The solubility product of barium sulphate is `1.5 xx 10^(-9)` at `18^(@)C` . Its solubility in water at `18^(@)C` is

Options

A. `1.5 xx 10^(-9)` mol `L^(-1)`
B. `1.5 xx 10^(-5) mol L^(-1)`
C. `3.9 xx 10^(-9) mol L^(-1)`
D. `3.9 xx 10^(-5) mol L^(-1)`

Explanation

To find the solubility of barium sulfate (BaSO₄) in water at 18°C, we can use the solubility product constant (Ksp) given as \(1.5 \times 10^{-9}\).

### Step-by-Step Solution:

1. **Write the Dissociation Equation**:
Barium sulfate dissociates in water as follows:
\[ \text{BaSO}_4 (s) \rightleftharpoons \text{Ba}^{2+} (aq) + \text{SO}_4^{2-} (aq) \]

2. **Define the Solubility**:
Let the solubility of BaSO₄ in water be \(S\) (in mol/L). At equilibrium, the concentrations of the ions will be:
- \([\text{Ba}^{2+}] = S\)
- \([\text{SO}_4^{2-}] = S\)

3. **Write the Expression for Ksp**:
The solubility product (Ksp) expression for BaSO₄ is given by:
\[ K_{sp} = [\text{Ba}^{2+}][\text{SO}_4^{2-}] = S \cdot S = S^2 \]

4. **Substitute the Value of Ksp**:
We know that \(K_{sp} = 1.5 \times 10^{-9}\). Therefore, we can set up the equation:
\[ S^2 = 1.5 \times 10^{-9} \]

5. **Solve for S**:
To find \(S\), take the square root of both sides:
\[ S = \sqrt{1.5 \times 10^{-9}} \]

6. **Calculate the Value**:
Performing the calculation:
\[ S \approx 3.87 \times 10^{-5} \text{ mol/L} \]

7. **Conclusion**:
The solubility of barium sulfate in water at 18°C is approximately:
\[ S \approx 3.87 \times 10^{-5} \text{ mol/L} \]

### Final Answer:
The solubility of barium sulfate in water at 18°C is approximately \(3.87 \times 10^{-5} \text{ mol/L}\).

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