Question

Calculate the equilibrium concentrations of all chemical species present in a 0.10 M H CO.(aq) solution. What is the percent ionization of the acid and resulting pH? Units Units [H2CO3) = Number [HCO3) = Number [CO32) = Number [H30+) = Number Units Units % ionization = Number pH = Number

Answer 1

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

[CO3^2-] = Ka2 = 4.7 x 10^-11 M

(Since, second dissociation gives [CO3^2-]= ka2 )

pH = - log [H3O+] = -log 2.12 x 10^-4 M = 4 - log 2.12

pH = 4 - 0.33 = 3.67