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Chemistry · EQUILIBRIUM
हिन्दीThe molar solubility of zirconium phosphate, \( \text{Zr}_3(\text{PO}_4)_4 \), in pure water is \( S\text{ mol L}^{-1} \). The solubility product constant \( K_{sp} \) in terms of \( S \) is:
Correct answer · Option 2
\( 6912 S^7 \)
Explanation
\( \text{Zr}_3(\text{PO}_4)_4(s) \rightleftharpoons 3\text{Zr}^{4+}(aq) + 4\text{PO}_4^{3-}(aq) \). At equilibrium, \( [\text{Zr}^{4+}] = 3S \) and \( [\text{PO}_4^{3-}] = 4S \). Thus, \( K_{sp} = (3S)^3(4S)^4 = 27 \times 256 \times S^7 = 6912 S^7 \).
