Calculate \(\left[\mathrm{OH}^{-}\right],\left[\mathrm{H}^{+}\right],\) and the \(\mathrm{pH}\) of 0.40\(M\) solutions of each of the following amines (the \(K_{\mathrm{b}}\) values are found in Table 14.3). a. aniline b. methylamine

The dissociation reactions of aniline and methylamine in water are as follows: For aniline: \(\mathrm{C_6H_5NH_2 + H_2O \rightleftharpoons C_6H_5NH_3^+ + OH^-}\) For methylamine: \(\mathrm{CH_3NH_2 + H_2O \rightleftharpoons CH_3NH_3^+ + OH^-}\)

Using the ICE (Initial, Change, Equilibrium) table method to identify initial concentrations and changes in concentrations during dissociation: For aniline: \[I:\ \mathrm{[C_6H_5NH_2] = 0.40\ M, [OH^-] = 0, [C_6H_5NH_3^+] = 0}\] \[C:\ \mathrm{[-x, +x, +x]}\] \[E:\ \mathrm{[0.40 - x\ M,\ x\ M,\ x\ M ]}\] For methylamine: \[I:\ \mathrm{[CH_3NH_2] = 0.40\ M, [OH^-] = 0, [CH_3NH_3^+] = 0}\] \[C:\ \mathrm{[-x, +x, +x]}\] \[E:\ \mathrm{[0.40 - x\ M,\ x\ M,\ x\ M ]}\]

For both amines, we write the equilibrium expression for \(K_{\mathrm{b}}\) using their concentrations at equilibrium: For aniline (Using Table 14.3, \(K_{\mathrm{b}} = 4.3 \times 10^{-10}\)): \[\frac{x^2}{(0.40 - x)} = 4.3 \times 10^{-10}\] For methylamine (Using Table 14.3, \(K_{\mathrm{b}} = 4.4 \times 10^{-4}\)): \[\frac{x^2}{(0.40 - x)} = 4.4 \times 10^{-4}\] Now, we will assume that the concentration \(x\) is negligible compared to the initial concentration (0.40 M), so we can simplify the expression: For aniline: \[\frac{x^2}{0.40} = 4.3 \times 10^{-10}\] \[x^2 = 1.72 \times 10^{-10}\] \[x = 4.15 \times 10^{-6}\] For methylamine: \[\frac{x^2}{0.40} = 4.4 \times 10^{-4}\] \[x^2 = 1.76 \times 10^{-4}\] \[x = 1.33 \times 10^{-2}\]

Now that we have the values of x, we can calculate the concentrations of \(\mathrm{OH}^{-}\), \(\mathrm{H}^{+}\), and \(\mathrm{pH}\) for each amine. For aniline: \(\mathrm{[OH^-]} = x = 4.15 \times 10^{-6}\ \mathrm{M}\) \(\mathrm{pOH} = -\log_{10}(4.15 \times 10^{-6}) = 5.38\) \(\mathrm{pH} = 14 - \mathrm{pOH} = 14 - 5.38 = 8.62\) \(\mathrm{[H^+]} = 10^{-\mathrm{pH}} = 10^{-8.62} = 2.4 \times 10^{-9}\ \mathrm{M}\) For methylamine: \(\mathrm{[OH^-]} = x = 1.33 \times 10^{-2}\ \mathrm{M}\) \(\mathrm{pOH} = -\log_{10}(1.33 \times 10^{-2}) = 1.88\) \(\mathrm{pH} = 14 - \mathrm{pOH} = 14 - 1.88 = 12.12\) \(\mathrm{[H^+]} = 10^{-\mathrm{pH}} = 10^{-12.12} = 7.6 \times 10^{-13}\ \mathrm{M}\) So, the final values are as follows: a. Aniline: - \(\mathrm{[OH^-]} = 4.15 \times 10^{-6}\ \mathrm{M}\) - \(\mathrm{[H^+]} = 2.4 \times 10^{-9}\ \mathrm{M}\) - \(\mathrm{pH} = 8.62\) b. Methylamine - \(\mathrm{[OH^-]} = 1.33 \times 10^{-2}\ \mathrm{M}\) - \(\mathrm{[H^+]} = 7.6 \times 10^{-13}\ \mathrm{M}\) - \(\mathrm{pH} = 12.12\)