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Step-by-Step Solution
1. Understanding the Problem
We have a sparingly soluble salt $MX_4$ that dissociates in water to give ions. The solubility product ($K_{sp}$) is defined in terms of the molar solubility $s$. We need to express $s$ in terms of $K_{sp}$.
2. Writing the Dissociation Equation
When $MX_4$ dissolves, it dissociates as follows:
$MX_4 \longrightarrow M^{4+} + 4 X^{-}
$
If the molar solubility (amount that dissolves) is $s$ mol L$^{-1}$, then:
$[M^{4+}] = s$
$[X^-] = 4s$
3. Expressing the Solubility Product
The solubility product $K_{sp}$ for the salt $MX_4$ is given by the product of the equilibrium concentrations of its ions, raised to their respective stoichiometric coefficients:
$K_{sp} = [M^{4+}] \times [X^-]^4
$
Substituting the expressions for the ion concentrations, we get:
$K_{sp} = (s)\,(4s)^4
$
4. Simplifying the Expression
Now simplify $(4s)^4$:
$(4s)^4 = 4^4 \, s^4 = 256\,s^4
$
Hence,
$K_{sp} = s \times 256\,s^4 = 256\,s^5
$
5. Solving for s
To find $s$ in terms of $K_{sp}$, we rearrange the equation:
$s^5 = \frac{K_{sp}}{256}
$
$s = \Bigl(\frac{K_{sp}}{256}\Bigr)^{\frac{1}{5}}
$
6. Conclusion
The molar solubility $s$ of the sparingly soluble salt $MX_4$ is given by:
$s = \left(\frac{K_{sp}}{256}\right)^{\frac{1}{5}}
$
