QJEEM21Ch94fd: JEE Main 2021 Chemistry - Class 11 JEEM21Ch94fd Solution: The solubility of CdSO 4 in water is 8.0 10 4 mol L 1 . Its…

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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)

Accepted Answer

In pure water

\therefore

K_sp = (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

K_sp = [Cd⁺²] [So

_4^{ - 2}

]

\Rightarrow

64

\times

10^-8 = S₁

\times

0.01

\Rightarrow

S₁ = 64

\times

10^-6

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