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Chemistry - Ionic Equilibrium Question with Solution | TestHub

ChemistryIonic EquilibriumSolubility ProductEasy2 minPYQ_2020

ChemistryEasysingle choice

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

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Answer:

Solution:

$C a_{3} (P O_{4})_{2} \to 3 C a^{2 +} + 2 (P O_{4})^{3 -}$
$S = \frac{10 W}{M}$ mol litre
$K_{s p}$ of $C a_{3} (P O_{4})_{2} = 108 S^{5}$
$= 108 {(\frac{10 W}{M})}^{5}$
$= 1 0^{7} {(\frac{W}{M})}^{5}$ (approximately).

Stream:NTA_ABHYAS_NEETSubject:ChemistryTopic:Ionic EquilibriumSubtopic:Solubility Product

⏱ 2mℹ️ Source: PYQ_2020

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