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 :

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

(B) $\left(\frac{\mathrm{K}_{\mathrm{sp}}}{5348}\right)^{\frac{1}{6}}$

(C) $\left(\frac{\mathrm{K}_{\mathrm{sp}}}{8435}\right)^{\frac{1}{7}}$

(D) $\left(\frac{\mathrm{K}_{\mathrm{sp}}}{9612}\right)^{\frac{1}{3}}$

JEE Main · 2025 · 22 Jan 2025 · Shift 2 · Equilibrium

The molar solubility(s) of zirconium phosphate with molecular formula

(Zr^{4 +})_{3} (PO_{4}^{3 -})_{4}

is given by relation :

Answer: (A)

The dissociation of zirconium phosphate is:

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

If the molar solubility is

s

, then at equilibrium

[Z r^{4 +}] = 3 s

and

[P O_{4}^{3 -}] = 4 s

.

The solubility product is:

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

Solving for

s

s = (\frac{K _{s p}}{6912})^{1/7}

JEE Main · 2025 · 22 Jan 2025 · Shift 2 · Single correct option