Solubility of Calcium phosphate (molecular mass, M) in water is W g per 100 mL at 25°C. Its solubility product at 25°C will be approximately

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$10^{9} {(\frac{W}{M})}^{5}$

$10^{3} {(\frac{W}{M})}^{5}$

$10^{5} {(\frac{W}{M})}^{5}$

$10^{7} {(\frac{W}{M})}^{5}$

answer is B.

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Detailed Solution

${Ca}_{3} {({PO}_{4})}_{2} (s) ⇌ \underset{S}{{Ca}_{3} {({PO}_{4})}_{2} (aq)} \to \underset{3 S}{3 {Ca}^{+ 2}} (aq) + 2 \underset{2 S}{{PO}_{4}} {}^{- 3}(aq)$

$Ksp = (3 s)^{3} \times (2 s)^{2} = 108 s^{5} = 108 \times {(\frac{w \times 10}{M})}^{5}$

$Ksp ≅ 10^{7} \times {(\frac{w}{5})}^{5}$

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