pH of a mixture containing 10 ml of 0.1M ${\mathrm{NH}}_4\mathrm{Cl}$  and 100 ml 0.1 M  ${\mathrm{NH}}_4\mathrm{OH}$  is (pKb of ${\mathrm{NH}}_4\mathrm{OH}=4.75)$

Millimole of  ${\mathrm{NH}}_4\mathrm{Cl}=10\times 0.1=1$

Millimole of  ${\mathrm{NH}}_4\mathrm{OH}=100\times 0.1=10$

$pOH=p{K}_b+\log \frac{\left[{\mathrm{NH}}_4\mathrm{Cl}\right]}{\left[{\mathrm{NH}}_4\mathrm{OH}\right]}$

$pOH=4.75+\log \frac{1}{10}$

$pOH=3.75$

$pH=14-pOH=14-3.75=10.25$