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

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 :
  1. $\left(\frac{\mathrm{K}_{\mathrm{sp}}}{9612}\right)^{\frac{1}{3}}$
  2. $\left(\frac{\mathrm{K}_{\mathrm{sp}}}{6912}\right)^{\frac{1}{7}}$
  3. $\left(\frac{\mathrm{K}_{\mathrm{sp}}}{8435}\right)^{\frac{1}{7}}$
  4. $\left(\frac{\mathrm{K}_{\mathrm{sp}}}{5348}\right)^{\frac{1}{6}}$

Solution

$\begin{array}{cc}\mathrm{Zr}_3\left(\mathrm{PO}_4\right)_4(\mathrm{~s}) \rightleftharpoons & 3 \mathrm{Zr}^{+4}(\mathrm{aq})+4 \mathrm{PO}_4^{-3}(\mathrm{aq}) \\ - & 3 \mathrm{~s} \quad 4 \mathrm{~s}\end{array}$
$\begin{aligned} & \mathrm{K}_{\text {sp }}=(3 \mathrm{~s})^3(4 \mathrm{~s})^4=6912 \mathrm{~s}^7 \\ & \mathrm{~s}=\left(\frac{\mathrm{K}_{\mathrm{sp}}}{6912}\right)^{1 / 7}\end{aligned}$

Asked in: JEE Main 2025 (22 Jan Shift 2)

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