The solubility product of a salt having general formula `MX_(2)` in water is `4xx10^(-12)`. The concentration of `M^(2+) ions` in the aqueous solution of the salt is:

Options

A. `2xx10^(-6)`M
B. `1xx10^(-4)`M
C. `1.6xx10^(-4)`M
D. `4.0xx10^(-6)`M

Explanation

To solve the problem, we need to determine the concentration of \( M^{2+} \) ions in a saturated solution of the salt \( MX_2 \) given its solubility product \( K_{sp} = 4 \times 10^{-12} \).

### Step-by-Step Solution:

1. **Write the Dissociation Equation**:
The salt \( MX_2 \) dissociates in water as follows:
\[ MX_2 \rightleftharpoons M^{2+} + 2X^{-} \]

2. **Define Solubility**:
Let the solubility of \( MX_2 \) be \( s \) mol/L. Then, at equilibrium:
- The concentration of \( M^{2+} \) ions will be \( s \).
- The concentration of \( X^{-} \) ions will be \( 2s \).

3. **Write the Expression for \( K_{sp} \)**:
The solubility product \( K_{sp} \) is given by the expression:
\[ K_{sp} = [M^{2+}][X^{-}]^2 \]
Substituting the concentrations from the dissociation:
\[ K_{sp} = (s)(2s)^2 = s \cdot 4s^2 = 4s^3 \]

4. **Substitute the Given \( K_{sp} \)**:
We know that \( K_{sp} = 4 \times 10^{-12} \). Therefore, we can set up the equation:
\[ 4s^3 = 4 \times 10^{-12} \]

5. **Solve for \( s^3 \)**:
Dividing both sides by 4:
\[ s^3 = 10^{-12} \]

6. **Calculate \( s \)**:
To find \( s \), take the cube root:
\[ s = (10^{-12})^{1/3} = 10^{-4} \]

7. **Determine the Concentration of \( M^{2+} \)**:
Since \( s \) represents the concentration of \( M^{2+} \) ions:
\[ [M^{2+}] = s = 10^{-4} \text{ mol/L} \]

### Final Answer:
The concentration of \( M^{2+} \) ions in the aqueous solution of the salt is:
\[ \boxed{1 \times 10^{-4} \text{ mol/L}} \]