If `K_(sp)` for `HgSO_(4)` is `6.4xx10^(-5)`, then solubility of this substance in mole per `m^(3)` is

Options

A. `8xx10^(-3)`
B. `6.4xx10^(-5)`
C. `8xx10^(-6)`
D. None of these

Explanation

To find the solubility of \( \text{HgSO}_4 \) in moles per cubic meter given its solubility product constant \( K_{sp} \), we can follow these steps:

### Step-by-Step Solution:

1. **Write the Dissociation Equation:**
The dissociation of \( \text{HgSO}_4 \) in water can be represented as:
\[ \text{HgSO}_4 (s) \rightleftharpoons \text{Hg}^{2+} (aq) + \text{SO}_4^{2-} (aq) \]

2. **Define Solubility:**
Let the solubility of \( \text{HgSO}_4 \) be \( S \) moles per cubic meter. Therefore, at equilibrium:
- The concentration of \( \text{Hg}^{2+} \) ions will be \( S \).
- 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:
\[ K_{sp} = [\text{Hg}^{2+}][\text{SO}_4^{2-}] \]
Substituting the concentrations:
\[ K_{sp} = S \times S = S^2 \]

4. **Substitute the Given \( K_{sp} \):**
We know that \( K_{sp} = 6.4 \times 10^{-5} \):
\[ S^2 = 6.4 \times 10^{-5} \]

5. **Solve for \( S \):**
To find \( S \), take the square root of both sides:
\[ S = \sqrt{6.4 \times 10^{-5}} \]

6. **Calculate \( S \):**
Performing the calculation:
\[ S \approx \sqrt{6.4} \times \sqrt{10^{-5}} \approx 2.5298 \times 10^{-2.5} \approx 8.0 \times 10^{-3} \text{ moles/m}^3 \]

### Final Answer:
The solubility of \( \text{HgSO}_4 \) is approximately:
\[ S \approx 8.0 \times 10^{-3} \text{ moles/m}^3 \]