The solubility of calcium phosphate in water is $x m o l L^{- 1}$ at $25^{o} C$ . Its solubility product is equal to

108 x^{2}

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36 x^{3}

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36 x^{5}

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108 x^{5}

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Solution

The correct option is C $108 x^{5}$
Let solubility be S,
$C a_{3} (P O_{4})_{2} \to 3 C a^{+ 2} + 2 P O_{4}^{- 3}$
$3 S$ $2 S$
Solubility product, $K_{s p} = (3 S)^{3} (2 S)^{2} = 108 S^{5}$

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