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JEE MAIN Chemistry QUESTION #1009
Question 1
The molar solubility (\(s\)) of zirconium phosphate with formula \((\text{Zr}^{4+})_3(\text{PO}_4^{3-})_4\) is given by the relation:
  • \(\left(\dfrac{K_{sp}}{9612}\right)^{1/3}\)
  • \(\left(\dfrac{K_{sp}}{6912}\right)^{1/7}\)✔️
  • \(\left(\dfrac{K_{sp}}{8435}\right)^{1/7}\)
  • \(\left(\dfrac{K_{sp}}{5348}\right)^{1/6}\)
Correct Answer Explanation
Dissolution: \(\text{Zr}_3(\text{PO}_4)_4 \rightleftharpoons 3\text{Zr}^{4+} + 4\text{PO}_4^{3-}\). If solubility \(= s\): \([\text{Zr}^{4+}]=3s\), \([\text{PO}_4^{3-}]=4s\). \(K_{sp} = (3s)^3(4s)^4 = 27s^3 \cdot 256s^4 = 6912\,s^7\). So \(s = \left(\dfrac{K_{sp}}{6912}\right)^{1/7}\).