JEE Main 2014 · 19 Apr · Shift 1 · Chemistry Q34

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ChemistrySingle correct+4 / -1Question 34 of 90

Zirconium phosphate $\ce{[Zr3(PO4)4]}$ dissociates into three zirconium cations of charge +4 and four phosphate anions of charge −3. If molar solubility of zirconium phosphate is denoted by S and its solubility product by $K_{sp}$ then which of the following relationship between S and $K_{sp}$ is correct ?

S=\{K_{sp}/(6912)^{1/7}\}

S=\{K_{sp}/144\}^{1/7}

S=(K_{sp}/6912)^{1/7}

S=\{K_{sp}/6912\}^7

Answer and worked solution

Official answer: C

$\ce{Zr3(PO4)4 <=> 3Zr^4+ + 4PO4^3-}$ , so
$K_{sp}=(3S)^3(4S)^4=27\times256\,S^7=6912\,S^7,\qquad S=\left(\frac{K_{sp}}{6912}\right)^{1/7}.$

Topics:Solubility product

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JEE Main 2014 · 19 Apr · Shift 1, Chemistry question 34