JEE MAIN - Chemistry (2025 - 22nd January Evening Shift - No. 11)
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 :
$\left(\frac{\mathrm{K}_{\mathrm{sp}}}{5348}\right)^{\frac{1}{6}}$
$\left(\frac{\mathrm{K}_{\mathrm{sp}}}{8435}\right)^{\frac{1}{7}}$
$\left(\frac{K_{s p}}{6912}\right)^{\frac{1}{7}}$
$\left(\frac{\mathrm{K}_{\mathrm{sp}}}{9612}\right)^{\frac{1}{3}}$
Explanation
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$\begin{aligned} & \mathrm{K}_{\mathrm{sp}}=(3 \mathrm{~s})^3 \cdot(4 \mathrm{~s})^4 \\\\ & \mathrm{~K}_{\mathrm{sp}}=6912 \mathrm{~s}^7 \\\\ & \mathrm{~s}=\left(\frac{\mathrm{K}_{\mathrm{sp}}}{6912}\right)^{\frac{1}{7}}\end{aligned}$Comments (0)
