A weak acid (HA) has a pKa of 4.734. If a solution of this acid has a pH of 4.273, what percentage of the acid is not ionized? (Assume all H in solution came from the ionization of HA.)
HA <------> H+ + A-
c x x
pH = -log [H+]
4.273 = -log [H+]
[H+] = 5.33*10^-5 M
so,
x = 5.33*10^-5 M
pKa = -log Ka
4.734 = -log Ka
Ka = 1.845*10^-5
Ka = x*x/c
1.845*10^-5 = (5.33*10^-5)*(5.33*10^-5)/c
c = 1.54*10^-4 M
This is the unionized acid present
percent of acid not ionized = (unionized acid
present)*100 / initial acid concentration
= 1.54*10^-4 *100 / (1.54*10^-4 + 5.33*10^-5)
=74.3 %
Answer: 74.3 %
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