Posted by Maria on Tuesday, December 20, 2011 at 4:42pm.
pH = 3.65 = -log(H^+).
(H^+) = 2.24E-4
Call benzoic acid (C6H5COOH) HBz.
...........HBz ==> H^+ + Bz^-
initial..2.24E-4...0.....0
change.....-x.......x.....x
equil....2.24E-4-x...x....x
Ka = (H^+)(Bz^-)/(HBz)
Substitute from the ICE chart above into the Ka expression and solve for Ka.
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