Solution :-
pH = 2.35
lets calculate the [H3O+] using the pH
[H3O+] = antilog [-pH]
          Â
= antilog [-2.35]
          Â
= 0.00447 M
HA + H2O ----- > H3O+ + A-
0.0175
MÂ Â Â Â Â Â Â Â Â Â Â
0Â Â Â Â Â Â Â Â Â Â Â Â Â
0
-x                       Â
+x          Â
+x
0.0175-x         Â
0.00447Â Â Â Â Â 0.00447
0.0175 - 0.00447 = 0.01303
ka= [H3O+] [A-] / [HA]
ka = [0.00447][0.00447]/[0.01303]
Ka= 1.53*10^-3
so the Ka of the acid is 1.53*10^-3
