Question

find the pH of an aqueous 0.050 M carbonic acid (H2CO3) solution. Ka1 = 4.3 x 10^-7, Ka2 = 5.2 x 10^-11.

Answer 1

There are two dissociation

H2CO3(aq) <-------> HCO3-(aq) + H+(aq) Ka1=4.3×10^-7

HCO3- (aq) <-------> CO32-(aq) + H+ (aq) Ka1 = 5.2×10^-11

The second dissociation very low and give very less contribution to pH calculation , so we can ignore second dissociation

For first dissociation

Ka1 = [ HCO3- ] [H+ ]/ [ H2CO3]

Initial Concentration

[ H2CO3 ]= 0.050M

[ H+ ] = 0

[ HCO3-] =0

at equilibrium

[ H2CO3 ] = 0.050 - X

[ HCO3-] = X

[ H+ ] = X

Therefore,

4.3×10^-7 = X^2/(0.05 - X)

X^2 + 4.3×10^-7X - 2.15 × 10^-8 = 0

X = 1.46×10^-4

[ H+ ] = 1.46 × 10^-4M

pH = -log[ H+ ]

= - log ( 1.46 ×10^-4)

= 3.84