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

Ca3PO42⇌2Ca2+(3s)+2PO43(2s) Ks=Ca2+3PO43−2=(3s)3(2s)2=108s5=(108)1.6×10−7M5=1.13×10−32M5

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

Ca3PO42⇌2Ca2+(3s)+2PO43(2s) Ks=Ca2+3PO43−2=(3s)3(2s)2=108s5=(108)1.6×10−7M5=1.13×10−32M5

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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

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$2.26 \times 10^{- 32} M^{5}$

$2.56 \times 10^{- 14} M^{5}$

$1.13 \times 10^{- 32} M^{5}$

$4.22 \times 10^{- 32} M^{5}$

answer is C.

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Detailed Solution

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