JEE Main 2025 Jan 22 Shift 2, Chemistry Q12: Solubility Equilibria Of Sparingly Soluble Salts

JEE Main2025Jan 22, Shift 2Chemistry

The molar solubility ( $s$ ) of zirconium phosphate with molecular formula $(Z r^{4 +})_{3} (P O_{4}^{3 -})_{4}$ is given by relation :

Solution

For the dissolution equilibrium:

$Z r_{3} (P O_{4})_{4} (s) ⇌ 3 Z r^{4 +} (a q) + 4 P O_{4}^{3 -} (a q)$

If $s$ is the molar solubility, then $[Z r^{4 +}] = 3 s$ and $[P O_{4}^{3 -}] = 4 s$ .

$K_{s p} = [Z r^{4 +}]^{3} [P O_{4}^{3 -}]^{4} = (3 s)^{3} (4 s)^{4} = 27 \cdot 256 \cdot s^{7} = 6912 s^{7}$ .

Solving for $s$ : $s = (\frac{K _{s p}}{6912})^{1/7}$ .

Concept behind this question

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