What is minimum concentration of `SO_(4)^(2-)` required to precipitate `BaSO_(4)` in solution containing `1 xx 10^(-4)` mole of `Ba^(2+)` ? (`K_(sp)` of `BaSO_(4) = 4 xx 10^(-10)`)

Options

A. `4xx10^(-10)M`
B. `2xx10^(-10)M`
C. `4xx10^(-6)M`
D. `2xx10^(-3)M`

Explanation

To find the minimum concentration of \( \text{SO}_4^{2-} \) required to precipitate \( \text{BaSO}_4 \) in a solution containing \( 1 \times 10^{-4} \) moles of \( \text{Ba}^{2+} \), we can use the solubility product constant (\( K_{sp} \)) of \( \text{BaSO}_4 \).

### Step-by-Step Solution:

1. **Write the Dissolution Equation**:
The dissolution of barium sulfate can be represented as:
\[ \text{BaSO}_4 (s) \rightleftharpoons \text{Ba}^{2+} (aq) + \text{SO}_4^{2-} (aq) \]

2. **Define the Solubility Product (\( K_{sp} \))**:
The solubility product expression for \( \text{BaSO}_4 \) is given by:
\[ K_{sp} = [\text{Ba}^{2+}][\text{SO}_4^{2-}] \]
where \( K_{sp} \) for \( \text{BaSO}_4 \) is given as \( 4 \times 10^{-10} \).

3. **Substitute Known Values**:
We know the concentration of \( \text{Ba}^{2+} \) is \( 1 \times 10^{-4} \) moles in a 1-liter solution. Therefore, the concentration of \( \text{Ba}^{2+} \) is:
\[ [\text{Ba}^{2+}] = 1 \times 10^{-4} \, \text{M} \]

4. **Set Up the Equation**:
To find the minimum concentration of \( \text{SO}_4^{2-} \) required to start precipitation, we set up the equation:
\[ K_{sp} = [\text{Ba}^{2+}][\text{SO}_4^{2-}] \]
Substituting the values we have:
\[ 4 \times 10^{-10} = (1 \times 10^{-4})[\text{SO}_4^{2-}] \]

5. **Solve for \( [\text{SO}_4^{2-}] \)**:
Rearranging the equation to solve for \( [\text{SO}_4^{2-}] \):
\[ [\text{SO}_4^{2-}] = \frac{K_{sp}}{[\text{Ba}^{2+}]} \]
Substituting the known values:
\[ [\text{SO}_4^{2-}] = \frac{4 \times 10^{-10}}{1 \times 10^{-4}} = 4 \times 10^{-6} \, \text{M} \]

6. **Conclusion**:
The minimum concentration of \( \text{SO}_4^{2-} \) required to precipitate \( \text{BaSO}_4 \) is:
\[ [\text{SO}_4^{2-}] > 4 \times 10^{-6} \, \text{M} \]