Question

Determine the pH of a 0.50 M solution of NH₄OCl.

Short answer

Answer

pH of the given solution comes out to be 8.4.

Step-by-step solution

 Calculation of Ka for Hydrolysis

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

Reaction involved:

$$\begin{gathered} {\mathrm{NH}}_{4\left(\mathrm{aq}\right)}^{+}\rightleftarrows {\mathrm{NH}}_{3\left(\mathrm{aq}\right)}+{\mathrm{H}}_{\left(\mathrm{aq}\right)}^{+} \\ {\mathrm{K}}_{\mathrm{a}}=5.6\times {10}^{-10} \\ {{\mathrm{OCI}}^{-}}_{(\mathrm{aq}}+{\mathrm{H}}_2\mathrm{O}\iff {\mathrm{HOCI}}_{\left(\mathrm{aq}\right)}+{{\mathrm{OH}}^{-}}_{\left(\mathrm{aq}\right)} \\ {\mathrm{K}}_{\mathrm{w}}=1\times {10}^{-14} \end{gathered}$$

We can here consider the overall reaction as :

$$\begin{gathered} {\mathrm{NH}}_{4\left(\mathrm{aq}\right)}^{+}+{\mathrm{OCI}}_{\left(\mathrm{aq}\right)}^{-}\leftrightarrows {\mathrm{NH}}_{3\left(\mathrm{aq}\right)}+{\mathrm{HOCI}}_{\left(\mathrm{aq}\right)} \\ \mathrm{K}=\frac{\left[{\mathrm{NH}}_{3\left(\mathrm{aq}\right)}\right]\left[{\mathrm{HOCI}}_{\left(\mathrm{aq}\right)}\right]}{\left[{\mathrm{NH}}_{4\left(\mathrm{aq}\right)}^{+}\right]\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{OCI}}_{\left(\mathrm{aq}\right)}^{-}\right]} \\ \mathrm{K}={\mathrm{K}}_{\mathrm{a}}{\mathrm{NH}}_{3\left(\mathrm{aq}\right)}\times \frac{1}{{\mathrm{K}}_{\mathrm{a}}{\left(\mathrm{HOCI}\right)}_{\left(\mathrm{aq}\right)}} \\ \mathrm{K}=\frac{5.6\times {10}^{-10}}{3.5\times {10}^{-8}} \\ \mathrm{K}=1.6\times {10}^{-2} \end{gathered}$$

Calculation of pH  for Given 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