Solubility of calcium phosphate (molecular mass, M ) in water is W g per 100 mL at 2 5 ∘ C . Its solubility product at 2 5 ∘ C will be approximately:

S = M 10 W m o l e /litre K SP of C a 3 ( P O 4 ) 2 = 108 S 5 . K SP = 108 ( M 10 W ) 5 = 1 0 7 ( M W ) 5 a pp ro x ima t e l y .

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Solubility of calcium phosphate (molecular mass, M ) in water is W g per 100 mL at $2 5^{\circ} C$ . Its solubility product at $2 5^{\circ} C$ will be approximately:

Options:

Answer:

Solution:

$S = \frac{10 W}{M} m o l e$ /litre

$K_{SP}$ of $C a_{3} (P O_{4})_{2} = 108 S^{5}$ .

$K_{SP} = 108 (\frac{10 W}{M})^{5} = 1 0^{7} (\frac{W}{M})^{5} a pp ro x ima t e l y$ .

Stream:JEESubject:ChemistryTopic:Ionic EquilibriumSubtopic:Solubility Product

⏱ 2mℹ️ Source: QB

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