JEE Main 2020 — Ionic Equilibrium Question with Solution
From: JEE Main 2020 (Online) 8th January Morning Slot
Question
The stoichiometry and solubility product of a salt with the solubility curve given below is, respectively :
This question includes a diagram. The text above accompanies the figure.
Choose an option
Show full solutionCorrect option: C
Correct answer
CXY2, 4 × 10–9 M3
Step-by-step explanation
From the given curve,
if [X] = 1 mM then [Y] = 2 mM
Salt is XY2
XY2(s) ⇌ X2+(aq.) + 2Y-(aq.)
ksp = [X2+][Y–]2
= (10–3) (2 × 10–3)2
= 4 × 10–9 M3
if [X] = 1 mM then [Y] = 2 mM
Salt is XY2
XY2(s) ⇌ X2+(aq.) + 2Y-(aq.)
ksp = [X2+][Y–]2
= (10–3) (2 × 10–3)2
= 4 × 10–9 M3
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