Taking $K_{s p}$ of $Ca X_{3} (PO X_{4}) X_{2}$ as $2.0 \times 1 0^{- 29}$ , calculate its molar solubility in water in units of $1 0^{- 7} mol L^{- 1}$ (two decimals; ignore hydrolysis of phosphate).

Numerical answer type, answer in x 10^-7 mol/L, tolerance 0.03.

Answer: 7.14 x 10^-7 mol/L

[Ca X^{2 +}] = 3 s

and

[PO X_{4} X^{3 -}] = 2 s

, so

K_{s p} = (3 s)^{3} (2 s)^{2} = 108 s^{5}

. Hence

s = (2.0 \times 1 0^{- 29} /108)^{1/5} = 7.1 \times 1 0^{- 7} M