If osmotic presssure of 1M aqueous solution of `H_(2)SO_(4)` at 500 K is 90.2 atm. Calculate `K_(a2)` of `H_(2)SO_(4)`. Give your answer after multiplying 1000 with `K_(a_(2))`. (Assumme ideal solution) [Given : `K_(a_(1))` of `H_(2)SO_(4)=oo`,R=0.082 L atm`//`mol-K]

Answer

3

Explanation

`{:(H_2SO_4,to,H^+,+,HSO_4^(-),,,H_2SO_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 implies 90.2=(2+x)xx0.082xx500implies x=0.2`
`:. [H^+]=1.2 M, [SO_4^(2-)]=0.2 M, " " [SO_4^(-)]=1-0.2=0.8`
`:.Ka_2=([H^+][SO_4^(2-)])/([HSO_4^(-)])=(1.2xx0.2)/0.8=0.3`