JEE Main 2025 22 January Evening Shift
Exam JEE Main
Year 2025
Date 22 January
Shift Evening Shift
Subject Chemistry
Chapter Ionic Equilibrium JEE Main 2025 · 22 January · Evening Shift · Chemistry · Ionic Equilibrium
The molar solubility(s) of zirconium phosphate with molecular...
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 :
Options A ( K s p 9612 ) 1 3 \left(\frac{\mathrm{K}_{\mathrm{sp}}}{9612}\right)^{\frac{1}{3}} ( 9612 K sp ) 3 1 B ( K s p 6912 ) 1 7 \left(\frac{\mathrm{K}_{\mathrm{sp}}}{6912}\right)^{\frac{1}{7}} ( 6912 K sp ) 7 1 C ( K s p 8435 ) 1 7 \left(\frac{\mathrm{K}_{\mathrm{sp}}}{8435}\right)^{\frac{1}{7}} ( 8435 K sp ) 7 1 D ( K s p 5348 ) 1 6 \left(\frac{\mathrm{K}_{\mathrm{sp}}}{5348}\right)^{\frac{1}{6}} ( 5348 K sp ) 6 1 Correct answer B. ( K s p 6912 ) 1 7 \left(\frac{\mathrm{K}_{\mathrm{sp}}}{6912}\right)^{\frac{1}{7}} ( 6912 K sp ) 7 1
Step-by-step solution
Z r 3 ( P O 4 ) 4 ( s ) ⇌ 3 Z r + 4 ( a q ) + 4 P O 4 − 3 ( a q ) − 3 s 4 s \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} Zr 3 ( PO 4 ) 4 ( s ) ⇌ − 3 Zr + 4 ( aq ) + 4 PO 4 − 3 ( aq ) 3 s 4 s K sp = ( 3 s ) 3 ( 4 s ) 4 = 6912 s 7 s = ( K s p 6912 ) 1 / 7 \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} K sp = ( 3 s ) 3 ( 4 s ) 4 = 6912 s 7 s = ( 6912 K sp ) 1/7 $
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