Question

$\mathrm{pH}$ of an aqueous solution of$0.1\mathrm{M}$ ${\mathrm{NH}}_4\mathrm{Cl}$ solution is : $({\mathrm{pK}}_{\mathrm{b}}{\mathrm{NH}}_4\mathrm{OH}=4.7)$.

Answer 1

Step 1: Analyzing the given conditions

${\mathrm{NH}}_4\mathrm{Cl}$(Ammonium chloride) is an acidic salt of strong acid(Hydrochloric acid) and weak base(Ammonium hydroxide).

Concentration of ${\mathrm{NH}}_4\mathrm{Cl}$ = $0.1\mathrm{M}$

${\mathrm{pK}}_{\mathrm{b}}=4.74$

Step 2: Calculating the pH of the ${\mathrm{NH}}_4\mathrm{Cl}$ solution

The formula for the calculation of pH of salt of strong acid and weak base can be represented as:

$\mathrm{pH}=-\frac{1}{2}\log {\mathrm{K}}_{\mathrm{w}}-\frac{1}{2}\log \mathrm{C}+\frac{1}{2}\log {\mathrm{K}}_{\mathrm{b}}$

$\mathrm{pH}=-\frac{1}{2}\log {10}^{(-14)}-\frac{1}{2}\log (0.1)+\frac{1}{2}\log (4.7)$

$$\begin{gathered} \mathrm{pH}=7+0.5-0.3360 \\ \mathrm{pH}=7.163 \end{gathered}$$

Therefore, $\mathrm{pH}$ of an aqueous solution of $0.1\mathrm{M}$ ${\mathrm{NH}}_4\mathrm{Cl}$ solution is $7.163$