The solubility of a saturated solution of calcium fluoride is 2 × 1 0 − 4 moles per litre. Its solubility product is

CaF₂(g) ⇌ Ca²⁺(aq) + 2F⁻(aq) s 2s K s p = ( s ) ( 2 s ) 2 = 4 s 3 K s p = 4 × ( 2 × 1 0 − 4 ) 3 K s p = 4 × ( 8 × 1 0 − 12 ) K s p = 32 × 1 0 − 12

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ChemistryIonic EquilibriumSolubility ProductMedium2 minQB

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The solubility of a saturated solution of calcium fluoride is $2 \times 1 0^{- 4}$ moles per litre. Its solubility product is

Answer:

Solution:

CaF₂(g) $⇌$ Ca²⁺(aq) + 2F⁻(aq)

s 2s

$K_{s p} = (s) (2 s)^{2} = 4 s^{3}$

$K_{s p} = 4 \times (2 \times 1 0^{- 4})^{3}$

$K_{s p} = 4 \times (8 \times 1 0^{- 12})$

$K_{s p} = 32 \times 1 0^{- 12}$

Stream:JEESubject:ChemistryTopic:Ionic EquilibriumSubtopic:Solubility Product

⏱ 2mℹ️ Source: QB

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