H2CO3 + H2O ⇄ H3O+ + HCO-
.18 0 0
-x +x +x
.18-x x x
x^2 ÷ .18 = Ka1 = 4.3 X 10^ -7
x= 2.78 X 10^ -4
HCO- + H2O ⇄ H3O+ + CO-
2.78 X 10^ -4 2.78 X 10^ -4 0
-x +x +x
2.78 X 10^ -4 -x 2.78 X 10^ -4+ x x
k a2 = 5.6 X 10^ -11 = (2.78 X 10^ -4 + x)(x) ÷ (2.78 X 10^ -4 - x)
x= 5.6 X 10^ -11
Total [ H3 O +] = 2.78 X 10^ -4 + 5.6 X 10^ -11 = 2.78 X 10^ -4
pH = - log (2.78 X 10^ -4 )= 3.555 ~ 3.56
The answer is B