The pH of a 0.29 M solution of carbonic acid (H2CO3) is measured to be 3.44. Calculate the acid dissociation constant Ka of carbonic acid. Round your answer to 2 significant digits. x10

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

b) mutualism

hope this : )