The solubility of Ca3PO42 in solution in 1.6×10−7molL−1. Its solubility product is

A
$2.56 \times 10^{- 14} M^{5}$
B
$4.22 \times 10^{- 32} M^{5}$
C
$1.13 \times 10^{- 32} M^{5}$
D
$2.26 \times 10^{- 32} M^{5}$

Solution:

$\begin{array}{l} {Ca}_{3} {({PO}_{4})}_{2} ⇌ \underset{(3 s)}{2 {Ca}^{2 +}} + \underset{(2 s)}{2 {PO}_{4}^{3}} \\ K_{s} = {[{Ca}^{2 +}]}^{3} {[{PO}_{4}^{3 -}]}^{2} = (3 s)^{3} (2 s)^{2} = 108 s^{5} = (108) {(1.6 \times 10^{- 7} M)}^{5} = 1.13 \times 10^{- 32} M^{5} \end{array}$

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The solubility of ${Ca}_{3} {({PO}_{4})}_{2}$ in solution in $1.6 \times 10^{- 7} {molL}^{- 1}$ . Its solubility product is

Solution:

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