Vikram K.

asked • 05/31/18# If osmotic pressure of 1M aqueous solution of H2SO4 at 500K is 90.2.calculate Ka2 for acid whereas Ka1is infinite

This is from solution chapter relating ionic equilibrium

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## 1 Expert Answer

J.R. S. answered • 05/31/18

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Ph.D. in Biochemistry--University Professor--Chemistry Tutor

π = iMRT

π = osmotic pressure = 90.2 atm (presumably atm)

i = van't off factor = ? for H

_{2}SO_{4}M = molarity = 1

R = gas constant = 0.0821 L-atm/mol-K

T = temperature in K = 500K

Solving for i: i = π/MRT = (90.2)/(1)(0.0821)(500) = 2.197

2 of the 2.197 can be accounted for by H

^{+}and HSO_{4}^{-}. The remaining 0.197 would be from H^{+}and SO_{4}^{2-}.H

_{2}SO_{4}==>H^{+}+ HSO_{4}^{-}K_{1}= infinite and thus i = 2 for H^{+}and HSO_{4}^{-}[HSO

_{4}^{-}] = 1 M since 100% dissociation of H_{2}SO_{4}HSO

_{4}^{-}==> H^{+}+ SO_{4}^{2-}K_{2}= ?K

_{2}= [H^{+}][SO_{4}^{2-}]/[HSO_{4}^{-}] = (0.0985)(0.0985)/1 = 9.7x10^{-3}## Still looking for help? Get the right answer, fast.

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J.R. S.

05/31/18