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If osmotic pressure of `1M` aqueous solution of `H_(2)SO_(4)` at `500K` is `90.2` atm. Calculate `K_(a2)` of `H_(2)SO_(4)`. Give your answer after multiplying `1000` with`K_(a2)`. (Assuming ideal solution). (Given: `K_(a1)` of `H_(2)SO_(4)` is `oo,R=0.082 l t-atm//mol-K)`.

Answer» Correct Answer - `300`
`{:(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`


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