Chemistry - Ionic Equilibrium Question with Solution | TestHub

ChemistryIonic EquilibriumSolubility ProductEasy2 minPYQ_2020
ChemistryEasysingle choice

Determine the solubility of silver chromate at 298 K given its Kspvalue is 1.1 × 10-12.

Options:

Answer:
A
Solution:

Ag 2 CrO 4 ⇌ 2Ag + + CrO 4 2 - ; K sp = 1.1 × 10 - 12
S                     2S            S

K sp = Ag + 2 · CrO 4 2 -

Ksp= [2S]2·[S] = 4S3

S3= Ksp4 = 1.1×10-124

⟹S=6.53×10-5

Stream:NTA_ABHYAS_NEETSubject:ChemistryTopic:Ionic EquilibriumSubtopic:Solubility Product
⏱ 2mℹ️ Source: PYQ_2020

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