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Question: The solubility of electrolytes\[M{X_1}\], \[M{X_2}\] and \[M{X_3}\] is \[1x{10^{ - 3}}moles{\text{ }...

The solubility of electrolytesMX1M{X_1}, MX2M{X_2} and MX3M{X_3} is 1x10−3moles per litre1x{10^{ - 3}}moles{\text{ }}per{\text{ }}litre. Hence their respective solubility products are:
A. 10−6,4×10−9,27×10−12{10^{ - 6}},4 \times {10^{ - 9}},27 \times {10^{ - 12}}
B. 10−9,4×10−9,32×10−12{10^{ - 9}},4 \times {10^{ - 9}},32 \times {10^{ - 12}}
C. 10−9,8×10−8,32×10−12{10^{^{ - 9}}},8 \times {10^{ - 8}},32 \times {10^{ - 12}}
D. none of these

Explanation

Solution

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)⇌AB(undissolved)⇌A++B−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=[A+][B−][AB]K = \dfrac{{[{A^ + }][{B^ - }]}}{{[AB]}}
As the concentration of undissolved or unionized ABAB is fixed at definite temperature, so [AB]=constan⁡t[AB] = cons\tan t . Thus the rate equation is
[A+][B−]=Ksp[{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=1×10−3s = 1 \times {10^{ - 3}}

ForMX1M{X_1} , the dissociation is MX1⇌M++X−M{X_1} \rightleftharpoons {M^ + } + {X^ - }
Thus the solubility product forMX1M{X_1} , Ksp=[M+][X−]Ksp = [{M^ + }][{X^ - }]
Ksp=s×s=[1×10−3][1×10−3]=10−6.Ksp = s \times s = [1 \times {10^{ - 3}}][1 \times {10^{ - 3}}] = {10^{ - 6}}.

ForMX2M{X_2} , the dissociation is MX2⇌M++2X−M{X_2} \rightleftharpoons {M^ + } + 2{X^ - }
Thus the solubility product forMX2M{X_2} , Ksp=[M+][X−]2Ksp = [{M^ + }]{[{X^ - }]^2}
Ksp=s×[2s]2=4s3=4[1×10−3]3=4×10−9.Ksp = s \times {[2s]^2} = 4{s^3} = 4{[1 \times {10^{ - 3}}]^3} = 4 \times {10^{ - 9}}.

ForMX3M{X_3} , the dissociation is MX3⇌M++3X−M{X_3} \rightleftharpoons {M^ + } + 3{X^ - }
Thus the solubility product forMX3M{X_3} , Ksp=[M+][X−]3Ksp = [{M^ + }]{[{X^ - }]^3}
Ksp=s×[3s]3=27s4=27[1×10−3]4=27×10−12.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 isMX1M{X_1}, MX2M{X_2} and MX3M{X_3} are 10−6,4×10−9,27×10−12.{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.