)$

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