Question

The solubility of electrolytes\[M{X_1}\], \[M{X_2}\] and \[M{X_3}\] is \[1x{10^{ - 3}}moles{\text{ }}per{\text{ }}litre\]. Hence their respective solubility products are:
A. ${10^{ - 6}},4 \times {10^{ - 9}},27 \times {10^{ - 12}}$
B. ${10^{ - 9}},4 \times {10^{ - 9}},32 \times {10^{ - 12}}$
C. ${10^{^{ - 9}}},8 \times {10^{ - 8}},32 \times {10^{ - 12}}$
D. none of these

Answer 1

Hint: An electrolyte is a compound which produces ions on dissolving in a solution. The product of the ion concentration is known as solubility product.

Complete step by step answer:
Let us first determine the solubility product of an electrolyte in a saturated solution.
A solution is said to be saturated when the solution remains in contact with the undissolved solute. The saturation stage is reached when no more solute goes into solution and hence the amount of solute in solution is constant.
The dissolved solute remains in equilibrium with the undissolved solute and the dissociated ions in solution. This is expressed as
$AB(dissolved) \rightleftharpoons AB(undissolved) \rightleftharpoons {A^ + } + {B^ - }$
The law of mass action states that the rate of a reaction is proportional to the concentrations of the products and reactants.
Thus rate of the reaction, $K = \dfrac{{[{A^ + }][{B^ - }]}}{{[AB]}}$
As the concentration of undissolved or unionized \[AB\] is fixed at definite temperature, so $[AB] = cons\tan t$ . Thus the rate equation is
\[[{A^ + }][{B^ - }] = Ksp\]
Hence solubility product is the product of the concentration of ions at a given temperature.
The solubility of the electrolytes is given as $s = 1 \times {10^{ - 3}}$

For\[M{X_1}\] , the dissociation is $M{X_1} \rightleftharpoons {M^ + } + {X^ - }$
Thus the solubility product for\[M{X_1}\] , $Ksp = [{M^ + }][{X^ - }]$
$Ksp = s \times s = [1 \times {10^{ - 3}}][1 \times {10^{ - 3}}] = {10^{ - 6}}.$

For\[M{X_2}\] , the dissociation is $M{X_2} \rightleftharpoons {M^ + } + 2{X^ - }$
Thus the solubility product for\[M{X_2}\] , $Ksp = [{M^ + }]{[{X^ - }]^2}$
$Ksp = s \times {[2s]^2} = 4{s^3} = 4{[1 \times {10^{ - 3}}]^3} = 4 \times {10^{ - 9}}.$

For\[M{X_3}\] , the dissociation is $M{X_3} \rightleftharpoons {M^ + } + 3{X^ - }$
Thus the solubility product for\[M{X_3}\] , $Ksp = [{M^ + }]{[{X^ - }]^3}$
$Ksp = s \times {[3s]^3} = 27{s^4} = 27{[1 \times {10^{ - 3}}]^4} = 27 \times {10^{ - 12}}.$

Therefore, the correct option is A. that is\[M{X_1}\], \[M{X_2}\] and \[M{X_3}\] are \[{10^{ - 6}},4 \times {10^{ - 9}},27 \times {10^{ - 12}}.\] .

Note: Solubility product should not be confused with ionic product. Ionic product is the product of the concentration of ions and is applicable to both saturated and unsaturated solutions. But solubility products are only applicable for saturated solutions. It maintains a dynamic equilibrium between the undissolved electrolyte and ions present in solution.