Calculate the solubility (in moles per liter) of \(\mathrm{Fe}(\mathrm{OH})_{3}\) \(\left(K_{\mathbb{\infty}}=4 \times 10^{-3 \mathrm{~s}}\right)\) in each of the following. a. water b. a solution buffered at \(\mathrm{pH}=5.0\) c. a solution buffered at \(\mathrm{pH}=11.0\)

For \(\mathrm{Fe}(\mathrm{OH})_{3}\), the solubility equilibrium is: \[\mathrm{Fe(OH)_3 (s)} \rightleftharpoons \mathrm{Fe^{3+} (aq)} + 3\mathrm{OH^- (aq)}\] The solubility product constant expression is: \[K_{sp} = [\mathrm{Fe^{3+}}][\mathrm{OH^-}]^3\]

In water, let the solubility of \(\mathrm{Fe}(\mathrm{OH})_{3}\) be \(s\). Then, the concentrations of ions in the solution are \([\mathrm{Fe^{3+}}] = s\) and \([\mathrm{OH^-}] = 3s\). We know that \(K_\infty = 4 \times 10^{37}\). We can find the solubility by solving the solubility product constant expression for \(s\): \[K_{sp} = [s][3s]^3 = 27s^4\] \[s^4 = \frac{K_\infty}{27}\] \[s = \sqrt[4]{\frac{4 \times 10^{37}}{27}}\] Now calculate the solubility: \[s = \sqrt[4]{\frac{4 \times 10^{37}}{27}} \approx 2.738 \times 10^9\] Thus, the solubility of \(\mathrm{Fe}(\mathrm{OH})_{3}\) in water is approximately \(2.738 \times 10^9\) moles per liter.

In a buffered solution at pH 5.0, the concentration of hydroxide ions can be found using the relationship between pH and pOH: \[pH + pOH = 14\] \[pOH = 14 - pH = 14 - 5 = 9\] The concentration of hydroxide ions is: \[[\mathrm{OH^-}] = 10^{-pOH} = 10^{-9}\] To find the new solubility, we can use the solubility product constant expression with the concentration of hydroxide ions we just found: \[K_{sp} = [\mathrm{Fe^{3+}}][\mathrm{OH^-}]^3 = [\mathrm{Fe^{3+}}](10^{-9})^3\] Now calculate the solubility: \[[\mathrm{Fe^{3+}}] = \frac{K_{sp}}{(10^{-9})^3} = \frac{4 \times 10^{37}}{10^{-27}} \approx 4 \times 10^{64}\] Thus, the solubility of \(\mathrm{Fe}(\mathrm{OH})_{3}\) in a solution buffered at pH 5.0 is approximately \(4 × 10^{64}\) moles per liter.

In a buffered solution at pH 11.0, we can again find the concentration of hydroxide ions using the relationship between pH and pOH: \[pOH = 14 - pH = 14 - 11 = 3\] The concentration of hydroxide ions is: \[[\mathrm{OH^-}] = 10^{-pOH} = 10^{-3}\] To find the new solubility, we can again use the solubility product constant expression with the concentration of hydroxide ions we just found: \[K_{sp} = [\mathrm{Fe^{3+}}][\mathrm{OH^-}]^3 = [\mathrm{Fe^{3+}}](10^{-3})^3\] Now calculate the solubility: \[[\mathrm{Fe^{3+}}] = \frac{K_{sp}}{(10^{-3})^3} = \frac{4 \times 10^{37}}{10^{-9}} \approx 4 \times 10^{46}\] Thus, the solubility of \(\mathrm{Fe}(\mathrm{OH})_{3}\) in a solution buffered at pH 11.0 is approximately \(4 × 10^{46}\) moles per liter.