pH = 2.40
[H+] = 10^-2.40 = 3.98 x 10^-3 M = x
C = 0.080 M
acid = HA
HA <-------------------------> H+ + A-
0.080 0 0 ---------------> initial
0.080-x x x ---------------> equilibrium
Ka = [H+][A-]/[HA]
Ka = (3.98 x 10^-3)^2 / (0.080 - 3.98 x 10^-3)
Ka = 1.58 x 10^-5 /(0.080 - 3.98 x 10^-3)
Ka = 2.078 x 10^-4
pKa = -log Ka = -log (2.078 x 10^-4 )
pKa = 3.68
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