If osmotic pressure of 1 M aqueous solution of `H_2SO_4` at 500 K is 90.2 atm.Calculate `Ka_2` of `H_2SO_4`.Give your answer the multiplying 10 with `Ka_2`(Assuming ideal solution)
(Given:`Ka_1` for `H_2SO_4` is `oo`, R=0.082 It-atm/mol-K).

Answer

`300`

Explanation

`{:(H_(2)SO_(4),rarr,H^(+),+,HSO_(4)^(-),,,HSO_(4)^(-)hArr,H^(+),+,SO_(4)^(2-)),(1,,,,,,1,1,,),(-,,1,,1,,1-x,1+x,,x):}`
Total concentration, `C_(T)=(1+X)+(1-X)+X=2+X`.
`pi=C_(T)RT rArr 90.2=(2+X)xx0.082xx500 rArr x=0.2`.
`therefore [H^(+)]=1.2M, [SO_(4)^(2-)]=0.2M, [HSO_(4)^(-)]=1-0.2=0.8`.
`therefore Ka_(2)=([H^(+)][SO_(4)^(2-)])/([HSO_(4)^(-)])=(1.2xx0.2)/(0.8)=0.3`
`1000 K_(a2)=1000xx0.3=300`