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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}

Answers (1)
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\\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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