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Question: Given, \(MY\) and \(N{Y_3}\), two nearly insoluble salts, have the same \({K_{sp}}\) values of \(6.2...

Given, MYMY and NY3N{Y_3}, two nearly insoluble salts, have the same Ksp{K_{sp}} values of 6.2×10−136.2 \times {10^{ - 13}} at room temperature. Which statement would be true in regards to MYMY and NY3N{Y_3}?
A) The molar solubility of MYMY and NY3N{Y_3} in water are identical.
B) The molar solubility of MYMY in water is less than that of NY3N{Y_3}.
C) The salts MYMYand NY3N{Y_3} are more soluble in 0.5 M KYKY than in pure water.
D) The addition of the salt KYKY to the solution of MYMY and NY3N{Y_3} will have no effect on their solubilities.

Explanation

Solution

Molar solubility is nothing but solubility. The solubility product, Ksp{K_{sp}}, is an equilibrium constant for a solid substance or salt dissolving in aqueous solution. Write the dissolution chemical reaction of MYMY and NY3N{Y_3} salts and then relate their Ksp{K_{sp}} values.

Complete answer:
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, Ksp{K_{sp}}, is an equilibrium constant for a solid substance or salt dissolving in aqueous solution. General dissolution reaction and hence Ksp{K_{sp}} expression of a salt is shown below:
AaBb(s)⇌aA+(aq)+bB−(aq){A_a}{B_b}(s) \rightleftharpoons a{A^ + }(aq) + b{B^ - }(aq)
Ksp=[A+]a[B−]b{K_{sp}} = {\left[ {{A^ + }} \right]^a}{\left[ {{B^ - }} \right]^b}
While calculating the Ksp{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 MYMY in water is:
MY⇌M++Y−MY \rightleftharpoons {M^ + } + {Y^ - }
Ksp of MY=[M+][Y−]{K_{sp}}{\text{ of }}MY = \left[ {{M^ + }} \right]\left[ {{Y^ - }} \right]
Let molar solubility of MYMY be SS, then
Ksp of MY=(S)(S)=S2{K_{sp}}{\text{ of }}MY = (S)(S) = {S^2}
Now it is given that MYMY have the Ksp{K_{sp}} values of 6.2×10−136.2 \times {10^{ - 13}}.
∴S2=6.2×10−13 S=6.2×10−13=7.88×10−7M \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 NY3N{Y_3} is:
NY3⇌N3++3Y−N{Y_3} \rightleftharpoons {N^{3 + }} + 3{Y^ - }
Let here molar solubility of NY3N{Y_3} be S′S', then
Ksp of NY3=[N3+][Y−]3=(S′)(3S′)3{K_{sp}}{\text{ of }}N{Y_3} = \left[ {{N^{3 + }}} \right]{\left[ {{Y^ - }} \right]^3} = (S'){(3S')^3}
Therefore, Ksp of NY3=27(S′)4{K_{sp}}{\text{ of }}N{Y_3} = 27{(S')^4}
But, Ksp{K_{sp}} value of NY3N{Y_3} is 6.2×10−136.2 \times {10^{ - 13}}
∴27(S′)4=6.2×10−13 ⇒S′=(6.2×10−1327)14=3.9×10−4M \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 MYMY is 7.88×10−7M7.88 \times {10^{ - 7}}M while molar solubility of NY3N{Y_3} is 3.9×10−4M3.9 \times {10^{ - 4}}M.
Thus, the molar solubility of MYMY in water is less than that of NY3N{Y_3}.

Hence, option B is correct.

Note:
Solubility product constant, Ksp{K_{sp}} represents the level at which a solute dissolves in solution. The higher is the Ksp{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.