The solubility in water of a sparingly soluble salt AB is 1.0 × 1 0 − 5 mol L − 1 . Its solubility product number will be

[ A ] = 1.0 × 1 0 − 5 , K sp = [ 2. B ] 2 [ A ] = [ 2 × 1 0 − 5 ] 2 [ 1.0 × 1 0 − 5 ] = 4 × 1 0 − 15 [ B ] = [ 1.0 × 1 0 − 5 ]

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Chemistry - Ionic Equilibrium Question with Solution | TestHub

ChemistryIonic EquilibriumSolubility ProductMedium2 minPYQ_2003

ChemistryMediumstatement

The solubility in water of a sparingly soluble salt $AB$ is $1.0 \times 1 0^{- 5} mol L^{- 1}$ . Its solubility product number will be

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Answer:

Solution:

$[A] = 1.0 \times 1 0^{- 5}, K_{sp} = [2. B]^{2} [A] = [2 \times 1 0^{- 5}]^{2} [1.0 \times 1 0^{- 5}] = 4 \times 1 0^{- 15} [B] = [1.0 \times 1 0^{- 5}]$

Stream:JEESubject:ChemistryTopic:Ionic EquilibriumSubtopic:Solubility Product

⏱ 2mℹ️ Source: PYQ_2003

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