Which of the following compounds has the lowest `Pb^(+2)` ion concentration?

Options

A. `PbCO_(3), K_(sp) = 1.5 xx 10^(-13)`
B. `Pb(OH)_(2), K_(Sp) = 4.2 xx 10^(-15)`
C. `PbSO_(4), K_(sp) = 1.3 xx 10^(-8)`
D. `PbI_(2), K_(sp) = 8.3 xx 10^(-9)`

Answer

To determine which compound has the lowest concentration of \( \text{Pb}^{2+} \) ions, we need to analyze the solubility of each given compound. The solubility product constant (\( K_{sp} \)) will help us find the concentration of \( \text{Pb}^{2+} \) ions for each compound.

### Step-by-Step Solution:

1. **Identify the Compounds and Their Dissociation:**
- **Option 1: \( \text{PbCO}_3 \)**
- Dissociation: \( \text{PbCO}_3 \rightleftharpoons \text{Pb}^{2+} + \text{CO}_3^{2-} \)
- \( K_{sp} = S^2 \)

- **Option 2: \( \text{Pb(OH)}_2 \)**
- Dissociation: \( \text{Pb(OH)}_2 \rightleftharpoons \text{Pb}^{2+} + 2\text{OH}^{-} \)
- \( K_{sp} = S \cdot (2S)^2 = 4S^3 \)

- **Option 3: \( \text{PbSO}_4 \)**
- Dissociation: \( \text{PbSO}_4 \rightleftharpoons \text{Pb}^{2+} + \text{SO}_4^{2-} \)
- \( K_{sp} = S^2 \)

- **Option 4: \( \text{PbI}_2 \)**
- Dissociation: \( \text{PbI}_2 \rightleftharpoons \text{Pb}^{2+} + 2\text{I}^{-} \)
- \( K_{sp} = S \cdot (2S)^2 = 4S^3 \)

2. **Calculate Solubility for Each Compound:**
- **For \( \text{PbCO}_3 \):**
- Given \( K_{sp} = 1.5 \times 10^{-13} \)
- \( S^2 = 1.5 \times 10^{-13} \)
- \( S = \sqrt{1.5 \times 10^{-13}} \approx 1.22 \times 10^{-7} \)
- Concentration of \( \text{Pb}^{2+} = S \approx 1.22 \times 10^{-7} \)

- **For \( \text{Pb(OH)}_2 \):**
- Given \( K_{sp} = 4.2 \times 10^{-15} \)
- \( 4S^3 = 4.2 \times 10^{-15} \)
- \( S^3 = 1.05 \times 10^{-15} \)
- \( S = \sqrt[3]{1.05 \times 10^{-15}} \approx 1.02 \times 10^{-5} \)
- Concentration of \( \text{Pb}^{2+} = S \approx 1.02 \times 10^{-5} \)

- **For \( \text{PbSO}_4 \):**
- Given \( K_{sp} = 1.3 \times 10^{-8} \)
- \( S^2 = 1.3 \times 10^{-8} \)
- \( S = \sqrt{1.3 \times 10^{-8}} \approx 3.61 \times 10^{-5} \)
- Concentration of \( \text{Pb}^{2+} = S \approx 3.61 \times 10^{-5} \)

- **For \( \text{PbI}_2 \):**
- Given \( K_{sp} = 8.3 \times 10^{-9} \)
- \( 4S^3 = 8.3 \times 10^{-9} \)
- \( S^3 = 2.075 \times 10^{-9} \)
- \( S = \sqrt[3]{2.075 \times 10^{-9}} \approx 1.27 \times 10^{-3} \)
- Concentration of \( \text{Pb}^{2+} = S \approx 1.27 \times 10^{-3} \)

3. **Compare the Concentrations:**
- \( \text{PbCO}_3: 1.22 \times 10^{-7} \)
- \( \text{Pb(OH)}_2: 1.02 \times 10^{-5} \)
- \( \text{PbSO}_4: 3.61 \times 10^{-5} \)
- \( \text{PbI}_2: 1.27 \times 10^{-3} \)

4. **Conclusion:**
- The compound with the lowest \( \text{Pb}^{2+} \) ion concentration is \( \text{PbCO}_3 \) with a concentration of approximately \( 1.22 \times 10^{-7} \).

### Final Answer:
**The compound with the lowest \( \text{Pb}^{2+} \) ion concentration is \( \text{PbCO}_3 \).**