The solubility product of a salt having general formula `MX_2` in water ` 4xx 10 ^(-12) ` .The concentration of `M^(2+) ` ions in the aqueous solution of the salt is

Options

A. ` 4.0xx 10 ^(-10) M`
B. ` 1.6 xx 10 ^(-4) M`
C. ` 1.0 xx 10^(-4) M`
D. ` 2.0 xx 10 ^(-6) M`

Explanation

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

### Step-by-Step Solution:

1. **Understanding the Dissociation of the Salt:**
The salt \( MX_2 \) dissociates in water as follows:
\[ MX_2 (s) \rightleftharpoons M^{2+} (aq) + 2X^{-} (aq) \]
From this equation, we can see that for every 1 mole of \( MX_2 \) that dissolves, it produces 1 mole of \( M^{2+} \) ions and 2 moles of \( X^{-} \) ions.

2. **Letting the Solubility be \( s \):**
Let the solubility of \( MX_2 \) in water be \( s \) moles per liter. Therefore:
- The concentration of \( M^{2+} \) ions will be \( s \) moles per liter.
- The concentration of \( X^{-} \) ions will be \( 2s \) moles per liter.

3. **Writing the Expression for \( K_{sp} \):**
The solubility product \( K_{sp} \) for the salt can be expressed as:
\[ K_{sp} = [M^{2+}][X^{-}]^2 \]
Substituting the concentrations we found:
\[ K_{sp} = (s)(2s)^2 = s \cdot 4s^2 = 4s^3 \]

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

5. **Solving for \( s \):**
Dividing both sides by 4:
\[ s^3 = 10^{-12} \]
Taking the cube root of both sides:
\[ s = (10^{-12})^{1/3} = 10^{-4} \text{ moles per liter} \]

6. **Finding the Concentration of \( M^{2+} \):**
Since the concentration of \( M^{2+} \) ions is equal to \( s \):
\[ [M^{2+}] = s = 1.0 \times 10^{-4} \text{ moles per liter} \]

### Final Answer:
The concentration of \( M^{2+} \) ions in the aqueous solution of the salt \( MX_2 \) is:
\[ 1.0 \times 10^{-4} \text{ M} \]