Question

The molar solubility of a sparingly soluble salt ${{M}_{x}}{{X}_{y}}$ in its saturated solution at a given temperature is S. The correct relation between S and ${{K}_{sp}}$(solubility product) of it is:
(A) $S={{\left( \dfrac{{{K}_{sp}}}{{{x}^{x}}.{{y}^{y}}} \right)}^{\dfrac{1}{x+y}}}$
(B) $S=\left( \dfrac{{{K}_{sp}}}{{{x}^{x}}.{{y}^{y}}} \right)$
(C) $S={{\left( \dfrac{{{K}_{sp}}}{{{x}^{x}}.{{y}^{y}}} \right)}^{x+y}}$
(D) $S=\dfrac{{{x}^{x}}.{{y}^{y}}}{{{K}_{sp}}}$

Answer 1

Hint: To solve this firstly write down the dissociation of the salt into corresponding ions. Find the concentration of the ions and write it in terms of molar solubility. Write the solubility product equation using these concentrations and then rearrange to get the correct answer.

Complete step by step solution:
In the question, the molar solubility of a sparingly soluble salt ${{M}_{x}}{{X}_{y}}$ in its saturated solution at a given temperature is S. We can write the dissociation of the salt in a solution as-
${{M}_{x}}{{X}_{y}}\rightleftharpoons x{{M}^{y+}}+y{{X}^{x-}}$
Following the stoichiometry, these will be the dissociated products. If y ‘X’ atoms are attached to M thus, dissociating M will gain a charge equal to the number of electrons lost which is y and similarly X will gain a charge of –x.
Now, the molar solubility is given to us which is S.
Therefore, concentration of M will be, $\left[ M \right]=xS$ and similarly,
Concentration of X will be, $\left[ X \right]=yS$
Now, we know that Ksp is the solubility product of a reaction. It is the measure for the degree of dissociation of the ions in a solution. It is written in terms of concentration of the ions-
\[Ksp=\left[ cation \right]\times \left[ anion \right]\]
I.e. Ksp is the product of the concentration of anions and cations present in the solution.
So for the dissociation of the above salt we can write that-
\[Ksp={{\left[ M \right]}^{x}}\times {{\left[ X \right]}^{y}}={{\left[ xS \right]}^{x}}\times {{\left[ yS \right]}^{y}}\]
If we separate the solubility, S from both the concentration, we can write that-\[Ksp=\left( {{x}^{x}}\times {{y}^{y}} \right){{S}^{x+y}}\]
Now, rearranging the equation we will get-
\[\begin{align}
 & {{S}^{x+y}}=\dfrac{{{K}_{sp}}}{\left( {{x}^{x}}\times {{y}^{y}} \right)} \\
& or,S={{\left[ \dfrac{{{K}_{sp}}}{\left( {{x}^{x}}\times {{y}^{y}} \right)} \right]}^{\dfrac{1}{x+y}}} \\
\end{align}\]
From the above calculations we can see that the solubility product and the molar solubility is related as- \[S={{\left[ \dfrac{{{K}_{sp}}}{\left( {{x}^{x}}\times {{y}^{y}} \right)} \right]}^{\dfrac{1}{x+y}}}\] .

Therefore, the correct answer is option [A] $S={{\left( \dfrac{{{K}_{sp}}}{{{x}^{x}}.{{y}^{y}}} \right)}^{\dfrac{1}{x+y}}}$.

Note: The solubility product can be calculated by using the concentration of the ions of the dissociated salt as we did in the above discussion. It is dependent on other surrounding factors like temperature, pressure and nature of the electrolyte. These factors affect the concentration of the ions and thus affect the solubility product too.