Question

Calculate [OH^-], [H⁺], and the pH of 0.40 M solutions of each of the following amines (the K_bvalues are found in Table 7.3).

a. aniline b. methylamine

Short answer

a) For aniline

$\left[{\mathrm{OH}}^{-}\right]=1.23\times {10}^{-5},\left[{\mathrm{H}}^{+}\right]=8.13\times {10}^{-10},{\mathrm{p}}^{\mathrm{H}}=9.1$

b) For Methylamine

$\left[{\mathrm{OH}}^{-}\right]=1.752\times {10}^{-2},\left[{\mathrm{H}}^{+}\right]=7.5\times {10}^{-13},{\mathrm{p}}^{\mathrm{H}}=12.1$

Step-by-step solution

Calculation of [OH-], [H+], and the pH of 0.40 M Solutions of Aniline

Aniline reacts with water as shown

$$\begin{gathered} {\mathrm{C}}_6{\mathrm{H}}_5{{\mathrm{NH}}_2}_{\left(\mathrm{aq}\right)}+{\mathrm{H}}_2\mathrm{O}\iff {\mathrm{C}}_6{\mathrm{H}}_5{{{\mathrm{NH}}_3}^{+}}_{\left(\mathrm{aq}\right)}+{{\mathrm{OH}}^{-}}_{\left(\mathrm{aq}\right)} \\ 0.4\mathrm{\_}00 \\ 0.4-\mathrm{x}-\mathrm{x}\mathrm{x} \end{gathered}$$

[OH^-] ion concentration can be calculated as

$$\begin{gathered} {\mathrm{K}}_{\mathrm{b}}=\frac{{\mathrm{x}}^2}{0.4-\mathrm{x}}=3.8\times {10}^{-10}({\mathrm{K}}_{\mathrm{b}}\mathrm{value}\mathrm{given}\mathrm{in}\mathrm{table}7.3) \\ {\mathrm{x}}^2=3.8\times {10}^{-10}\times 0.4(\mathrm{Taken},0.4\times ~>>\mathrm{x}) \\ \mathrm{x}=1.23\times {10}^{-5} \end{gathered}$$

[H⁺] ion concentration can be calculated as

$$\begin{gathered} {\mathrm{K}}_{\mathrm{w}}=\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]=1\times {10}^{-14} \\ \left[{\mathrm{H}}^{+}\right]=\frac{{\mathrm{K}}_{\mathrm{w}}}{\left[{\mathrm{OH}}^{-}\right]}=\frac{1\times {10}^{-14}}{1.23\times {10}^{-5}}=8.13\times {10}^{-10} \end{gathered}$$

Now pH can be calculated as

${\mathrm{p}}^{\mathrm{H}}=-\log \left[{\mathrm{H}}^{+}\right]=-\log (8.13\times {10}^{-10})=9.1$

Calculation of [OH-], [H+], and the pH of 0.40 M Solutions of Methylamine 

Methylamine reacts with water as shown

$$\begin{gathered} {\mathrm{CH}}_3{{\mathrm{NH}}_2}_{\left(\mathrm{aq}\right)}+{\mathrm{H}}_2\mathrm{O}\iff {\mathrm{CH}}_3{{{\mathrm{NH}}_3}^{+}}_{\left(\mathrm{aq}\right)}+{{\mathrm{OH}}^{-}}_{\left(\mathrm{aq}\right)} \\ 0.4\mathrm{\_}00 \\ 0.4-\mathrm{x}-\mathrm{x}\mathrm{x} \end{gathered}$$

[OH^-] ion concentration can be calculated as

$$\begin{gathered} {\mathrm{K}}_{\mathrm{b}}=\frac{{\mathrm{x}}^2}{0.4-\mathrm{x}}=4.38\times {10}^{-4}({\mathrm{K}}_{\mathrm{b}}\mathrm{value}\mathrm{given}\mathrm{in}\mathrm{table}7.3) \\ {\mathrm{x}}^2=4.38\times {10}^{-4}\times 0.4 \\ \mathrm{x}=1.752\times {10}^{-2} \end{gathered}$$

[H⁺] ion concentration can be calculated as

localid="1650275459888" $$\begin{gathered} {\mathrm{K}}_{\mathrm{w}}=\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]=1\times {10}^{-14} \\ \left[{\mathrm{H}}^{+}\right]=\frac{{\mathrm{K}}_{\mathrm{w}}}{\left[{\mathrm{OH}}^{-}\right]}=\frac{1\times {10}^{-14}}{1.752\times {10}^{-2}}=7.5\times {10}^{-13} \end{gathered}$$

Now pH can be calculated as

localid="1650275474018" ${\mathrm{p}}^{\mathrm{H}}=-\log \left[{\mathrm{H}}^{+}\right]=-\log (7.5\times {10}^{-13})=12.1$