Q

# Help me answer: - Equilibrium - JEE Main-3

If solubility product of $Zr_{3}(PO_{4})_{4}$ is denoted by $K_{sp}$ and its molar solubility is denoted by $S$, then which of the following relation between $S$ and $K_{sp}$ is correct ?

• Option 1)

$S=\left ( \frac{K_{sp}}{216} \right )^{1/7}$

• Option 2)

$S=\left ( \frac{K_{sp}}{6912} \right )^{1/7}$

• Option 3)

$S=\left ( \frac{K_{sp}}{144} \right )^{1/6}$

• Option 4)

$S=\left ( \frac{K_{sp}}{929} \right )^{1/9}$

Views

$\\Zr_{3}(PO_{4})_{4}\rightleftharpoons 3Zr^{4+}(aq)+4PO_{4}^{3-}(aq)\\(5)\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; (35)\; \; \; \; \; \; \; \; \; \; (45)$

$K_{sp}=[Zr^{4+}]^{3}[PO{_{4}}^{3-}]^{4}$

$= (35)^{3} \; \; \; \; (45)^{4}=69125^{2}$

$S=\left ( \frac{K_{sp}}{6912} \right )^{1/2}$

Option 1)

$S=\left ( \frac{K_{sp}}{216} \right )^{1/7}$

Option 2)

$S=\left ( \frac{K_{sp}}{6912} \right )^{1/7}$

Option 3)

$S=\left ( \frac{K_{sp}}{144} \right )^{1/6}$

Option 4)

$S=\left ( \frac{K_{sp}}{929} \right )^{1/9}$

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