Question

Calculate the pH of a 0.050 M (C₂H₅)₂NH solution (K_b = 1.3 × 10^-3).

Short answer

pH of a 0.050 M (C₂H₅)₂NH solution is 11.93.

Step-by-step solution

Calculation of [OH-] of Diethylamine

Diethylamino reacts with water as shown

$\begin{array}{l}{\left({\mathrm{C}}_2{\mathrm{H}}_5\right)}_2{\text{(NH}}_{\text{(aq)}}{\text{+H}}_{\text{2}}\text{O}\rightleftarrows {\left({\mathrm{C}}_2{\mathrm{H}}_5\right)}_2{\text{\hspace{0.17em}NH}}_2{{}^{\text{+}}}_{\text{(aq)}}{{\text{+\hspace{0.17em}OH}}^{\text{-}}}_{\text{(aq)}}\\ 0.05-00\\ 0.05-\text{x\hspace{0.17em}\hspace{1em}\hspace{1em}}-\text{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{0.17em}x\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}x}\end{array}$

[OH^-] ion concentration can be calculated as

$\begin{array}{l}{\text{K}}_{\text{b}}=\frac{{\text{x}}^2}{0.05-\text{x}}\\ {\text{K}}_{\text{b}}=1.3\times {10}^{-3}\\ {\text{x}}^2=1.3\times {10}^{-3}\times 0.05\\ \text{x}=8.6\times {10}^{-3}\end{array}$

Calculating pH from [OH-]

[H⁺] ion concentration can be calculated as

$\begin{array}{l}{\text{K}}_{\text{w}}=\left[{\text{H}}^{\text{+}}{\text{][OH}}^{\text{-}}\right]=1\times {10}^{-14}\\ [{\text{H}}^{\text{+}}\text{]}=\frac{{\text{K}}_{\text{w}}}{{\text{[OH}}^{\text{-}}]}=\frac{1\times {10}^{-14}}{8.6\times {10}^{-3}}=1.163\times {10}^{-12}\end{array}$

Now pH can be calculated as

$\begin{array}{l}{\text{pH\hspace{0.17em}=\hspace{0.17em}-log[H}}^{\text{+}}\text{]}\\ {\text{pH\hspace{0.17em}=\hspace{0.17em}-log(1.163×10}}^{\text{-12}}\text{)}\\ \text{pH\hspace{0.17em}=\hspace{0.17em}11.93}\end{array}$