If the solubility of `Ca_(3)(PO_(4))_(2)` in water is 'X' mol `L_(-1)`, its solubility product in `mol^(5)L^(-5)` is

Options

A. `6x^(5)`
B. `36x^(5)`
C. `64x^(5)`
D. `108x^(5)`

Explanation

`Ca_(3)(PO_(4))^(2)rArr3Ca^(2)+2PO_(4)^(3-)`
`K_(sp)=[Ca^(2+)]^(3)[PO_(4)^(3-)]^(2)=[3x]^(3)[2x]^(2)=[27x^(3)][4x^(2)]`
`=108x^(5)`