Answer: 4.55x10^-7
Explanation:
H2CO3 + H20 <==> H3O+ + HCO3-
Desigining an ICE table, we have:
Initial conc. of H2CO3 = 0.29 M
Initial conc. of H3O+ = 0
Initial conc. of HCO3- = 0
Change in conc. of H2CO3 = - x
Change in conc. of H3O+ = x
Change in conc. of HCO3- = x
Equilibrium conc. of H2CO3 = 0.29 - x
Equilibrium conc. of H3O+ = x
Equilibrium conc. of HCO3- = x
but pH = 3.44
pH = - log[H30+]
[H30+] = 10^-3.44 = 3.63x10^-4
[H30+] = 3.63x10^-4
Ka = [H3O+].[HCO3-] / [H2CO3]
[H30+] = x = 3.63x10^-4
[HCO3-] = 3.63x10^-4
[H2CO3] = 0.29 - x = 0.29 - 3.63x10^-4 = 0.289637
Ka = (3.63x10^-4)(3.63x10^-4)/0.289637 = 4.55x10^-7
