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Ionic Equilibrium question

2019 · 8 Apr · Shift 1 · Q19
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Ionic Equilibrium question

2019 · 8 Apr · Shift 1 · Q19

JEE MainChemistryIonic EquilibriumMCQ+4 / −1
If solublity product of Zr3(PO4)4Zr_3(PO_4)_4Zr3​(PO4​)4​ is denoted by Ksp and its molar solubility is denoted by S, then which of the following relation between S and Ksp is correct ?
  1. A
    S=(Ksp929)1/9S = {\left( {{{{K_{sp}}} \over {929}}} \right)^{1/9}}S=(929Ksp​​)1/9
  2. B
    S=(Ksp6912)1/7S = {\left( {{{{K_{sp}}} \over {6912}}} \right)^{1/7}}S=(6912Ksp​​)1/7
  3. C
    S=(Ksp144)1/6S = {\left( {{{{K_{sp}}} \over {144}}} \right)^{1/6}}S=(144Ksp​​)1/6
  4. D
    S=(Ksp216)1/7S = {\left( {{{{K_{sp}}} \over {216}}} \right)^{1/7}}S=(216Ksp​​)1/7
View written solutionFree

Correct answer: B

  1. Write the dissolution equilibrium

For Zr3(PO4)4(s)⇌3Zr4++4PO43−Zr_3(PO_4)_4(s) \rightleftharpoons 3Zr^{4+} + 4PO_4^{3-}Zr3​(PO4​)4​(s)⇌3Zr4++4PO43−​

If the molar solubility is SSS, then at equilibrium:

  • [Zr4+]=3S[Zr^{4+}] = 3S[Zr4+]=3S
  • [PO43−]=4S[PO_4^{3-}] = 4S[PO43−​]=4S
  1. Write the solubility product expression

By definition, Ksp=[Zr4+]3[PO43−]4K_{sp} = [Zr^{4+}]^3[PO_4^{3-}]^4Ksp​=[Zr4+]3[PO43−​]4

Substitute the concentrations: Ksp=(3S)3(4S)4K_{sp} = (3S)^3(4S)^4Ksp​=(3S)3(4S)4

  1. Simplify

Ksp=33⋅44⋅S3+4K_{sp} = 3^3 \cdot 4^4 \cdot S^{3+4}Ksp​=33⋅44⋅S3+4 Ksp=27⋅256⋅S7K_{sp} = 27 \cdot 256 \cdot S^7Ksp​=27⋅256⋅S7 Ksp=6912 S7K_{sp} = 6912\,S^7Ksp​=6912S7

  1. Solve for SSS

S7=Ksp6912S^7 = \frac{K_{sp}}{6912}S7=6912Ksp​​

Therefore, S=(Ksp6912)1/7S = \left(\frac{K_{sp}}{6912}\right)^{1/7}S=(6912Ksp​​)1/7

  1. Match with the options

This corresponds to Option B.

Comparison with stored answer: Stored correct answer is B, which matches the derived answer.

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