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

First, we need to recognize that the \(\mathrm{Co}\left(\mathrm{H}_{2}\mathrm{O}\right)_{6}^{3+}\) ions undergo hydrolysis in the solution, which means they react with water to release \(\mathrm{H}^{+}\) ions. The balanced equation for this reaction is: \[\mathrm{Co}\left(\mathrm{H}_{2}\mathrm{O}\right)_{6}^{3+} \left(aq\right) + \mathrm{H}_{2}\mathrm{O} \left(l\right) \rightleftarrows \mathrm{Co}\left(\mathrm{H}_{2}\mathrm{O}\right)_{5}\left(\mathrm{OH}\right)^{2+} \left(aq\right) + \mathrm{H_{3}O}^{+} \left(aq\right)\]

Now we need to set up the equilibrium expression for the reaction, using the given \(K_{\mathrm{a}}\) value: \[K_{\mathrm{a}} = \dfrac{[\mathrm{Co}\left(\mathrm{H}_{2}\mathrm{O}\right)_{5}\left(\mathrm{OH}\right)^{2+}][\mathrm{H_{3}O}^{+}]}{[\mathrm{Co}\left(\mathrm{H}_{2}\mathrm{O}\right)_{6}^{3+}]}\] Since we're given the initial concentration of \(\mathrm{CoCl}_{3}\) (which is the same as the initial concentration of \(\mathrm{Co}\left(\mathrm{H}_{2}\mathrm{O}\right)_{6}^{3+}\)), we can solve for the equilibrium concentration of \(\mathrm{H}^{+}\) ions. Let's call this concentration \(x\). At equilibrium, the concentrations are as follows: \([\mathrm{Co}\left(\mathrm{H}_{2}\mathrm{O}\right)_{6}^{3+}] = 0.10 \mathrm{M} - x\) \([\mathrm{Co}\left(\mathrm{H}_{2}\mathrm{O}\right)_{5}\left(\mathrm{OH}\right)^{2+}] = x\) \([\mathrm{H_{3}O}^{+}] = x\) Now substitute these values into the equilibrium expression: \[1.0 \times 10^{-5} = \dfrac{x^2}{0.10\mathrm{M} - x}\] We can simplify this expression by assuming that the amount of dissociation is small and \(x << 0.10\mathrm{M}\). Therefore, we can approximate that \(0.10\mathrm{M} - x \approx 0.10\mathrm{M}\). So, \[1.0 \times 10^{-5} \approx \dfrac{x^2}{0.10\mathrm{M}}\] Now, we can solve for \(x\). \[x = \sqrt{1.0 \times 10^{-5} \times 0.10\mathrm{M}} \approx 1.0 \times 10^{-3}\mathrm{M}\]

Now that we have calculated the concentration of \(\mathrm{H}^{+}\) ions in the solution, we can use this value to calculate the pH: \[pH = -\log{[\mathrm{H}^{+}]}\] \[pH = -\log{(1.0 \times 10^{-3}\mathrm{M})} \approx 3.00\] Hence, the pH of the 0.10 M \(\mathrm{CoCl}_3\) solution is approximately 3.00.