Question

Calculate the pH of the following solutions:

$\begin{array}{l}\mathbf{a}.\mathbf{1}.\mathbf{2}\mathbf{M}{\mathbf{CaBr}}_2\\ \mathbf{b}.\mathbf{0}.\mathbf{84}\mathbf{M}{\mathbf{C}}_6{\mathbf{H}}_5{\mathbf{NH}}_3{\mathbf{NO}}_3({\mathbf{K}}_b\mathbf{for}{\mathbf{C}}_6{\mathbf{H}}_5{\mathbf{NH}}_2=\mathbf{3}.\mathbf{8}\times {\mathbf{10}}^{-\mathbf{10}})\\ \mathbf{c}.\mathbf{0}.\mathbf{57}\mathbf{M}{\mathbf{KC}}_7{\mathbf{H}}_5{\mathbf{O}}_2({\mathbf{K}}_a\mathbf{for}{\mathbf{HC}}_7{\mathbf{H}}_5{\mathbf{O}}_2=\mathbf{6}.\mathbf{4}\times {\mathbf{10}}^{-\mathbf{5}})\end{array}$

Short answer

1. pH of 1.2 M CaBr₂ =7
2. pH of 0.84 M C₆H₅NH₃NO₃ =2.33
3. pH of 0.57 M KC₇H₅O₂ =9

Step-by-step solution

pH of 1.2 M CaBr2

As CaBr₂ is a salt of a strong acid (HBr) and strong base Ca(OH)₂, the pH of the solution will be 7; thus, the solution will be neutral.

pH of 0.84 M C6H5NH3NO3

On dissociation, C₆H₅NH₃NO₃gives C₆H₅NH₃⁺and NO₃^-; out of these two, C₆H₅NH₃⁺further hydrolyzes as:

$\begin{array}{l}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{{\text{NH}}_{\text{3}}}^{\text{+}}{\text{+H}}_{\text{2}}\text{O\hspace{0.17em}}\rightleftharpoons {\text{\hspace{0.17em}C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{NH}}_{\text{2}}{\text{+\hspace{0.17em}H}}_{\text{3}}{\text{O}}^{\text{+}}\\ \mathrm{Initial}0.8400\\ \text{Change\hspace{1em}\hspace{1em}}-\mathrm{x}+\mathrm{x}+\mathrm{x}\\ \text{Equilibrium\hspace{0.17em}}0.84-\mathrm{x}\mathrm{x}\mathrm{x}\end{array}$

$\begin{array}{c}{\mathrm{K}}_{\mathrm{a}}=\frac{\left[{\text{C}}_6{\text{H}}_{\text{5}}{\mathrm{NH}}_2\right]\left[\mathrm{O}{\text{H}}^{-}\right]}{\left[{\text{C}}_6{\text{H}}_{\text{5}}{{\text{NH}}_{\text{3}}}^{+}\right]}\\ 2.63\times {10}^{-5}=\frac{{\mathrm{x}}^2}{0.84-\mathrm{x}}\\ \mathrm{x}=4.7\times {10}^{-3}\end{array}$

$\begin{array}{c}\mathrm{pH}=-\log \left[{\text{H}}^{\text{+}}\right]\\ =-\log (4.7\times {10}^{-3})\\ =2.33\end{array}$

pH of 0.57 M KC7H5O2

On dissociation, KC₇H₅O₂gives K⁺and C₇H₅O₂^-; out of these two, C₇H₅O₂^-further hydrolyses as:

$\begin{array}{l}{\text{C}}_7{\text{H}}_{\text{5}}{{\text{O}}_2}^{-}{\text{+H}}_{\text{2}}\text{O\hspace{0.17em}}\rightleftharpoons {\text{\hspace{0.17em}HC}}_7{\text{H}}_{\text{5}}{\text{O}}_{\text{2}}{\text{+\hspace{0.17em}OH}}^{\text{-}}\\ \mathrm{Initial}0.5700\\ \text{Change\hspace{1em}\hspace{1em}}-\mathrm{x}+\mathrm{x}+\mathrm{x}\\ \text{Equilibrium\hspace{0.17em}}0.57-\mathrm{x}\mathrm{x}\mathrm{x}\end{array}$

$\begin{array}{c}{\mathrm{K}}_{\mathrm{b}}=\frac{\left[{\text{HC}}_7{\text{H}}_{\text{5}}{\text{O}}_{\text{2}}\right]\left[\mathrm{O}{\text{H}}^{-}\right]}{\left[{\text{C}}_7{\text{H}}_{\text{5}}{{\text{O}}_{\text{2}}}^{-}\right]}\\ 1.56\times {10}^{-10}=\frac{{\mathrm{x}}^2}{0.57-\mathrm{x}}\\ \mathrm{x}=9.42\times {10}^{-6}\end{array}$

$\begin{array}{c}\left[{\text{H}}^{\text{+}}\right]=\frac{{\mathrm{K}}_{\mathrm{w}}}{\left[{\text{OH}}^{-}\right]}\\ =\frac{1\times {10}^{-14}}{9.42\times {10}^{-6}}\\ =1.06\times {10}^{-9}\end{array}$

$\begin{array}{c}\mathrm{pH}=-\log \left[{\text{H}}^{\text{+}}\right]\\ =-\log (1.06\times {10}^{-9})\\ =9\end{array}$