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) M`
B. `1.5 xx 10^(-5) M`
C. `3.87 xx 10^(-9) M `
D. `3.87 xx 10^(-5) M`

Explanation

To find the solubility of barium sulfate (BaSO₄) in water at 18°C, we can use the concept of solubility product (Ksp). The solubility product is defined as the product of the molar concentrations of the ions in a saturated solution of the salt, each raised to the power of their coefficients in the balanced equation.

### Step-by-Step Solution:

1. **Write the dissociation equation for BaSO₄:**
\[ \text{BaSO}_4 (s) \rightleftharpoons \text{Ba}^{2+} (aq) + \text{SO}_4^{2-} (aq) \]

2. **Define the solubility (S):**
Let the solubility of BaSO₄ in water be \( S \) moles per liter. Therefore, at equilibrium:
- The concentration of \( \text{Ba}^{2+} \) ions = \( S \)
- The concentration of \( \text{SO}_4^{2-} \) ions = \( S \)

3. **Write the expression for the solubility product (Ksp):**
\[ K_{sp} = [\text{Ba}^{2+}][\text{SO}_4^{2-}] = S \times S = S^2 \]

4. **Substitute the given Ksp value:**
We know that \( K_{sp} \) for BaSO₄ at 18°C is \( 1.5 \times 10^{-9} \):
\[ S^2 = 1.5 \times 10^{-9} \]

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 of S:**
\[ S \approx 3.87 \times 10^{-5} \text{ moles per liter} \]

7. **Round the answer:**
Rounding to one decimal place, we can express the solubility as:
\[ S \approx 3.9 \times 10^{-5} \text{ moles per liter} \]

### Final Answer:
The solubility of barium sulfate (BaSO₄) in water at 18°C is approximately \( 3.9 \times 10^{-5} \) moles per liter.