The solubility product of `BaSO_(4)" is " 1.5 xx 10^(-9)` The precipitation in a `0.01 M Ba^(+2)` solution will start , on adding `H_(2)SO_(4)` of concentration

Options

A. `10^(-9)`M
B. `10^(-8)M`
C. `10^(-7)M`
D. `10^(-6)M`

Answer

To solve the problem, we need to determine the concentration of \( H_2SO_4 \) at which precipitation of \( BaSO_4 \) will start in a \( 0.01 \, M \) \( Ba^{2+} \) solution, given that the solubility product (\( K_{sp} \)) of \( BaSO_4 \) is \( 1.5 \times 10^{-9} \).

### Step-by-Step Solution:

1. **Write the Dissociation Equation for \( BaSO_4 \)**:
\[ BaSO_4 (s) \rightleftharpoons Ba^{2+} (aq) + SO_4^{2-} (aq) \]

2. **Define the Solubility Product Expression**:
The solubility product (\( K_{sp} \)) is given by:
\[ K_{sp} = [Ba^{2+}][SO_4^{2-}] \]

3. **Substitute Known Values**:
We know that the concentration of \( Ba^{2+} \) is \( 0.01 \, M \). Let the concentration of \( SO_4^{2-} \) be \( S \). Thus, we can write:
\[ K_{sp} = (0.01)(S) \]
Given \( K_{sp} = 1.5 \times 10^{-9} \), we can set up the equation:
\[ 1.5 \times 10^{-9} = (0.01)(S) \]

4. **Solve for \( S \)**:
Rearranging the equation to find \( S \):
\[ S = \frac{1.5 \times 10^{-9}}{0.01} = 1.5 \times 10^{-7} \, M \]

5. **Relate \( S \) to \( H_2SO_4 \)**:
Since \( H_2SO_4 \) dissociates into \( 2H^+ \) and \( SO_4^{2-} \), the concentration of \( SO_4^{2-} \) produced by \( H_2SO_4 \) is equal to its concentration. Therefore, the concentration of \( H_2SO_4 \) required to start precipitation is also \( S \):
\[ [H_2SO_4] = S = 1.5 \times 10^{-7} \, M \]

### Conclusion:
The precipitation of \( BaSO_4 \) will start when the concentration of \( H_2SO_4 \) is \( 1.5 \times 10^{-7} \, M \).

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