If solubility product of $\mathrm{Zr}_{3}(\mathrm{PO}_{4})_{4}$ is denoted by $\mathrm{K}_{\mathrm{SP}}$ and its molar solubility is denoted by S, then which of the following relation between S and $\mathrm{K}_{\mathrm{SP}}$ is correct?

(A) $\mathrm{S}=\left(\frac{\mathrm{K}_{\mathrm{SP}}}{6912}\right)^{1/7}$

(B) $\mathrm{S}=\left(\frac{\mathrm{K}_{\mathrm{SP}}}{144}\right)^{1/6}$

(C) $\mathrm{S}=\left(\frac{\mathrm{K}_{\mathrm{sp}}}{929}\right)^{1/9}$

(D) $\mathrm{S}=\left(\frac{\mathrm{K}_{\mathrm{SP}}}{216}\right)^{1/7}$

JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Equilibrium

If solubility product of

Zr_{3} (PO_{4})_{4}

is denoted by

K_{SP}

and its molar solubility is denoted by S, then which of the following relation between S and

K_{SP}

is correct?

Answer: (A)

The salt dissolves as

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

, so at molar solubility

S

the ion concentrations are

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

and

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

. The solubility product is

K_{s p} = (3 S)^{3} (4 S)^{4} = 27 \times 256 S^{7} = 6912 S^{7}

, so rearranging gives

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

JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Single correct option