Question

Determine the pH of a 0.50 M solution of NH₄OCl.

Short answer

pH of the given solution comes out to be 8.4.

Step-by-step solution

 Calculation of Kafor Hydrolysis

Species present are NH₄⁺, OCl^- and H₂O.

Reaction involved:

$\begin{array}{l}{\text{NH}}_{\text{4}}{{}^{\text{+}}}_{\text{(aq)}}\rightleftarrows {{\text{\hspace{0.17em}NH}}_{\text{3}}}_{\text{(aq)}}{{\text{+\hspace{0.17em}H}}^{\text{+}}}_{\text{(aq)}}\\ {\text{\hspace{1em}K}}_{\text{a}}{\text{=5.6×10}}^{\text{-10}}\end{array}$

$\begin{array}{l}{{\text{OCl}}^{\text{-}}}_{\text{(aq)}}{\text{+H}}_{\text{2}}\text{O}\rightleftarrows {\text{\hspace{0.17em}HOCl}}_{\text{(aq)}}{{\text{+\hspace{0.17em}OH}}^{\text{-}}}_{\text{(aq)}}\\ {\text{\hspace{1em}K}}_{\text{b}}{\text{=2.86×10}}^{\text{-7}}\end{array}$

$\begin{array}{l}{\text{H}}_{\text{2}}{\text{O+H}}_{\text{2}}\text{O}\rightleftarrows {\text{\hspace{0.17em}H}}_{\text{3}}{{\text{O}}^{\text{+}}}_{\text{(aq)}}{{\text{+\hspace{0.17em}OH}}^{\text{-}}}_{\text{(aq)}}\\ {\text{\hspace{1em}K}}_{\text{w}}{\text{=1×10}}^{\text{-14}}\end{array}$

We can here consider the overall reaction as :

$\begin{array}{l}{\text{NH}}_{\text{4}}{{}^{\text{+}}}_{\text{(aq)}}{{\text{+OCl}}^{\text{-}}}_{\text{(aq)}}\rightleftarrows {{\text{\hspace{0.17em}NH}}_{\text{3}}}_{\text{(aq)}}{\text{+\hspace{0.17em}HOCl}}_{\text{(aq)}}\\ \text{K}=\frac{\left[{{\text{NH}}_3}_{\text{(aq)}}\right]\left[{\text{HOCl}}_{\text{(aq)}}\right]}{\left[{\text{NH}}_{\text{4}}{{}^{\text{+}}}_{\text{(aq)}}{{\text{][OCl}}^{\text{-}}}_{\text{(aq)}}\right]}\\ \text{K}=\frac{\left[H^{+}\right]\left[{\text{NH}}_3\right]}{[{\text{NH}}_{\text{4}}{{}^{\text{+}}}_{\text{(aq)}}\text{]}}\times \frac{\left[{\text{HOCl}}_{\text{(aq)}}\right]}{\left[H^{+}\right]{{\text{[OCl}}^{\text{-}}}_{\text{(aq)}}]}\end{array}$

$\begin{array}{l}\text{K}=K_a{{\text{(NH}}_3}_{\text{(aq)}})\times \frac{1}{K_a{\text{(HOCl}}_{\text{(aq)}})}\\ \text{K}=\frac{{\text{5.6×10}}^{\text{-10}}}{{\text{3.5×10}}^{\text{-8}}}\\ \text{K}=1.6{\text{×10}}^{\text{-2}}\end{array}$

Calculation of pH  forGiven Salt

Initial Concentration NH₄⁺ = 0.50, At equilibrium Concentration NH₄⁺ = 0.50-x

Initial Concentration OCl^- = 0.50, At equilibrium Concentration OCl^- = 0.50-x

Initial Concentration NH₃= 0, At equilibrium Concentration NH₃= x

Initial Concentration HOCl = 0, At equilibrium Concentration HOCl = x

$\begin{array}{l}{\text{K}}_{\text{a}}\text{=}\frac{{\text{[x]}}^{\text{2}}}{{\text{[0.5-x]}}^{\text{2}}}\\ {\text{K}}_{\text{a}}{\text{=1.6×10}}^{\text{-2}}\\ \frac{\text{x}}{\text{0.5-x}}\text{=0.126}\\ {\text{x=5.6×10}}^{\text{-2}}\end{array}$

${\text{[OCl}}^{\text{-}}\text{]}=0.5-\text{x}=0.444$

$\text{[HOCl]}=\text{x}=5.6\times {10}^{-2}$

$\begin{array}{l}{\text{K}}_{\text{a}}=\frac{{\text{[H}}^{\text{+}}{\text{][OCl}}^{\text{-}}\text{]}}{\text{[HOCl]}}=\frac{{\text{[H}}^{\text{+}}\text{](}0.444)}{5.6\times {10}^{-2}}=3.5\times {10}^{-8}\\ {\text{[H}}^{\text{+}}\text{]}=\frac{3.5\times {10}^{-8}\times 5.6\times {10}^{-2}}{0.444}=4.4\times {10}^{-9}\end{array}$

${\text{pH=\hspace{0.17em}-log[H}}^{\text{+}}{\text{]=\hspace{0.17em}-log(4.4×10}}^{\text{-9}}\text{)=\hspace{0.17em}8.4}$