Calculate the \(\mathrm{pH}\) of a \(0.10 \mathrm{M} \mathrm{CoCl}_{3}\) solution. The \(K_{\mathrm{a}}\) value for \(\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}\) is \(1.0 \times 10^{-5}\).

We start by writing the dissociation reaction for \(\mathrm{Co}\left(\mathrm{H}_{2}\mathrm{O}\right)_{6}^{3+}\): \[\mathrm{Co}\left(\mathrm{H}_{2}\mathrm{O}\right)_{6}^{3+} \rightleftharpoons \mathrm{Co}\left(\mathrm{H}_{2}\mathrm{O}\right)_{5}\mathrm{OH}^{2+} + \mathrm{H}^{+}\] Now, we can write down the equilibrium expression using the given \(K_{\mathrm{a}}\) value: \[K_\mathrm{a} = \dfrac{[\mathrm{Co}\left(\mathrm{H}_{2}\mathrm{O}\right)_{5}\mathrm{OH}^{2+}][\mathrm{H}^{+}]}{[\mathrm{Co}\left(\mathrm{H}_{2}\mathrm{O}\right)_{6}^{3+}]}\]

Initially, the concentration of \(\mathrm{Co}\left(\mathrm{H}_{2}\mathrm{O}\right)_{6}^{3+}\) is \(0.10 \mathrm{M}\). Before dissociation occurs, we have \([\mathrm{Co}\left(\mathrm{H}_{2}\mathrm{O}\right)_{5}\mathrm{OH}^{2+}] = [\mathrm{H}^{+}] = 0\). Letting \(x\) represent the change in concentration of \(\mathrm{H}^{+}\), the equilibrium concentrations will be: \[[\mathrm{Co}\left(\mathrm{H}_{2}\mathrm{O}\right)_{5}\mathrm{OH}^{2+}] = [\mathrm{H}^{+}] = x\] \[[\mathrm{Co}\left(\mathrm{H}_{2}\mathrm{O}\right)_{6}^{3+}] = 0.10 - x\] Since we expect \(x\) to be much smaller than \(0.10\), we can make the simplification that \(0.10 - x \approx 0.10\). Then, we can substitute these concentrations back into the equilibrium expression and solve for \(x\): \[K_\mathrm{a} = \dfrac{x \cdot x}{0.10} = 1.0 \times 10^{-5}\] \[x^2 = 1.0 \times 10^{-5} \cdot 0.10\] \[x = \sqrt{1.0 \times 10^{-6}} = 1 \times 10^{-3}\] Thus, we find that \([\mathrm{H}^{+}] \approx 1 \times 10^{-3} \mathrm{M}\).

Finally, we can use the concentration of \([\mathrm{H}^{+}]\) to calculate the \(\mathrm{pH}\) using the formula: \[\mathrm{pH} = -\log{[\mathrm{H}^{+}]}\] \[\mathrm{pH} = -\log{(1 \times 10^{-3})} = 3\] The \(\mathrm{pH}\) of the \(0.10 \mathrm{M}\) \(\mathrm{CoCl}_{3}\) solution is approximately \(3\).