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Ionic Equilibrium question

2008 · Shift 2 · Q10
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  5. /2008 · Shift 2 · Q10

Ionic Equilibrium question

2008 · Shift 2 · Q10

JEE AdvancedChemistryIonic EquilibriumMCQ+3 / −1
Solubility product constants (K sp{}_{sp}sp​) of salts of types MX, MX 2{}_22​ and M 3{}_33​ X at temperature T are 4.0 ×\times× 10 −8{}^{-8}−8, 3.2 ×\times× 10 −14{}^{-14}−14 and 2.7 ×\times× 10 −15{}^{-15}−15, respectively. Solubilities (mol dm −3{}^{-3}−3) of the salts at temperature 'T' are in the order:
  1. A
    MX > MX 2{}_22​ > M 3{}_33​ X
  2. B
    M 3{}_33​ X > MX 2{}_22​ > MX
  3. C
    MX 2{}_22​ > M 3{}_33​ X > MX
  4. D
    MX > M 3{}_33​ X > MX 2{}_22​
View written solutionFree

Correct answer: D

The problem requires us to calculate the molar solubility (S) for three different types of salts (MX, MX₂, and M₃X) from their given solubility product constants (Ksp) and then compare these solubilities.

Step 1: Calculate the solubility of salt MX

For a salt of type MX, the dissolution equilibrium is: MX(s)⇌M+(aq)+X−(aq)MX(s) \rightleftharpoons M^+(aq) + X^-(aq)MX(s)⇌M+(aq)+X−(aq) If the molar solubility is 'S' mol dm⁻³, then at equilibrium, [M+]=S[M^+] = S[M+]=S and [X−]=S[X^-] = S[X−]=S. The solubility product expression is: Ksp=[M+][X−]=(S)(S)=S2K_{sp} = [M^+][X^-] = (S)(S) = S^2Ksp​=[M+][X−]=(S)(S)=S2 Given KspK_{sp}Ksp​ for MX is 4.0×10−84.0 \times 10^{-8}4.0×10−8. S2=4.0×10−8S^2 = 4.0 \times 10^{-8}S2=4.0×10−8 SMX=4.0×10−8=2.0×10−4 mol dm−3S_{MX} = \sqrt{4.0 \times 10^{-8}} = 2.0 \times 10^{-4} \text{ mol dm}^{-3}SMX​=4.0×10−8​=2.0×10−4 mol dm−3

Step 2: Calculate the solubility of salt MX₂

For a salt of type MX₂, the dissolution equilibrium is: MX2(s)⇌M2+(aq)+2X−(aq)MX_2(s) \rightleftharpoons M^{2+}(aq) + 2X^-(aq)MX2​(s)⇌M2+(aq)+2X−(aq) If the molar solubility is 'S' mol dm⁻³, then at equilibrium, [M2+]=S[M^{2+}] = S[M2+]=S and [X−]=2S[X^-] = 2S[X−]=2S. The solubility product expression is: Ksp=[M2+][X−]2=(S)(2S)2=4S3K_{sp} = [M^{2+}][X^-]^2 = (S)(2S)^2 = 4S^3Ksp​=[M2+][X−]2=(S)(2S)2=4S3 Given KspK_{sp}Ksp​ for MX₂ is 3.2×10−143.2 \times 10^{-14}3.2×10−14. 4S3=3.2×10−144S^3 = 3.2 \times 10^{-14}4S3=3.2×10−14 S3=3.2×10−144=0.8×10−14=8.0×10−15S^3 = \frac{3.2 \times 10^{-14}}{4} = 0.8 \times 10^{-14} = 8.0 \times 10^{-15}S3=43.2×10−14​=0.8×10−14=8.0×10−15 SMX2=8.0×10−153=2.0×10−5 mol dm−3S_{MX_2} = \sqrt[3]{8.0 \times 10^{-15}} = 2.0 \times 10^{-5} \text{ mol dm}^{-3}SMX2​​=38.0×10−15​=2.0×10−5 mol dm−3

Step 3: Calculate the solubility of salt M₃X

For a salt of type M₃X, the dissolution equilibrium is: M3X(s)⇌3M+(aq)+X3−(aq)M_3X(s) \rightleftharpoons 3M^+(aq) + X^{3-}(aq)M3​X(s)⇌3M+(aq)+X3−(aq) If the molar solubility is 'S' mol dm⁻³, then at equilibrium, [M+]=3S[M^+] = 3S[M+]=3S and [X3−]=S[X^{3-}] = S[X3−]=S. The solubility product expression is: Ksp=[M+]3[X3−]=(3S)3(S)=27S4K_{sp} = [M^+]^3[X^{3-}] = (3S)^3(S) = 27S^4Ksp​=[M+]3[X3−]=(3S)3(S)=27S4 Given KspK_{sp}Ksp​ for M₃X is 2.7×10−152.7 \times 10^{-15}2.7×10−15. 27S4=2.7×10−1527S^4 = 2.7 \times 10^{-15}27S4=2.7×10−15 S4=2.7×10−1527=0.1×10−15=1.0×10−16S^4 = \frac{2.7 \times 10^{-15}}{27} = 0.1 \times 10^{-15} = 1.0 \times 10^{-16}S4=272.7×10−15​=0.1×10−15=1.0×10−16 SM3X=1.0×10−164=1.0×10−4 mol dm−3S_{M_3X} = \sqrt[4]{1.0 \times 10^{-16}} = 1.0 \times 10^{-4} \text{ mol dm}^{-3}SM3​X​=41.0×10−16​=1.0×10−4 mol dm−3

Step 4: Compare the solubilities

Now we compare the calculated solubilities:

  • SMX=2.0×10−4S_{MX} = 2.0 \times 10^{-4}SMX​=2.0×10−4 mol dm⁻³
  • SM3X=1.0×10−4S_{M_3X} = 1.0 \times 10^{-4}SM3​X​=1.0×10−4 mol dm⁻³
  • SMX2=2.0×10−5S_{MX_2} = 2.0 \times 10^{-5}SMX2​​=2.0×10−5 mol dm⁻³

Comparing the values: 2.0×10−4>1.0×10−4>2.0×10−52.0 \times 10^{-4} > 1.0 \times 10^{-4} > 2.0 \times 10^{-5}2.0×10−4>1.0×10−4>2.0×10−5 Therefore, the order of solubilities is: SMX>SM3X>SMX2S_{MX} > S_{M_3X} > S_{MX_2}SMX​>SM3​X​>SMX2​​ This corresponds to option D.

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