Question

Solubility product expression of salt $M{X_4}$, which is sparingly soluble with a solubility $s$ can be given as:
A) $256{s^5}$
B) $16{s^3}$
C) $5s$
D) $25{s^4}$

Answer 1

The solubility product of a sparingly soluble salt at a given temperature is defined as, the product of molar concentration of its ions in a saturated solution, with each concentration term raised to the power equal to number of ions present in the chemical reaction representing the equilibrium of dissociation of one molecule of the salt.

Complete answer:
From the definition we understood what a solubility product is.
Now, applying this to a reaction
${A_x}{B_y} \rightleftharpoons {\text{x}}{{\text{A}}^{{\text{y}} + }} + {\text{y}}{{\text{B}}^{{\text{x}} - }}$
To find the solubility product we can use the equation,
Solubility product $\left( {{K_{sp}}} \right) = {\left[ {{A^{y + }}} \right]^x}{\left[ {{B^{x - }}} \right]^y}$ $\left( {{K_{sp}}} \right) = {\left[ {{A^{y + }}} \right]^x}{\left[ {{B^{x - }}} \right]^y}$
Where $[A]$ represents its concentration.
Here in question we have to find the solubility product of $M{X_4}$.
Given that solubility of $M{X_4}$ is $s\;mol/L$.
Its solubility reaction will be
$M{X_4} \rightleftharpoons {{\text{M}}^{4 + }} + 4{{\text{X}}^ - }$
From the stoichiometry of the reaction, we can deduct that,
1 mole of $M{X_4}$ gives 1 mole of ${{\text{M}}^{4 + }}$ and 4 mole of ${{\text{X}}^ - }$
So it is given that solubility of $M{X_4}$ is $s\;mol/L$
Therefore the solubility of ${{\text{M}}^{4 + }}$ will be $s\;mol/L$ and solubility of ${{\text{X}}^ - }$ will be $4s\;mol/L$.
Since we now know the solubility of its ions, we can find the solubility product $\left( {{K_{sp}}} \right)$ of it.
Solubility product $(K_{sp}) = [{M^{(4 + )}}]{[{X^ - }]^4}$
We know that $\left[ {{{\text{M}}^{4 + }}} \right] = s$ and ${\left[ {{X^ - }} \right]^4} = 4s$
Substituting that in the equation we get,
Solubility product $(K_{sp}) = (s){(4s)^4}$
$\Rightarrow \;{K_{sp}} = 256{s^5}$
So the solubility product of salt $M{X_4}$ is $256{s^5}$.

So option (a) is correct.

Note: Both ionic product and solubility product are represented by the same type of expression with the difference that ionic product is applicable to solution of any concentration whether saturated or unsaturated whereas solubility product is applicable to saturated solution only. Thus, we can say that solubility products are the ionic products of saturated solution.