Calculate the solubility (in moles per liter) of \(\mathrm{Fe}(\mathrm{OH})_{3}\) $\left(K_{\mathrm{sp}}=4 \times 10^{-38}\right)$ in each of the following. a. water b. a solution buffered at pH \(=5.0\) c. a solution buffered at pH\(=11.0\)

The dissolution of iron (III) hydroxide, \(\mathrm{Fe}(\mathrm{OH})_{3}\) in water can be represented as: \[\mathrm{Fe(OH)_3 (s)} \rightleftharpoons \mathrm{Fe^{3+} (aq)} + 3\mathrm{OH^{-} (aq)}\] This balanced equation shows that for every mole of \(\mathrm{Fe}(\mathrm{OH})_{3}\) that dissolves, one mole of \(\mathrm{Fe^{3+}}\) ions and 3 moles of \(\mathrm{OH^{-}}\) ions are produced.

For the balanced dissolution equation, the solubility product constant is given by: \[K_{\text{sp}} = [\mathrm{Fe^{3+}}][\mathrm{OH^{-}}]^3\]

Let \(x\) be the solubility of \(\mathrm{Fe}(\mathrm{OH})_{3}\) in moles per liter. Then, \[[\mathrm{Fe^{3+}}] = x\] \[[\mathrm{OH^{-}}] = 3x\]

Substitute the values of \([\mathrm{Fe^{3+}}]\) and \([\mathrm{OH^{-}}]\) in the \(K_{sp}\) expression: \[K_{\text{sp}} = x(3x)^3 = 27x^4\] Given \(K_{sp} = 4 \times 10^{-38}\), we can find the solubility in water: \[x^4 = \frac{4 \times 10^{-38}}{27}\] Solve for x: \[x = \sqrt[4]{\frac{4 \times 10^{-38}}{27}}\] \[x \approx 2.22 \times 10^{-10}\ \text{mol/L}\]

First, calculate the hydroxide ion concentration, \([\mathrm{OH^{-}}]\), using the pH value: \[\text{pH} + \text{pOH} = 14\] \[5 + \text{pOH} = 14\] \[\text{pOH} = 9\] \[ [\mathrm{OH^{-}}] = 10^{-9}\] Now, substitute the \([\mathrm{OH^{-}}]\) value in the \(K_{sp}\) expression: \[K_{sp} = [\mathrm{Fe^{3+}}][(10^{-9})]^3\] \[4 \times 10^{-38} = [\mathrm{Fe^{3+}}] \times 10^{-27}\] Solve for \([\mathrm{Fe^{3+}}]\): \[[\mathrm{Fe^{3+}}] \approx 4 \times 10^{-11}\ \text{mol/L}\]

Using the pH value, calculate the hydroxide ion concentration, \([\mathrm{OH^{-}}]\): \[\text{pH} + \text{pOH} = 14\] \[11 + \text{pOH} = 14\] \[\text{pOH} = 3\] \[ [\mathrm{OH^{-}}] = 10^{-3}\] Now, substitute the \([\mathrm{OH^{-}}]\) value in the \(K_{sp}\) expression: \[K_{sp} = [\mathrm{Fe^{3+}}][(10^{-3})]^3\] \[4 \times 10^{-38} = [\mathrm{Fe^{3+}}] \times 10^{-9}\] Solve for \([\mathrm{Fe^{3+}}]\): \[[\mathrm{Fe^{3+}}] \approx 4 \times 10^{-29}\ \text{mol/L}\] #Conclusion# To summarize, the solubility of \(\mathrm{Fe}(\mathrm{OH})_{3}\) in different solutions is as follows: a. water - \(2.22 \times 10^{-10}\) mol/L b. solution buffered at pH = 5.0 - \(4 \times 10^{-11}\) mol/L c. solution buffered at pH = 11.0 - \(4 \times 10^{-29}\) mol/L