Question
The molar solubility $$\left( {{\text{in}}\,mol\,{L^{ - 1}}} \right)$$ of a sparingly soluble salt $$M{X_4}$$ is $$'s'.$$ The corresponding solubility product is $${K_{sp}}.$$ $$'s'$$ is given in term of $${K_{sp}}$$ by the relation :
A.
$$s = {\left( {256{K_{sp}}} \right)^{\frac{1}{5}}}$$
B.
$$s = {\left( {128{K_{sp}}} \right)^{\frac{1}{4}}}$$
C.
$$s = {\left( {\frac{{{K_{sp}}}}{{128}}} \right)^{\frac{1}{4}}}$$
D.
$$s = {\left( {\frac{{{K_{sp}}}}{{256}}} \right)^{\frac{1}{5}}}$$
Answer :
$$s = {\left( {\frac{{{K_{sp}}}}{{256}}} \right)^{\frac{1}{5}}}$$
Solution :
\[\begin{align}
& M{{X}_{4}}\rightleftharpoons {{\underset{S}{\mathop{M}}\,}^{4+}}+\underset{4S}{\mathop{4{{X}^{-}}}}\, \\
& {{K}_{sp}}=\left[ s \right]{{\left[ 4s \right]}^{4}}=256\,{{s}^{5}}\,\,\,\therefore \,\,s={{\left( \frac{{{K}_{sp}}}{256} \right)}^{\frac{1}{5}}} \\
\end{align}\]