Chemistry - Ionic Equilibrium Question with Solution | TestHub

ChemistryIonic EquilibriumSolubility ProductEasy2 minPYQ_2020
ChemistryEasysingle choice

A silver chromate solution at298 Khas itsKspvalue =1.1× 10 −12 . Determine the solubility of the solution in mol L −1

Options:

Answer:
A
Solution:

Ag 2 CrO 4 s ⇌ 2Ag + 2s + CrO 4 2− s ;  K sp =1.1× 10 −12 ​ 
K sp = Ag + 2 · CrO 4 2 -

K sp  =  [2s] 2 .  s  = 4 s 3

s 3 = K sp 4

= 1.1× 10 −12 4

​s = 275× 10 −15 3 ​  

=6.5× 10 −5  mol  L −1

⇒s=6.53× 10 −5

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

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