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Step-by-Step Solution
1. Understanding the Problem
We are given a sparingly soluble salt with general formula $MX_2$. Its solubility product ($K_{sp}$) in water is $4 \times 10^{-12}$. We need to find the concentration of $M^{2+}$ ions in the saturated solution of this salt.
2. Writing the Dissociation Equation
When $MX_2$ dissolves slightly in water, it dissociates according to the following equation:
$MX_2 \rightleftharpoons M^{2+} + 2X^{-}$
Let $s$ be the solubility of $MX_2$, i.e., the molar amount (in mol/L) of $MX_2$ that dissolves to reach equilibrium.
3. Expressing $K_{sp}$ in Terms of $s$
From the dissociation equation, the concentration of $M^{2+}$ will be $s$, while the concentration of $X^{-}$ will be $2s$. Therefore, the solubility product is given by:
$K_{sp} = [M^{2+}][X^{-}]^2 = (s)\bigl(2s\bigr)^2 = s \times 4s^2 = 4s^3$
4. Solving for $s$
We know $K_{sp} = 4 \times 10^{-12}$. Thus,
$4 \times 10^{-12} = 4s^3$
Dividing both sides by 4,
$10^{-12} = s^3$
Therefore,
$s = 10^{-4}$
5. Determining the Concentration of $M^{2+}$
Since $[M^{2+}] = s$,
$[M^{2+}] = 1.0 \times 10^{-4}\,\text{M}$
Hence, the concentration of $M^{2+}$ ions in the solution is $1.0 \times 10^{-4}\,\text{M}$.
