from the pH find the [H3O+]
[H3O+] = 10-pH = 10-2.69 = 0.00204
construct the ICE table
lets mono protic acid = HA
        HA (aq) + H2O
(l) <-----> H3O+ (aq) + A- (l)
IÂ Â Â Â Â Â Â Â
0.0129Â 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 Â
0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
0
CÂ Â Â Â Â
-x                                       Â
+x           Â
+x
EÂ Â
0.0129-x                             Â
+x             Â
+x
from the ICE table concentration of H3O+ = x = 0.00204
concentration of H3O+ = [A-] = x = 0.00204
equilibrium concentration of HA = 0.0129 - 0.00204 = 0.01086
M
Ka = [x] [x] / [0.0129 -x]
Ka = [0.0024] [0.0024] / [0.01086]
Ka = 5.3 x 10^-4
