JEE Main 2021ChemistryIonic EquilibriumSolubility Product And Common Ion EffectmediumMCQ
JEE Main 2021 — Ionic Equilibrium Question with Solution
From: JEE Main 2021 (Online) 25th February Morning Shift
Question
The solubility of AgCN in a buffer solution of pH = 3 is x. The value of x is : [Assume : No cyano complex is formed; Ksp(AgCN) = 2.2 × 10−16 and Ka(HCN) = 6.2 × 10−10]
Choose an option
▸Show full solutionCorrect option: A
Correct answer
A1.9 × 10−5
Step-by-step explanation
Let solubility is x
Ksp×Ka1=[Ag+][CN−]×[H+][CN−][HCN]
2.2×10−16×6.2×10−101=10−3[S][S]
S2=6.22.2×10−9
S2=3.55×10−10
S=3.55×10−10
S=1.88×10−5=1.9×10−5
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