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 $\mathrm{Zr_3(PO_4)_4}$ is denoted by $K_{sp}$ and its molar solubility is denoted by S, then which of the following relations between S and $K_{sp}$ is correct?

Options

a
$S = \left(\frac{K_{sp}}{144}\right)^{1/6}$
b
$S = \left(\frac{K_{sp}}{6912}\right)^{1/7}$
✓
c
$S = \left(\frac{K_{sp}}{929}\right)^{1/9}$
d
$S = \left(\frac{K_{sp}}{216}\right)^{1/7}$

Correct Answerb

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

Detailed Solution

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

Step 2 — Write $K_{sp}$ : $K_{sp} = (3S)^3(4S)^4 = 27S^3 \cdot 256S^4 = 6912S^7$

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

Answer: Option (2)

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