Solution:
Dissociation and ICE table of H2CO3 is written as,
H2CO3 + H2O. = H3O+ + HCO3-
0.10 ------------------ 0 ----------0 (initial)
0.10 - X --------------X --------X (final)
Thus,
Ka1 = X .X / (0.10 -X)
Since, H2CO3 is a weak acid, hence X is neglected from denominator.
Ka1 = X2 / 0.10
X2 = 0.10 x Ka1 = 0.10 x 4.5 x 10-7 = 4.5 x 10-8
X = √ 4.5 x 10-8 = 2.12 x 10-4 M
Therefore,
[H3O+] = X = 2.12 x 10-4 M
[HCO3-] = X = 2.12 x 10-4 M
[H2CO3] = 0.10 - X = 0.10 - 2.12 x 10-4 M = 9.98 x 10-2 M
[CO32-] = Ka2 = 4.7 x 10-11 M
(Since, second dissociation gives [CO32-]= ka2 )
pH = - log [H3O+] = -log 2.12 x 10-4 M = 4 - log 2.12
pH = 4 - 0.33 = 3.67
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