lets mono protic acid = HA
construct ICE table
    HA (aq) + H2O (l) <----> H3O+
(aq) + A- (aq)
IÂ Â Â Â Â Â
0.34Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
0
CÂ Â Â Â
-x                               Â
+x               Â
+x
E =
0.34-x                          Â
+x                 Â
+x
Ka = [H3O+] [A-] / [HA]
3.4 x 10^-6 = [x] [x] /[0.34 -x]
x2 + x 3.4 * 10^-6 - 1.156 x 10^-6 = 0
solve the quadratic equation
x = 0.001073 M = [H3O+]
pH = -log[H3O+] = -log[0.001073M]
pH = 2.97
