Question

$K_{sp}$ of $M(\mathrm{OH})_{x}$ is $27 \times 10^{-12}$ and its solubility in water is $10^{-3} \mathrm{~mol}$ per litre. Find the value of $x$.

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Answer

To solve for the value of $x$ in the dissolution of $M(\mathrm{OH})x$, given $K{sp} = 27 \times 10^{-12}$ and solubility $S = 10^{-3}$ mol/L, let's go through the steps systematically:

  1. Write the Dissociation Equation: $$ M(\mathrm{OH})_x \rightarrow M^{x+} + x ,\mathrm{OH}^- $$

  2. Express Ionic Concentrations in Terms of Solubility:

    • The concentration of $M^{x+}$ ions will be $S , \text{mol/L}$.
    • The concentration of $\mathrm{OH}^-$ ions will be $xS , \text{mol/L}$.
  3. Set Up the $K_{sp}$ Expression: $$ K_{sp} = [M^{x+}] \cdot [\mathrm{OH}^-]^x = S \cdot (xS)^x $$

    • Substitute $[M^{x+}] = S$ and $[\mathrm{OH}^-] = xS$: $$ K_{sp} = S \cdot (xS)^x = S \cdot x^x \cdot S^x = x^x \cdot S^{x+1} $$
  4. Insert Known Values: $$ 27 \times 10^{-12} = x^x \cdot (10^{-3})^{x+1} $$

  5. Simplify the $K_{sp}$ Equation: $$ 27 \times 10^{-12} = x^x \cdot 10^{-3(x+1)} $$

  6. Compare Terms:

    • The term $10^{-3(x+1)}$ indicates that our exponent must match, so: $$ 27 \times 10^{-12} = x^x \cdot 10^{-3x-3} $$ $$ 27 = x^x $$
    • We infer that $x^x = 27$. Testing common values:
      • $3^3 = 27$

Conclusion:

Since $3^3 = 27$ satisfies the equation, we find that $x = 3$.

Thus, the value of $x$ is ( \boxed{3} ).


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