Solid `Na_(2)SO_(4)` is slowly added to a solution which is 0.020 M in `Ba(NO_(3))_(2)` and 0.020 M is `Pb(NO_(3))_(2)`. Assume that there is no increase in volume on adding `Na_(2)SO_(4)`. There preferential precipitation takes place. What is the concentration of `Ba^(2+)` when `PbSO_(4)` starts to precipitate? `[K_(sp)(BaSO_(4))=1.0xx10^(-10) and K_(sp)(PbSO_(4))=1.6xx10^(-8)]`

Options

A. `5.0xx10^(-9)M`
B. `8.0xx10^(-7)M`
C. `1.25xx10^(-4)M`
D. `1.95 xx10^(-8)M`

Answer

To solve the problem, we need to determine the concentration of \( \text{Ba}^{2+} \) when \( \text{PbSO}_4 \) starts to precipitate. We will use the solubility product constants (\( K_{sp} \)) for both \( \text{BaSO}_4 \) and \( \text{PbSO}_4 \).

### Step-by-Step Solution:

1. **Write the Dissociation Reactions:**
- For \( \text{PbSO}_4 \):
\[ \text{PbSO}_4 (s) \rightleftharpoons \text{Pb}^{2+} + \text{SO}_4^{2-} \]
- For \( \text{BaSO}_4 \):
\[ \text{BaSO}_4 (s) \rightleftharpoons \text{Ba}^{2+} + \text{SO}_4^{2-} \]

2. **Write the Expression for \( K_{sp} \):**
- For \( \text{PbSO}_4 \):
\[ K_{sp} = [\text{Pb}^{2+}][\text{SO}_4^{2-}] = 1.6 \times 10^{-8} \]
- For \( \text{BaSO}_4 \):
\[ K_{sp} = [\text{Ba}^{2+}][\text{SO}_4^{2-}] = 1.0 \times 10^{-10} \]

3. **Calculate the Concentration of \( \text{SO}_4^{2-} \) When \( \text{PbSO}_4 \) Starts to Precipitate:**
- Given that the concentration of \( \text{Pb}^{2+} \) is \( 0.020 \, M \):
\[ K_{sp} = [\text{Pb}^{2+}][\text{SO}_4^{2-}] \]
\[ 1.6 \times 10^{-8} = (0.020)[\text{SO}_4^{2-}] \]
- Rearranging gives:
\[ [\text{SO}_4^{2-}] = \frac{1.6 \times 10^{-8}}{0.020} = 8.0 \times 10^{-7} \, M \]

4. **Calculate the Concentration of \( \text{Ba}^{2+} \) Using \( K_{sp} \) for \( \text{BaSO}_4 \):**
- Now, using the \( K_{sp} \) expression for \( \text{BaSO}_4 \):
\[ K_{sp} = [\text{Ba}^{2+}][\text{SO}_4^{2-}] \]
\[ 1.0 \times 10^{-10} = [\text{Ba}^{2+}](8.0 \times 10^{-7}) \]
- Rearranging gives:
\[ [\text{Ba}^{2+}] = \frac{1.0 \times 10^{-10}}{8.0 \times 10^{-7}} = 1.25 \times 10^{-4} \, M \]

### Final Answer:
The concentration of \( \text{Ba}^{2+} \) when \( \text{PbSO}_4 \) starts to precipitate is \( 1.25 \times 10^{-4} \, M \).