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A 0.037 M solution of a weak acid (HA) has a pH of 4.49. What is the Ka of the acid?

Answer:
pH = 4.49
[H+] = 10^-pH = 3.23 x 10^-5
The equilibrium is
HA <-----> H+ + A-
initial concentration
0.037
at equilibrium
0.037-3.23 x 10^-5 . . . .3.23 x 10^-5 . . . .3.23 x 10^-5
Ka = ( 3.35 x 10^-5)^2 / 0.037 - 3.23 x 10^-5 =
= 3 x 10^-8

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