In pure water,
Ksp=S2=(8×10−4)2
=64×10−8
In 0.01MH2SO4
xH2SO4(aq)→x2H(aq)++(x+0.01)SO42−(aq.)
Ksp=x(x+0.01)
=64×10−8
x+0.01≅0.01M
So, x(0.01)=64×10−8
x=64×10−6M
JEE Main 2021 — Chemistry Physical Chemistry
The solubility of CdSO4 in water is 8.0×10−4molL−1. Its solubility in 0.01MH2SO4 solution is ___ ×10−6molL−1 (Round off to the Nearest integer) (Assume that solubility is much less than 0.01M )
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