pH of a solution made by mixing 50 mL of 0.2 M NH₄Cl and 75 mL of 0.1 M NaOH is [pK_b of NH₃(aq) = 4.74. log 3 = 0.47]

$$\begin{gathered} \stackrel{+}{\mathrm{N}}{\mathrm{H}}_4 + \mathrm{NaOH} ⟶ {\mathrm{NH}}_4\mathrm{OH}+\mathrm{NaCl} \\ \mathrm{mmol} 50 \times 0.2 75 \times 0.1 \\ =10 =75 \\ \mathrm{mmol} 10-7.5 - 7.5 \\ =2.5 \end{gathered}$$

$\Rightarrow$This will result in a basic buffer.

$\begin{array}{l}\Rightarrow \mathrm{pOH}={\mathrm{pK}}_{\mathrm{b}}+\log \frac{\left[\text{ Salt }\right]}{\left[\text{ Base }\right]}\\ =4.74+\log \frac{2.5}{7.5}=4.27\\ \Rightarrow \mathrm{pH}=14-4.27=9.73\end{array}$