Question

1. Calculate the pH of a 0.326 M H₂SO₃. Calculate the [SO₃^2-] in the solution.

Answer 1

H2SO3 <---> H+ (aq) + HSO3- (aq)

at equilibrium [H2SO3] = 0.326-X , [H+] =[HSO3-] = X

Ka1 = [H+] [HSO3-] /[H2SO3]               wher Ka1 = 0.0139

0.0139 = (X)(X) / ( 0.326-X)

X = [H+] = 0.0607 =[HSO3-] = [H+]

pH =-log [H+] = -log ( 0.0607) = 1.2

HSO3 - <----> H+(aq) + SO32- (aq)

at equilibrium [HSO3-] = 0.0607-X , [H+] =[SO32-] = X

Ka2 = [H+] [SO32-] /[HSO3-]

6.73 x 10^-8 = ( X) (X) / ( 0.0607-X)

X = [SO32-] = 6.39 x 10^-5 M   

( H+ froms Ka1 and Ka2 but value from ka2 is very less , hence H+ from ka1 is considered)