JEE MAIN - Chemistry (2020 - 8th January Morning Slot - No. 18)
The stoichiometry and solubility product of a salt with the solubility curve given below is, respectively :
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XY, 2 × 10–6 M3
XY2, 1 × 10–9 M3
XY2, 4 × 10–9 M3
X2Y, 2 × 10–9 M3
Explanation
From the given curve,
if [X] = 1 mM then [Y] = 2 mM
$$ \therefore $$ 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
$$ \therefore $$ 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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