JEE Challenger
More from Equilibrium
ChemMedAdv 2008 • Paper 2

Order of Solubilities from Solubility Product Constants

Solubility product constants (KspK_{sp}) of salts of types MX\text{MX}, MX2\text{MX}_2 and M3X\text{M}_3\text{X} at temperature 'TT' are 4.0×10−84.0 \times 10^{-8}, 3.2×10−143.2 \times 10^{-14} and 2.7×10−152.7 \times 10^{-15}, respectively. Solubilities (mol dm−3\text{mol dm}^{-3}) of the salts at temperature 'TT' are in the order

Options

A

MX>MX2>M3X\text{MX} > \text{MX}_2 > \text{M}_3\text{X}

B

M3X>MX2>MX\text{M}_3\text{X} > \text{MX}_2 > \text{MX}

C

MX2>M3X>MX\text{MX}_2 > \text{M}_3\text{X} > \text{MX}

D

MX>M3X>MX2\text{MX} > \text{M}_3\text{X} > \text{MX}_2

Correct

Topics & Concepts

EquilibriumSolubility Product

Step-by-Step Solution

To determine the order of solubilities of the given salts at temperature TT, we calculate the molar solubility (SS) for each type of salt using its corresponding solubility product constant (KspK_{sp}).

1. For salt of type MX\text{MX}:

The dissociation equilibrium in an aqueous solution is given by:
MX(s)⇌M+(aq)+X−(aq)\text{MX}(s) \rightleftharpoons \text{M}^+(aq) + \text{X}^-(aq)

If S1S_1 is the molar solubility of MX\text{MX}:
[M+]=S1and[X−]=S1[\text{M}^+] = S_1 \quad \text{and} \quad [\text{X}^-] = S_1

The expression for the solubility product is:
Ksp(MX)=[M+][X−]=S12K_{sp}(\text{MX}) = [\text{M}^+][\text{X}^-] = S_1^2

Given Ksp(MX)=4.0×10−8K_{sp}(\text{MX}) = 4.0 \times 10^{-8}:
S12=4.0×10−8S_1^2 = 4.0 \times 10^{-8}
S1=4.0×10−8=2.0×10−4 mol dm−3S_1 = \sqrt{4.0 \times 10^{-8}} = 2.0 \times 10^{-4} \text{ mol dm}^{-3}

2. For salt of type MX2\text{MX}_2:

The dissociation equilibrium in an aqueous solution is given by:
MX2(s)⇌M2+(aq)+2X−(aq)\text{MX}_2(s) \rightleftharpoons \text{M}^{2+}(aq) + 2\text{X}^-(aq)

If S2S_2 is the molar solubility of MX2\text{MX}_2:
[M2+]=S2and[X−]=2S2[\text{M}^{2+}] = S_2 \quad \text{and} \quad [\text{X}^-] = 2S_2

The expression for the solubility product is:
Ksp(MX2)=[M2+][X−]2=S2⋅(2S2)2=4S23K_{sp}(\text{MX}_2) = [\text{M}^{2+}][\text{X}^-]^2 = S_2 \cdot (2S_2)^2 = 4S_2^3

Given Ksp(MX2)=3.2×10−14K_{sp}(\text{MX}_2) = 3.2 \times 10^{-14}:
4S23=3.2×10−144S_2^3 = 3.2 \times 10^{-14}
S23=0.8×10−14=8.0×10−15S_2^3 = 0.8 \times 10^{-14} = 8.0 \times 10^{-15}
S2=(8.0×10−15)1/3=2.0×10−5 mol dm−3S_2 = \left(8.0 \times 10^{-15}\right)^{1/3} = 2.0 \times 10^{-5} \text{ mol dm}^{-3}

3. For salt of type M3X\text{M}_3\text{X}:

The dissociation equilibrium in an aqueous solution is given by:
M3X(s)⇌3M+(aq)+X3−(aq)\text{M}_3\text{X}(s) \rightleftharpoons 3\text{M}^+(aq) + \text{X}^{3-}(aq)

If S3S_3 is the molar solubility of M3X\text{M}_3\text{X}:
[M+]=3S3and[X3−]=S3[\text{M}^+] = 3S_3 \quad \text{and} \quad [\text{X}^{3-}] = S_3

The expression for the solubility product is:
Ksp(M3X)=[M+]3[X3−]=(3S3)3⋅S3=27S34K_{sp}(\text{M}_3\text{X}) = [\text{M}^+]^3[\text{X}^{3-}] = (3S_3)^3 \cdot S_3 = 27S_3^4

Given Ksp(M3X)=2.7×10−15K_{sp}(\text{M}_3\text{X}) = 2.7 \times 10^{-15}:
27S34=2.7×10−1527S_3^4 = 2.7 \times 10^{-15}
S34=0.1×10−15=1.0×10−16S_3^4 = 0.1 \times 10^{-15} = 1.0 \times 10^{-16}
S3=(1.0×10−16)1/4=1.0×10−4 mol dm−3S_3 = \left(1.0 \times 10^{-16}\right)^{1/4} = 1.0 \times 10^{-4} \text{ mol dm}^{-3}

Conclusion:

Comparing the calculated molar solubilities:
S1(MX)=2.0×10−4 mol dm−3S_1 (\text{MX}) = 2.0 \times 10^{-4} \text{ mol dm}^{-3}
S3(M3X)=1.0×10−4 mol dm−3S_3 (\text{M}_3\text{X}) = 1.0 \times 10^{-4} \text{ mol dm}^{-3}
S2(MX2)=2.0×10−5 mol dm−3S_2 (\text{MX}_2) = 2.0 \times 10^{-5} \text{ mol dm}^{-3}

Thus, the order of solubilities is:
MX>M3X>MX2\text{MX} > \text{M}_3\text{X} > \text{MX}_2

This corresponds to Option D.

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