What will be the unit in which the solubility product of $C a_{3} (P O_{4})_{2}$ is expressed?

m o l d m^{- 3}

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m o l^{2} d m^{- 6}

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m o l^{3} d m^{- 9}

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m o l^{5} d m^{- 15}

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Solution

The correct option is D $m o l^{5} d m^{- 15}$
If solubility of $C a_{3} (P O_{4})_{2}$ is $S$ ,

$C a_{3} (P O_{4})_{2} \to 3 C a^{+ 2} + 2 P O_{4}^{- 3}$
3S 2S

Solubility product $K_{s p} = (3 S)^{3} (2 S)^{2} = 108 S^{5}$

So units of solubility product $= (m o l / l i t)^{5} = m o l^{5} d m^{- 15}$

Hence, option D is correct.

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