A 0.020 M solution of a weak acid HA has a pH of 3.50. What is the percent ionization of HA in the solution?
For the solution described above, what is the Ka?
use:
pH = -log [H+]
3.5 = -log [H+]
[H+] = 3.162*10^-4 M
HA dissociates as:
HA
-----> H+ + A-
2*10^-2
0 0
2*10^-2 -
x
x x
Here:
x = [H+] = 3.162*10^-4 M
1)
% dissociation = x * 100 / initial concentration
= (3.162*10^-4) * 100 / 0.020
= 1.58 %
Answer: 1.58 %
2)
Ka = [H+][A-]/[HA]
Ka = x*x/(c-x)
Ka = 3.162*10^-4*3.162*10^-4/(0.02-3.162*10^-4)
Ka = 5.08*10^-6
Answer: 5.08*10^-6
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