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Q.

The solubility of calcium phosphate in water is $xx mol L^{- 1} at 25^{°} C$ .Its solubility product is equal to

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a

$36 x^{s}$

b

$36 x^{3}$

c

$108 x^{s}$

d

$108 x^{2}$

answer is D.

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Detailed Solution

${Ca}_{3} {({PO}_{4})}_{2} ⇌ 3 {Ca}^{+ 2} + 2 {PO}_{4}^{- 3}$

3s 2s

$K_{Sp} = {(3 s)}^{3} . {(2 S)}^{2} = 27 S^{3} . 4 S^{2} = 108 S^{5} [S = x]$

$= 108 x^{5}$

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