Question

What is the percent ionization of carbonic in 0.1441 M carbonic acid solution, H2CO3? H2CO3(aq) + H2O(l) « HCO3−(aq) + H3O+(aq) Ka1 = 4.3000e-7 HCO3−(aq) + H2O(l) « CO32−(aq) + H3O+(aq) Ka2 = 5.6000e-11

Answer 1

H₂CO₃ + H₂O <-------> H₃O⁺ + HCO₃^-   Ka1 = 4.3 X 10^-7

Ka1 = 4.3 X 10^-7 = [H₃O⁺][HCO₃^- ] /[H₂CO₃ ]

4.3 X 10^-7 = x² / 0.1441 - x

x² + 4.3 X 10^-7x - 6.2 X 10^-8 = 0 this is a quadrtic equation

solving

x = 2.5 X 10^-4M, H₃O⁺ = 2.5 X 10^-4M, HCO₃^- = 2.5 X 10^-4M

HCO₃^- + H₂O <-------> H₃O⁺ + CO₃^2-      Ka2 = 4.3 X 10^-11

Ka2 = 5.6X 10^-11 = x²/ 2.5 X 10^-4

Assuming x << 2.5 X 10^-4

x² = 5.6X 10^-11 X 2.5 X 10^-4 = 1.4 X 10^-14

x = 1.2 X 10^-7

total H₃O⁺ = 2.5 X 10^-4 + 1.2 X 10^-7 = 2.5012 X 10^-4 M

% dissociation = ([H₃O⁺ ] / [H₂CO₃ ]) X 100

% dissociation = (2.5012 X 10^-4 / 0.1441) X 100 = 0.17%