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 find 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. **Identify the \( K_{sp} \) values:**
- For \( \text{BaSO}_4 \), \( K_{sp} = 1.0 \times 10^{-10} \)
- For \( \text{PbSO}_4 \), \( K_{sp} = 1.6 \times 10^{-8} \)

2. **Determine when \( \text{PbSO}_4 \) starts to precipitate:**
- The precipitation of \( \text{PbSO}_4 \) occurs when the ionic product \( Q \) equals \( K_{sp} \).
- The ionic product \( Q \) for \( \text{PbSO}_4 \) can be expressed as:
\[ Q = [\text{Pb}^{2+}][\text{SO}_4^{2-}] \]
- Given that the concentration of \( \text{Pb}^{2+} \) is 0.020 M, we can set up the equation:
\[ Q = (0.020)[\text{SO}_4^{2-}] = K_{sp} = 1.6 \times 10^{-8} \]

3. **Calculate the concentration of \( \text{SO}_4^{2-} \) when \( \text{PbSO}_4 \) starts to precipitate:**
\[ 0.020[\text{SO}_4^{2-}] = 1.6 \times 10^{-8} \]
\[ [\text{SO}_4^{2-}] = \frac{1.6 \times 10^{-8}}{0.020} = 8.0 \times 10^{-7} \, \text{M} \]

4. **Determine the concentration of \( \text{Ba}^{2+} \) when \( \text{PbSO}_4 \) starts to precipitate:**
- The ionic product \( Q \) for \( \text{BaSO}_4 \) is given by:
\[ Q = [\text{Ba}^{2+}][\text{SO}_4^{2-}] \]
- Setting \( Q \) equal to \( K_{sp} \):
\[ K_{sp} = 1.0 \times 10^{-10} = [\text{Ba}^{2+}](8.0 \times 10^{-7}) \]
- Solving for \( [\text{Ba}^{2+}] \):
\[ [\text{Ba}^{2+}] = \frac{1.0 \times 10^{-10}}{8.0 \times 10^{-7}} = 1.25 \times 10^{-4} \, \text{M} \]

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

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