${C a}_{3} {(P O_{4})}_{2} (s) ⇋ 3 {C a}^{2 +} (a q) + 2 P O_{4}^{3 -} (a q)$
Calculate the solubility product.

K_{s p} = {[{C a}^{2 +}]}^{3} {[P O_{4}^{3 -}]}^{2}

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K_{s p} = {[{C a}^{2 +}]}^{2} {[P O_{4}^{3 -}]}^{3}

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K_{s p} = [{C a}^{2 +}] [P O_{4}^{3 -}] / [{C a}_{3} {(P O_{4})}_{2}]

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K_{s p} = {[{C a}^{2 +}]}^{3} {[P O_{4}^{3 -}]}^{2} / [{C a}_{3} {(P O_{4})}_{2}]

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K_{s p} = {[{C a}^{2 +}]}^{2} {[P O_{4}^{3 -}]}^{3} / [{C a}_{3} {(P O_{4})}_{2}]

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Solution

The correct option is A $K_{s p} = {[{C a}^{2 +}]}^{3} {[P O_{4}^{3 -}]}^{2}$
${C a}_{3} {(P O_{4})}_{2} (s) ⇋ 3 {C a}^{2 +} (a q) + 2 P O_{4}^{3 -} (a q)$
The expression for the solubility product is $K_{s p} = {[{C a}^{2 +}]}^{3} {[P O_{4}^{3 -}]}^{2}$ .
The solubility product is the product of ionic concentrations raised to appropriate stoichiometric coefficients.

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Similar questions

What is the correct equilibrium expression (K_sp) for the reaction below:
Ca₃(PO₄)_2(s) ⇌ 3Ca²⁺_(aq) + 2PO₄^2-_(aq)

{C a}_{3} {({P O}_{4})}_{2}

C a_{3} (P O_{4})_{2}

2.0 \times 10^{- 8}

C a^{2 +}

1.6 \times 10^{- 5}

P O_{4}^{3 -}

(K_{s p})

C a_{3} (P O_{4})_{2}

{C a}_{3} {({P O}_{4})}_{2}

y

C a, {C a}^{+}, {C a}^{2 +}, {C a}^{3 +}, {C a}^{-}, {C a}^{2 -}

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