`M_(2)SO_(4) (M^(o+)` is a monovalent metal ion) has a `K_(sp)` of `3.2 xx 10^(-6)`at `298 K`. The maximum concentration of `SO_(4)^(2-)` ion that could be attained in a saturated solution of this solid at `298 K` is

Options

A. `3 xx 10^(-3)M`
B. `7 xx 10^(-2)M`
C. `2.89 xx 10^(-4)M`
D. `2xx10^(-2)M`

Explanation

To find the maximum concentration of \( \text{SO}_4^{2-} \) ions in a saturated solution of \( \text{M}_2\text{SO}_4 \) at 298 K, we can follow these steps:

### Step-by-Step Solution:

1. **Write the Dissociation Reaction**:
The dissociation of \( \text{M}_2\text{SO}_4 \) in water can be represented as:
\[ \text{M}_2\text{SO}_4 (s) \rightleftharpoons 2 \text{M}^+ (aq) + \text{SO}_4^{2-} (aq) \]

2. **Define Solubility**:
Let the solubility of \( \text{M}_2\text{SO}_4 \) be \( S \) mol/L. According to the stoichiometry of the dissociation:
- The concentration of \( \text{M}^+ \) ions will be \( 2S \).
- The concentration of \( \text{SO}_4^{2-} \) ions will be \( S \).

3. **Write the Expression for \( K_{sp} \)**:
The solubility product constant \( K_{sp} \) is given by:
\[ K_{sp} = [\text{M}^+]^2 [\text{SO}_4^{2-}] \]
Substituting the concentrations in terms of \( S \):
\[ K_{sp} = (2S)^2 \cdot (S) = 4S^3 \]

4. **Substitute the Given \( K_{sp} \)**:
We know that \( K_{sp} = 3.2 \times 10^{-6} \). Therefore, we can set up the equation:
\[ 4S^3 = 3.2 \times 10^{-6} \]

5. **Solve for \( S^3 \)**:
Rearranging the equation gives:
\[ S^3 = \frac{3.2 \times 10^{-6}}{4} = 0.8 \times 10^{-6} \]

6. **Calculate \( S \)**:
To find \( S \), we take the cube root:
\[ S = \sqrt[3]{0.8 \times 10^{-6}} \]

7. **Final Calculation**:
Evaluating the cube root:
\[ S \approx 2 \times 10^{-2} \text{ mol/L} \]

### Conclusion:
Thus, the maximum concentration of \( \text{SO}_4^{2-} \) ions that could be attained in a saturated solution of \( \text{M}_2\text{SO}_4 \) at 298 K is:
\[ \boxed{2 \times 10^{-2} \text{ mol/L}} \]