The solubility of calcium phosphate in water is $x m o l h^{- 1}$ at $25^{o} C$ . Its solubility product is equal to:

108 x^{2}

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36 x^{3}

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36 x^{5}

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108 x^{5}

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Solution

The correct option is D $108 x^{5}$
Calcium phosphate is $C a_{3} (P O_{4})_{2}$ .
$C a_{3} (P O_{4})_{2} ⇌ 3 C a^{2 +} + 2 P O_{4}^{3 -}$
$3 x$ $2 x$
$K_{s p} = [3 x]^{3} [2 x]^{2} = 108 x^{5}$

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Similar questions

x m o l L^{- 1}

25^{o} C

= M

25^{o} C

w g / 100 m L

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Solubility of calcium phosphate (molecular mass, M) in water is W g per 100 mL at 25°C. Its solubility product at 25°C will be approximately -

1.5625 \times 10^{- 10}

25^{o} C

g L^{- 1}

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