The molar solubility(s) of zirconium phosphate with molecular formula ( Zr 4+ ) 3 ( PO 4 3- ) 4 is given by relation :

Chemistry · Ionic Equilibrium

JEE Main 2025 — 22 January, Evening Shift — Question 35

The molar solubility(s) of zirconium phosphate with molecular formula $\left(\mathrm{Zr}^{4+}\right)_{3}\left(\mathrm{PO}_{4}^{3-}\right)_{4}$ is given by relation :

Option A:
$\left(\frac{K_{s p}}{6912}\right)^{\frac{1}{7}}$
Correct
Option B:
$\left(\frac{\mathrm{K}_{\mathrm{sp}}}{5348}\right)^{\frac{1}{6}}$
Option C:
$\left(\frac{\mathrm{K}_{\mathrm{sp}}}{8435}\right)^{\frac{1}{7}}$
Option D:
$\left(\frac{\mathrm{K}_{\mathrm{sp}}}{9612}\right)^{\frac{1}{3}}$

Answer: A

Step-by-step solution

$\mathrm{Zr}_{3}\left(\mathrm{PO}_{4}\right)_{4}(\mathrm{~s}) \rightleftharpoons 3 \mathrm{Zr}^{+4}(\mathrm{aq})+4 \mathrm{PO}_{4}{ }^{-3}(\mathrm{aq})$

$\mathrm{K}_{\mathrm{sp}}=(3 \mathrm{~s})^{3}(4 \mathrm{~s})^{4}=6912 \mathrm{~s}^{7}$

$\mathrm{s}=\left(\frac{\mathrm{K}_{\mathrm{sp}}}{6912}\right)^{1 / 7}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 22 January, Evening Shift
Subject: Chemistry
Chapter: Ionic Equilibrium
Topic: Sparingly Soluble Salts, Solubility Product & Precipitation Conditions