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# Given, $MY$ and $N{Y_3}$, two nearly insoluble salts, have the same ${K_{sp}}$ values of $6.2 \times {10^{ - 13}}$ at room temperature. Which statement would be true in regards to $MY$ and $N{Y_3}$?A) The molar solubility of $MY$ and $N{Y_3}$ in water are identical.B) The molar solubility of $MY$ in water is less than that of $N{Y_3}$.C) The salts $MY$and $N{Y_3}$ are more soluble in 0.5 M $KY$ than in pure water.D) The addition of the salt $KY$ to the solution of $MY$ and $N{Y_3}$ will have no effect on their solubilities.  Verified
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Hint: Molar solubility is nothing but solubility. The solubility product, ${K_{sp}}$, is an equilibrium constant for a solid substance or salt dissolving in aqueous solution. Write the dissolution chemical reaction of $MY$ and $N{Y_3}$ salts and then relate their ${K_{sp}}$ values.

Solubility, usually means molar solubility, of a salt is expressed as the concentration of ions or the dissolved salt in a saturated solution. The solubility product, ${K_{sp}}$, is an equilibrium constant for a solid substance or salt dissolving in aqueous solution. General dissolution reaction and hence ${K_{sp}}$ expression of a salt is shown below:
${A_a}{B_b}(s) \rightleftharpoons a{A^ + }(aq) + b{B^ - }(aq)$
${K_{sp}} = {\left[ {{A^ + }} \right]^a}{\left[ {{B^ - }} \right]^b}$
While calculating the ${K_{sp}}$ value, it is necessary that concentration of each ion must be raised to the power of its stoichiometric coefficient.
- Now, dissolution reaction of the salt $MY$ in water is:
$MY \rightleftharpoons {M^ + } + {Y^ - }$
${K_{sp}}{\text{ of }}MY = \left[ {{M^ + }} \right]\left[ {{Y^ - }} \right]$
Let molar solubility of $MY$ be $S$, then
${K_{sp}}{\text{ of }}MY = (S)(S) = {S^2}$
Now it is given that $MY$ have the ${K_{sp}}$ values of $6.2 \times {10^{ - 13}}$.
$\begin{gathered} \therefore {S^2} = 6.2 \times {10^{ - 13}} \\ S = \sqrt {6.2 \times {{10}^{ - 13}}} = 7.88 \times {10^{ - 7}}M \\ \end{gathered}$
- Dissolution reaction of salt $N{Y_3}$ is:
$N{Y_3} \rightleftharpoons {N^{3 + }} + 3{Y^ - }$
Let here molar solubility of $N{Y_3}$ be $S'$, then
${K_{sp}}{\text{ of }}N{Y_3} = \left[ {{N^{3 + }}} \right]{\left[ {{Y^ - }} \right]^3} = (S'){(3S')^3}$
Therefore, ${K_{sp}}{\text{ of }}N{Y_3} = 27{(S')^4}$
But, ${K_{sp}}$ value of $N{Y_3}$ is $6.2 \times {10^{ - 13}}$
$\begin{gathered} \therefore 27{(S')^4} = 6.2 \times {10^{ - 13}} \\ \Rightarrow S' = {\left( {\dfrac{{6.2 \times {{10}^{ - 13}}}}{{27}}} \right)^{\dfrac{1}{4}}} = 3.9 \times {10^{ - 4}}M \\ \end{gathered}$
Now, from the above calculation, we get that molar solubility of $MY$ is $7.88 \times {10^{ - 7}}M$ while molar solubility of $N{Y_3}$ is $3.9 \times {10^{ - 4}}M$.
Thus, the molar solubility of $MY$ in water is less than that of $N{Y_3}$.

Hence, option B is correct.

Note:
Solubility product constant, ${K_{sp}}$ represents the level at which a solute dissolves in solution. The higher is the ${K_{sp}}$ value, the more soluble a substance is. A key point to note is that the relation between the solubility and solubility product depends on the stoichiometry of the dissolution reaction.