NH₄OH and NH₄Cl constitute to form a basic buffer.

pOH = pK_b + log ${{\left[ {Salt} \right]} \over {\left[ {Base} \right]}}$

We know, pOH+ pH = 14 or pOH = 14 – pH

$\therefore$ 14 - pH - $\log {{\left[ {Salt} \right]} \over {\left[ {Base} \right]}}$ = pK_b

$\Rightarrow$ 14 - 9.25 - $\log {{0.1} \over {0.1}}$ = pK_b

$\Rightarrow$ 14 – 9.25 – 0 = pK_b

$\Rightarrow$ pK_b = 4.75