If te solubility product of `BaSO_(4)` is `1.5 xx 10^(-9)` in water, its solubility in moles per litre, is

Options

A. `1.5 xx 10^(-9)`
B. `3.9 xx 10^(-5)`
C. `7.5 xx 10^(-5)`
D. `1.5 xx 10^(-5)`

Explanation

To determine the solubility of \( \text{BaSO}_4 \) in moles per liter given its solubility product (\( K_{sp} \)), we can follow these steps:

### Step-by-Step Solution:

1. **Write the Dissolution Reaction**:
When barium sulfate (\( \text{BaSO}_4 \)) dissolves in water, it dissociates into barium ions (\( \text{Ba}^{2+} \)) and sulfate ions (\( \text{SO}_4^{2-} \)):
\[ \text{BaSO}_4 (s) \rightleftharpoons \text{Ba}^{2+} (aq) + \text{SO}_4^{2-} (aq) \]

2. **Define Solubility**:
Let the solubility of \( \text{BaSO}_4 \) be \( S \) moles per liter. At equilibrium, the concentration of \( \text{Ba}^{2+} \) ions will be \( S \) and the concentration of \( \text{SO}_4^{2-} \) ions will also be \( S \).

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

4. **Substitute the Given \( K_{sp} \)**:
We know that \( K_{sp} \) for \( \text{BaSO}_4 \) 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 \( S \)**:
First, calculate the square root:
\[ S = \sqrt{1.5} \times \sqrt{10^{-9}} = \sqrt{1.5} \times 10^{-4.5} \]
Approximating \( \sqrt{1.5} \) gives approximately \( 1.22 \):
\[ S \approx 1.22 \times 10^{-4.5} = 1.22 \times 10^{-5} \, \text{moles per liter} \]

### Final Answer:
The solubility of \( \text{BaSO}_4 \) in water is approximately \( 1.22 \times 10^{-5} \) moles per liter.

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