Calculate the $pH$ of the solution in which $0.2 \, M \, NH_{4}Cl$ and $0.1 \, M \, NH_{3}$ are present. The $pK_{b}$ of ammonia solution is $4.75$.

(N/A) The given solution is a basic buffer containing a weak base $(NH_{3})$ and its salt with a strong acid $(NH_{4}Cl)$.
Using the Henderson-Hasselbalch equation for basic buffers:
$pOH = pK_{b} + \log \left( \frac{[Salt]}{[Base]} \right)$
Given:
$pK_{b} = 4.75$
$[Salt] = [NH_{4}Cl] = 0.2 \, M$
$[Base] = [NH_{3}] = 0.1 \, M$
Substituting the values:
$pOH = 4.75 + \log \left( \frac{0.2}{0.1} \right)$
$pOH = 4.75 + \log(2)$
$pOH = 4.75 + 0.301 = 5.051$
Since $pH + pOH = 14$ at $25^{\circ}C$:
$pH = 14 - 5.051 = 8.949 \approx 8.95$