A weak acid (HA) has a pKa of 4.161. If a solution of this acid has a pH of 4.140, what percentage of the acid is not ionized? (Assume all H in solution came from the ionization of HA.)
pH = 4.140
pKa = 4.161
HA ---------------------> H+ A-
[H+] = 10^-pH = 7.244 x 10^-5 M
[A-] = 7.244 x 10^-5 M
Ka = 6.90 x 10^-5
HA ---------------------> H+ + A-
C 0 0 ------------------> initial
-x +x +x ------------------> dissociation
C-x x x ----------------------> equilibrium
x = [H+] = [A-] = 7.244 x 10^-5 M
Ka = [H+][A-]/[HA]
Ka = x^2 / C- x
6.90 x 10^-5 = x^2 / C- x
6.90 x 10^-5 = (7.244 x 10^-5 )^2 / C- 7.244 x 10^-5 M
by solving this
C =1.485 x 10^-4 M -----------------------> initial concentration
percentage of the acid is ionized = (x /C) x 100 = ( 7.244 x 10^-5 / 1.485 x 10^-4 ) x 100
= 48.79 %
percentage of the acid is not ionized = 100 - 48.79 %
= 51.21 %
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