What will be the osmotic pressure of `0.1 M` monobasic acid its `pH` is `2` at `25^(@)C`?

Answer

`2.69` atm

Explanation

`HAhArrH^(+)+A^(-)`
[H^(+)]=Calpha=10^(-2)~
`therefore alpha=(10^(-2))/(0.1)=10^(-1)=0.1`
`therefore` Total particles in solution `=[1+a=1+0.1=1.1`
Then `p=C.S.T(1+alpha)`
`=0.1xx0.0821xx298xx1.1`
`=2.69` atm