JEE Main 25 Feb 2021 Shift 2 Solved Paper

Let, solubility of Ca(OH)2 in pure water =Smol∕L
‌Ca(OH)2⇌Ca2++2OH−
‌Smol∕L‌‌2×S(mol∕L)
=‌‌[Ca2+][OH−]2=S×(2S)2=4S3(mol∕L)
The expression of Ksp can also be written as,
Ksp‌‌=xx⋅yy⋅Sx+y
‌‌=11⋅22⋅S1+2
‌‌=4S3
Ksp‌‌=xx⋅yy⋅Sx+y
‌‌=11⋅22⋅S1+2
‌‌=4S3‌‌[∵‌ For ‌Ca(OH)2:x=1,y=2]
x and y are the coefficients of cations and anions respectively
S‌‌=(‌

Ksp

4

)1∕3=(‌

5.5×10−6

4

)1∕3
‌‌=1.11×10−2mol∕L