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Chemistry - Ionic Equilibrium Question with Solution | TestHub

ChemistryIonic EquilibriumSolubility ProductMedium2 minPYQ_2018

ChemistryMediumsingle choice

The solubility of ${BaSO}_{4}$ in water is $2 .42 \times 10^{- 3} {gL}^{- 1}$ at $298 K .$ The value of its solubility product $(K_{sp})$ will be

(Given- the molar mass of ${BaSO}_{4} = 233 g {mol}^{- 1}$ )NEET 2018

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Answer:

Solution:

Solubility of ${BaSO}_{4} = 2 .42 \times 10^{- 3} {gL}^{- 1}$

$∴ s = \frac{2 .42 \times 10^{- 3}}{233} = 1 .038 \times 10^{- 5} mol L^{- 1}$

$K_{sp} = s^{2} = {(1 .038 \times 10^{- 5})}^{2}$

$= 1 .08 \times 10^{- 10} {mol}^{2} L^{- 2}$

Stream:NEETSubject:ChemistryTopic:Ionic EquilibriumSubtopic:Solubility Product

⏱ 2mℹ️ Source: PYQ_2018

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