OnlineCBT
Chemistry · EQUILIBRIUM
हिन्दीIf the molar solubility of zirconium phosphate, \( \text{Zr}_3(\text{PO}_4)_4 \), in pure water is \( S \text{ mol L}^{-1} \), its solubility product \( K_{\text{sp}} \) is given by:
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Correct answer · Option A
\( 6912 S^7 \)
Explanation
\( [\text{Zr}^{4+}] = 3S \) and \( [\text{PO}_4^{3-}] = 4S \). Therefore, \( K_{\text{sp}} = (3S)^3 (4S)^4 = 27 \times 256 \times S^7 = 6912 S^7 \).
