Chemistry - Ionic Equilibrium Question with Solution | TestHub

ChemistryIonic EquilibriumMiscellaneous/MixedEasy2 minPYQ_2020

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The Solubility of $A B_{2}$ is $0 .05 g$ per $100 mL$ at $25^{o} C .$ Calculate $K_{sp}$ of $A B_{2}$ at $25^{o} C?$ [Atomic mass of $A = 20 amu,$ atomic mass of $B = 40 amu$ ]

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Answer:

Solution:

$= 0 .05 g/100 mL$
$= 0 .5 g/L$
$= \frac{0.5}{100} = 5 \times 10^{- 3} mol/L$
$K_{s p} = 4 s^{3} = 4 \times {(5 \times 1 0^{- 3})}^{3} = 5 \times 1 0^{- 7}$

Stream:NTA_ABHYASSubject:ChemistryTopic:Ionic EquilibriumSubtopic:Miscellaneous/Mixed

⏱ 2mℹ️ Source: PYQ_2020

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