The `K_(sp)` for `X_(2) SO_(4)` at `25^(@) (X^(+)` is a monovalent ion) is `3.2 xx 10^(-5)` The maximum concentration of `X^(+)` that could be attained in a saturated solution of this solid at `25^@C` is

Options

A. `4xx10^(-2)`M
B. `2.89xx10^(-4)`M
C. `3xx10^(-3)M`
D. `6xx10^(-3)` M

Answer

To solve the problem, we need to determine the maximum concentration of the monovalent ion \(X^+\) in a saturated solution of \(X_2SO_4\) given its solubility product (\(K_{sp}\)).

### Step-by-Step Solution:

1. **Write the Dissociation Equation**:
The dissociation of \(X_2SO_4\) in water can be represented as:
\[ X_2SO_4 (s) \rightleftharpoons 2X^+ (aq) + SO_4^{2-} (aq) \]

2. **Define Solubility**:
Let the solubility of \(X_2SO_4\) be \(S\) mol/L. When \(X_2SO_4\) dissolves, it produces:
- \(2S\) mol/L of \(X^+\)
- \(S\) mol/L of \(SO_4^{2-}\)

3. **Write the Expression for \(K_{sp}\)**:
The solubility product (\(K_{sp}\)) is given by the expression:
\[ K_{sp} = [X^+]^2 \cdot [SO_4^{2-}] \]
Substituting the concentrations in terms of \(S\):
\[ K_{sp} = (2S)^2 \cdot (S) = 4S^2 \cdot S = 4S^3 \]

4. **Substitute the Given \(K_{sp}\)**:
We know that \(K_{sp} = 3.2 \times 10^{-5}\):
\[ 4S^3 = 3.2 \times 10^{-5} \]

5. **Solve for \(S\)**:
Rearranging the equation gives:
\[ S^3 = \frac{3.2 \times 10^{-5}}{4} = 0.8 \times 10^{-5} = 8.0 \times 10^{-6} \]
Now, take the cube root to find \(S\):
\[ S = \sqrt[3]{8.0 \times 10^{-6}} = 2 \times 10^{-2} \, \text{mol/L} \]

6. **Calculate Maximum Concentration of \(X^+\)**:
Since the concentration of \(X^+\) is \(2S\):
\[ [X^+] = 2S = 2 \times (2 \times 10^{-2}) = 4 \times 10^{-2} \, \text{mol/L} \]

### Final Answer:
The maximum concentration of \(X^+\) that could be attained in a saturated solution of \(X_2SO_4\) at \(25^\circ C\) is:
\[ \boxed{4 \times 10^{-2} \, \text{mol/L}} \]