JEE MAIN - Chemistry (2021 - 18th March Evening Shift - No. 21)
The solubility of CdSO4 in water is 8.0 $$\times$$ 10$$-$$4 mol L$$-$$1. Its solubility in 0.01 M H2SO4 solution is __________ $$\times$$ 10$$-$$6 mol L$$-$$1. (Round off to the Nearest Integer). (Assume that solubility is much less than 0.01 M)
Answer
64
Explanation
In pure water
_18th_March_Evening_Shift_en_21_3.png)
$$ \therefore $$ Ksp = (s)2 = (8 $$\times$$ 10-4)2 = 64 $$\times$$ 10-8
In H2SO4 solution,
_18th_March_Evening_Shift_en_21_4.png)
As S1 < < 0.01 so, S1 + 0.01 $$ \simeq $$ 0.01
$$ \therefore $$ Ksp = [Cd+2] [So$$_4^{ - 2}$$]
$$ \Rightarrow $$ 64 $$\times$$ 10-8 = S1 $$\times$$ 0.01
$$ \Rightarrow $$ S1 = 64 $$\times$$ 10-6
$$ \therefore $$ Ksp = (s)2 = (8 $$\times$$ 10-4)2 = 64 $$\times$$ 10-8
In H2SO4 solution,
As S1 < < 0.01 so, S1 + 0.01 $$ \simeq $$ 0.01
$$ \therefore $$ Ksp = [Cd+2] [So$$_4^{ - 2}$$]
$$ \Rightarrow $$ 64 $$\times$$ 10-8 = S1 $$\times$$ 0.01
$$ \Rightarrow $$ S1 = 64 $$\times$$ 10-6
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