Problem 52 - Chemistry The Central Science

Question

\(\begin{array}{llll} \text { (a) The molar solubility of } \mathrm{PbBr}_{2} & \text { at } 25^{\circ} \mathrm{C} & \text { is }\end{array}\) \(1.0 \times 10^{-2} \mathrm{~mol} / \mathrm{L} .\) 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 \(\mathrm{Ca}(\mathrm{OH})_{2}\)

Step-by-Step Solution

Answer

(a) The solubility-product constant for \(\mathrm{PbBr}_2\) is \(K_{sp} = 4.0 \times 10^{-6}\). (b) The solubility-product constant for \(\mathrm{AgIO}_3\) is \(K_{sp} = 8.36 \times 10^{-8}\). (c) The pH of a saturated solution of \(\mathrm{Ca(OH)}_{2}\) is 12.18.

Step 1: (a) Calculate the solubility-product constant for \(\mathrm{PbBr}_2\)

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 \(K_{sp}\): \(K_{sp} = 4.0 \times 10^{-6}\)

Step 2: (b) Calculate the solubility-product constant for \(\mathrm{AgIO}_3\)

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 \(K_{sp}\): \(K_{sp} = 8.36 \times 10^{-8}\)

Step 3: (c) Calculate the pH of saturated \(\mathrm{Ca(OH)}_{2}\) solution

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: \(pH = 14 - pOH = 14 - 1.82 = 12.18\)

Key Concepts

Molar Solubility

Molar solubility refers to the number of moles of a solute that can dissolve in a liter of solution to form a saturated solution. It helps us understand how much of a substance can be dissolved before the solution becomes saturated.

In the case of \(\text{PbBr}_2\), to find the molar solubility, we start with the dissolution equation, which shows how the solid dissolves in water to form its ions:

• \(\text{PbBr}_{2(s)} \rightleftharpoons \text{Pb}^{2+}_{(aq)} + 2 \text{Br}^-_{(aq)}\)

The molar solubility tells us that for every 1 mole of \(\text{PbBr}_2\) that dissolves, we will get 1 mole of \(\text{Pb}^{2+}\) and 2 moles of \(\text{Br}^-\).

Knowing the molar solubility of \(\text{PbBr}_2\) is \(1.0 \times 10^{-2}\) mol/L helps us calculate the concentrations of these ions in a saturated solution for further calculations, such as finding the Ksp.

Ksp Calculation

The solubility-product constant, \(K_{sp}\), is a measure of the solubility of a compound. It's determined from the concentrations of the dissolved ions in a saturated solution.

For \(\text{PbBr}_2\), the \(K_{sp}\) expression is based on its dissolution reaction:

• \(K_{sp} = [\text{Pb}^{2+}][\text{Br}^-]^2\)

Here,

• \([\text{Pb}^{2+}] = 1.0 \times 10^{-2} \text{ M}\)
• \([\text{Br}^-] = 2 \times 1.0 \times 10^{-2} \text{ M}\)

Substituting these values into the equation, we calculate \(K_{sp} = (1.0 \times 10^{-2})(2 \times 1.0 \times 10^{-2})^2\).

After performing the calculations, we find \(K_{sp} = 4.0 \times 10^{-6}\). This value provides insight into the solubility properties of \(\text{PbBr}_2\) under the given conditions.

Saturated Solution pH

The pH of a saturated solution is influenced by its ion concentrations, such as hydroxide ions (\(\text{OH}^-\)), when dealing with compounds like \(\text{Ca(OH)}_{2}\).

From the dissolution of \(\text{Ca(OH)}_{2}\), we have:

• \(\text{Ca(OH)}_{2(s)} \rightleftharpoons \text{Ca}^{2+}_{(aq)} + 2 \text{OH}^-_{(aq)}\)

The \(K_{sp}\) expression becomes \([\text{Ca}^{2+}][\text{OH}^-]^2\).
By solving the equation \(5.5 \times 10^{-6} = s(2s)^2\), we determine the molar solubility \(s = 7.61 \times 10^{-3} \text{ M}\).

This provides us \([\text{OH}^-] = 1.52 \times 10^{-2} \text{ M}\), which we use to find the pOH: \(pOH = -\log_{10}(1.52 \times 10^{-2}) = 1.82\).

Finally, the pH is calculated as \(pH = 14 - 1.82 = 12.18\), indicating a strongly basic solution.