Q28 / 155 · WBJEE 2016 · Chemistry

The molar solubility (in mol $\mathrm{L}^{-1}$ ) of a sparingly soluble salt $M X_{4}$ is ${ }^{\prime} S$. The corresponding solubility product is $K_{\mathrm{sp}} . S$ in terms of $'K_{\text {ip }}$' s given by the relation

  • A
    $S=\left(\frac{K_{\mathrm{s} \rho}}{128}\right)^{1 / 4}$
  • Correct answer
    B
    $S=\left(\frac{K_{s p}}{256}\right)^{1 / 5}$
  • C
    $S=\left(256 K_{s p}\right)^{1 / 5}$
  • D
    $S=\left(128 K_{\mathrm{sp}}\right)^{1 / 4}$

Solution

$\because$ For a compound $A_{x}B_{y}$ $$ K_{s p}=x^{x} \cdot y^{y} \cdot S^{x+y} . \text { (where, } S=\text { solubility }) $$ For $M x_{4}$ $$ \begin{array}{l} K_{\mathrm{sp}}=1 \times(4)^{4} \cdot \mathrm{S}^{1+4} \\ K_{\mathrm{sp}}=256 \times \mathrm{S}^{5} \end{array} $$ $\therefore S=\left(\frac{K_{s}}{256}\right)^{1 / 5}$

Hence, the correct answer is option: 'B'.

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