\(\begin{array}{llll} \text { (a) The molar solubility of } \mathrm{PbBr}_{2} & \text { at } 25^{\circ} \mathrm{C} & \text { is }\end{array}\) \(1.0 \times 10^{-2} \mathrm{~mol} / \mathrm{L} .\) Calculate \(K_{s p} .(\mathbf{b})\) If \(0.0490 \mathrm{~g}\) of \(\mathrm{AgIO}_{3}\) dis- solves per liter of solution, calculate the solubility-product constant. (c) Using the appropriate \(K_{s p}\) value from Appendix \(\mathrm{D},\) calculate the \(\mathrm{pH}\) of a saturated solution of \(\mathrm{Ca}(\mathrm{OH})_{2}\)

1. Write the balanced dissolution reaction: \(\mathrm{PbBr}_{2(s)} \rightleftharpoons \mathrm{Pb}^{2+}_{(aq)} + 2 \mathrm{Br}^-_{(aq)}\) 2. Write the expression for solubility-product constant: \(K_{sp} = [\mathrm{Pb}^{2+}][\mathrm{Br}^-]^2\) 3. The molar solubility of \(\mathrm{PbBr}_2\) is \(1.0 \times 10^{-2} \mathrm{mol}/\mathrm{L}\). For each 1 mole of \(\mathrm{PbBr}_{2}\) dissolved, we have 1 mole of \(\mathrm{Pb}^{2+}\) and 2 moles of \(\mathrm{Br}^{-}\). Therefore, we have: \([\mathrm{Pb}^{2+}] = 1.0 \times 10^{-2} \mathrm{M}\) and \([\mathrm{Br}^-] = 2 \times 1.0 \times 10^{-2} \mathrm{M}\) 4. Substitute the concentrations into the \(K_{sp}\) expression: \(K_{sp} = (1.0 \times 10^{-2})(2 \times 1.0 \times 10^{-2})^2\) 5. Calculate the \(K_{sp}\): \(K_{sp} = 4.0 \times 10^{-6}\)

1. Write the balanced dissolution reaction: \(\mathrm{AgIO}_{3(s)} \rightleftharpoons \mathrm{Ag}^+_{(aq)} + \mathrm{IO}_3^-_{(aq)}\) 2. Write the expression for solubility-product constant: \(K_{sp} = [\mathrm{Ag}^+][\mathrm{IO}_3^-]\) 3. Calculate the molar solubility: Molar mass of \(\mathrm{AgIO}_{3} = 169.87 \mathrm{g/mol}\), so molar solubility = \(0.0490 \mathrm{g}/1 \mathrm{L} \cdot \frac{1 \text{ mole}}{169.87 \mathrm{g}} = 2.89 \times 10^{-4} \mathrm{M}\) 4. Since the dissolution reaction is a 1:1 ratio, \([\mathrm{Ag}^+] = [\mathrm{IO}_3^-] = 2.89 \times 10^{-4} \mathrm{M}\) 5. Substitute the concentrations into the \(K_{sp}\) expression: \(K_{sp} = 2.89 \times 10^{-4} \cdot 2.89 \times 10^{-4}\) 6. Calculate the \(K_{sp}\): \(K_{sp} = 8.36 \times 10^{-8}\)

1. From Appendix D, find the \(K_{sp}\) of \(\mathrm{Ca(OH)}_{2}\): \(K_{sp} = 5.5 \times 10^{-6}\) 2. Write the balanced dissolution reaction: \(\mathrm{Ca(OH)}_{2(s)} \rightleftharpoons \mathrm{Ca}^{2+}_{(aq)} + 2 \mathrm{OH}^-_{(aq)}\) 3. Write the expression for solubility-product constant: \(K_{sp} = [\mathrm{Ca}^{2+}][\mathrm{OH}^-]^2\) 4. Since \([\mathrm{Ca}^{2+}] = s\), and \([\mathrm{OH}^-] = 2s\), we can rewrite the \(K_{sp}\) expression as: \(K_{sp} = s(2s)^2\) 5. Substitute the \(K_{sp}\) value and solve for s: \(5.5 \times 10^{-6} = s(2s)^2\) 6. Calculate the molar solubility "s": \(s = 7.61 \times 10^{-3} \mathrm{M}\) 7. Calculate the concentration of \(\mathrm{OH}^-\): \([\mathrm{OH}^-] = 2s = 2 \times 7.61 \times 10^{-3} \mathrm{M} = 1.52 \times 10^{-2} \mathrm{M}\) 8. Calculate the pOH value: \(pOH = -\log_{10}([\mathrm{OH}^-]) = -\log_{10}(1.52 \times 10^{-2}) = 1.82 \) 9. Calculate the pH value: \(pH = 14 - pOH = 14 - 1.82 = 12.18\)