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 products (\( K_{sp} \)) of the given compounds. The lower the solubility, the lower the concentration of \( \text{Pb}^{2+} \) ions in solution.

### Step-by-Step Solution:

1. **Identify the Compounds**: The compounds we are analyzing are:
- \( \text{PbCO}_3 \)
- \( \text{Pb(OH)}_2 \)
- \( \text{PbSO}_4 \)
- \( \text{PbI}_2 \)

2. **Write the Dissociation Equations**:
- For \( \text{PbCO}_3 \):
\[ \text{PbCO}_3 (s) \rightleftharpoons \text{Pb}^{2+} (aq) + \text{CO}_3^{2-} (aq) \]
- For \( \text{Pb(OH)}_2 \):
\[ \text{Pb(OH)}_2 (s) \rightleftharpoons \text{Pb}^{2+} (aq) + 2 \text{OH}^{-} (aq) \]
- For \( \text{PbSO}_4 \):
\[ \text{PbSO}_4 (s) \rightleftharpoons \text{Pb}^{2+} (aq) + \text{SO}_4^{2-} (aq) \]
- For \( \text{PbI}_2 \):
\[ \text{PbI}_2 (s) \rightleftharpoons \text{Pb}^{2+} (aq) + 2 \text{I}^{-} (aq) \]

3. **Write the Expression for \( K_{sp} \)**:
- For \( \text{PbCO}_3 \):
\[ K_{sp} = [\text{Pb}^{2+}][\text{CO}_3^{2-}] = S^2 \]
- For \( \text{Pb(OH)}_2 \):
\[ K_{sp} = [\text{Pb}^{2+}][\text{OH}^{-}]^2 = 4S^3 \]
- For \( \text{PbSO}_4 \):
\[ K_{sp} = [\text{Pb}^{2+}][\text{SO}_4^{2-}] = S^2 \]
- For \( \text{PbI}_2 \):
\[ K_{sp} = [\text{Pb}^{2+}][\text{I}^{-}]^2 = 4S^3 \]

4. **Substitute the Given \( K_{sp} \) Values**:
- \( K_{sp}(\text{PbCO}_3) = 1.5 \times 10^{-13} \)
- \( K_{sp}(\text{Pb(OH)}_2) = 4.2 \times 10^{-15} \)
- \( K_{sp}(\text{PbSO}_4) = 1.3 \times 10^{-8} \)
- \( K_{sp}(\text{PbI}_2) = 8.3 \times 10^{-9} \)

5. **Calculate the Solubility (S)**:
- For \( \text{PbCO}_3 \):
\[ S^2 = 1.5 \times 10^{-13} \implies S = \sqrt{1.5 \times 10^{-13}} \approx 1.22 \times 10^{-7} \]
- For \( \text{Pb(OH)}_2 \):
\[ 4S^3 = 4.2 \times 10^{-15} \implies S^3 = 1.05 \times 10^{-15} \implies S \approx 1.02 \times 10^{-5} \]
- For \( \text{PbSO}_4 \):
\[ S^2 = 1.3 \times 10^{-8} \implies S = \sqrt{1.3 \times 10^{-8}} \approx 3.61 \times 10^{-5} \]
- For \( \text{PbI}_2 \):
\[ 4S^3 = 8.3 \times 10^{-9} \implies S^3 = 2.075 \times 10^{-9} \implies S \approx 1.26 \times 10^{-3} \]

6. **Compare the Solubilities**:
- \( \text{PbCO}_3 \): \( 1.22 \times 10^{-7} \)
- \( \text{Pb(OH)}_2 \): \( 1.02 \times 10^{-5} \)
- \( \text{PbSO}_4 \): \( 3.61 \times 10^{-5} \)
- \( \text{PbI}_2 \): \( 1.26 \times 10^{-3} \)

7. **Conclusion**: The compound with the lowest solubility, and thus the lowest concentration of \( \text{Pb}^{2+} \) ions, is \( \text{PbCO}_3 \).

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