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

Answer

D

Explanation

`Ca_(3)(pO_(4))_(2)` dissociates as follows,
`Ca_(3)(pO_(4))_(2) rarr 3Ca^(2+) + 2PO_(4)^(3-)`
X 3X 2X
`K_(sp) = [Ca^(2+)]^(3)[PO_(4)^(3-)]^(2)`
`=(3X)^(3)(2X)^(2)`
`=(27X)^(3)(4X)^(2)`
`=(108X)^(5)`