The solubility product of `BaSO_(4)" at " 25^(@) C " is " 1.0 xx 10^(-9)`. What would be the concentration of `H_(2)SO_(4)` necessary to precipitate `BaSO_(4)` from a solution of `0.01" M Ba"^(+2)` ions

Options

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

Answer

To solve the problem, we need to determine the concentration of \( H_2SO_4 \) necessary to precipitate \( BaSO_4 \) from a solution containing \( 0.01 \, M \) \( Ba^{2+} \) ions, given that the solubility product \( K_{sp} \) of \( BaSO_4 \) is \( 1.0 \times 10^{-9} \).

### Step-by-Step Solution:

1. **Write the Dissociation Equation**:
The dissociation of \( BaSO_4 \) in water can be represented as:
\[ BaSO_4 (s) \rightleftharpoons Ba^{2+} (aq) + SO_4^{2-} (aq) \]

2. **Define the Solubility Product Expression**:
The solubility product \( K_{sp} \) for \( BaSO_4 \) is given by:
\[ K_{sp} = [Ba^{2+}][SO_4^{2-}] \]

3. **Substitute Known Values**:
We know that the concentration of \( Ba^{2+} \) ions is \( 0.01 \, M \). Let the concentration of \( SO_4^{2-} \) ions at equilibrium be \( S \). Thus, we can write:
\[ K_{sp} = (0.01)(S) = 1.0 \times 10^{-9} \]

4. **Solve for \( S \)**:
Rearranging the equation to find \( S \):
\[ S = \frac{1.0 \times 10^{-9}}{0.01} = 1.0 \times 10^{-7} \, M \]

5. **Relate \( S \) to \( H_2SO_4 \)**:
The dissociation of sulfuric acid \( H_2SO_4 \) in water is:
\[ H_2SO_4 (aq) \rightarrow 2H^+ (aq) + SO_4^{2-} (aq) \]
From this, we can see that one mole of \( H_2SO_4 \) produces one mole of \( SO_4^{2-} \). Therefore, the concentration of \( H_2SO_4 \) required to achieve a \( SO_4^{2-} \) concentration of \( S \) is also \( S \).

6. **Final Concentration of \( H_2SO_4 \)**:
Thus, the concentration of \( H_2SO_4 \) necessary to precipitate \( BaSO_4 \) is:
\[ [H_2SO_4] = S = 1.0 \times 10^{-7} \, M \]

### Conclusion:
The concentration of \( H_2SO_4 \) necessary to precipitate \( BaSO_4 \) from a solution of \( 0.01 \, M \) \( Ba^{2+} \) ions is \( 1.0 \times 10^{-7} \, M \).