JEE Main · 2019 · Shift-IhardIEQ-018

If solubility product of Zr3(PO4)4 is denoted by Ksp and its molar solubility is denoted by S, then which of the…

Ionic Equilibrium · Class 11 · JEE Main Previous Year Question

Question

If solubility product of Zr3(PO4)4\mathrm{Zr_3(PO_4)_4} is denoted by KspK_{sp} and its molar solubility is denoted by S, then which of the following relations between S and KspK_{sp} is correct?

Options
  1. a

    S=(Ksp144)1/6S = \left(\frac{K_{sp}}{144}\right)^{1/6}

  2. b

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

    ✓
  3. c

    S=(Ksp929)1/9S = \left(\frac{K_{sp}}{929}\right)^{1/9}

  4. d

    S=(Ksp216)1/7S = \left(\frac{K_{sp}}{216}\right)^{1/7}

Correct Answerb

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

Detailed Solution

Step 1 — Dissociation of Zr3(PO4)4\mathrm{Zr_3(PO_4)_4}: Zr3(PO4)4⇌3Zr4++4PO43−[Zr4+]=3S,[PO43−]=4S\mathrm{Zr_3(PO_4)_4 \rightleftharpoons 3Zr^{4+} + 4PO_4^{3-}} [\mathrm{Zr^{4+}}] = 3S, \quad [\mathrm{PO_4^{3-}}] = 4S

Step 2 — Write KspK_{sp}: Ksp=(3S)3(4S)4=27S3⋅256S4=6912S7K_{sp} = (3S)^3(4S)^4 = 27S^3 \cdot 256S^4 = 6912S^7

Step 3 — Solve for S: S7=Ksp6912⇒S=(Ksp6912)1/7S^7 = \frac{K_{sp}}{6912} \Rightarrow S = \left(\frac{K_{sp}}{6912}\right)^{1/7}

Answer: Option (2)

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