If the molar solubility (in $\mathrm{mol} \mathrm{L}^{-1}$ ) of a sparingly soluble salt $A B_4$ is $S$ and…

If the molar solubility (in $\mathrm{mol} \mathrm{L}^{-1}$ ) of a sparingly soluble salt $A B_4$ is $S$ and the corresponding solubility product is $K_{\text {sp }}$, then $s$ in terms of $K_{\mathrm{sp}}$ is given by the relation
  1. $S=\left(\frac{K_{\mathrm{sp}}}{128}ight)^{1 / 4}$
  2. $S=\left(\frac{K_{\mathrm{sp}}}{256}ight)^{1 / 5}$
  3. $S=\left(256 K_{\mathrm{sp}}ight)^{1 / 5}$
  4. $S=\left(128 K_{\mathrm{sp}}ight)^{1 / 4}$

Solution

For the reaction, $\begin{array}{ccc}A B_4(s) \stackrel{(a q)}{ightleftharpoons} & A^{4+}(a q)+4 B^{-}(a q) \\ - & S & 4 S\end{array}$ Molar solubility (in $\mathrm{mol} / \mathrm{L}$ ) of a sparingly soluble salt $A B_4$ is ' $S$ ' and corresponding solubility product is $K_{\mathrm{sp}}$. $S$ in term of $K_{\mathrm{sp}}$ is given by above relation is $$ \begin{aligned} & K_{\text {sp }}=\left[A^{4+}ight]\left[B^{-}ight]^4 \\ & K_{\text {sp }}=S \times(4 S)^4 \\ & K_{\text {sp }}=256 S^5 \end{aligned} $$ Hence, the molar solubility, $S=\left(\frac{K_{\mathrm{sp}}}{256}ight)^{\frac{1}{5}}$

Asked in: JEE-TOPICTESTS-CHEMISTRY

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