Question

Consider a 0.67-M solution of ${\mathbf{C}}_2{\mathbf{H}}_5{\mathbf{NH}}_2$ (${\mathbf{K}}_b=\mathbf{5}.\mathbf{6}\times {\mathbf{10}}^{-\mathbf{4}}$ ).

a. Which of the following are major species in the solution?

$\begin{array}{l}\mathbf{i}.{\mathbf{C}}_2{\mathbf{H}}_5{\mathbf{NH}}_2\\ \mathbf{ii}.{\mathbf{H}}^{+}\\ \mathbf{iii}.{\mathbf{OH}}^{-}\\ \mathbf{iv}.{\mathbf{H}}_2\mathbf{O}\\ \mathbf{v}.{\mathbf{C}}_2{\mathbf{H}}_5{\mathbf{NH}}_3{{}^{+}}_{}\end{array}$

b. Calculate the pH of this solution

Short answer

1. As ethylamine is a weak base, the major species in the solution are: ${\mathrm{C}}_2{\mathrm{H}}_5{\mathrm{NH}}_2\mathrm{and}\mathrm{water}.$
2. The pH of the solution = 12.3

Step-by-step solution

Major Species Present in Solution

The major species present are C₂H₅NH₂ and water, and the reaction that takes place is:

${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{{\text{NH}}_2}_{\text{(aq)}}{\text{+H}}_2{\text{O}}_{\text{(l)}}\rightleftharpoons {\text{\hspace{0.17em}C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{NH}}_3{{}^{\text{+}}}_{\text{(aq)}}{{\text{+\hspace{0.17em}OH}}^{\text{-}}}_{\text{(aq)}}$

c. So the major species present in the solution are ${\mathrm{C}}_2{\mathrm{H}}_5{\mathrm{NH}}_2\mathrm{and}\mathrm{water}.$

pH Calculation

C₂H₅NH₂dissociates in water as:

$\begin{array}{l}{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{{\text{NH}}_2}_{\text{(aq)}}{\text{+H}}_2{\text{O}}_{\text{(l)}}\rightleftarrows {\text{\hspace{0.17em}C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{NH}}_3{{}^{\text{+}}}_{\text{(aq)}}{{\text{+\hspace{0.17em}OH}}^{\text{-}}}_{\text{(aq)}}\\ \mathrm{Initial}0.6700\\ \text{Change\hspace{1em}\hspace{1em}}-\mathrm{x}+\mathrm{x}+\mathrm{x}\\ \text{Equilibrium\hspace{0.17em}}0.67-\mathrm{x}\mathrm{x}\mathrm{x}\end{array}$

$\begin{array}{c}{\mathrm{K}}_{\mathrm{b}}=\frac{{\text{[C}}_{\text{2}}{\text{H}}_{\text{5}}{{\text{NH}}_3}^{\text{+}}\left]\right[\mathrm{O}{\text{H}}^{-}]}{{\text{[C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{NH}}_2]}\\ 5.6\times {10}^{-4}=\frac{{\mathrm{x}}^2}{0.67-\mathrm{x}}\\ \mathrm{x}=1.94\times {10}^{-2}\end{array}$

$\begin{array}{c}\left[{\text{H}}^{+}\right]=\frac{{\mathrm{K}}_{\mathrm{w}}}{\left[\mathrm{O}{\text{H}}^{-}\right]}\\ =\frac{1\times {10}^{-14}}{1.94\times {10}^{-2}}\\ =5.15\times {10}^{-13}\end{array}$

${\text{pH=\hspace{0.17em}-log[H}}^{\text{+}}\text{]=-log(}5.15\times {10}^{-13}\text{)=12.3}$