Question

Chloroacetic acid, HC2H202CI, is a stronger monoprotic acid than acetic acid. In a 0.10 M solution, the pH is 1.96. Incorrect. Did you use the correct equilibrium expression? Did you use the correct number of significant figures? Calculate the Ka for chloroacetic acid 0.0013 Ka

I can't figure this problem out. Please help!

Answer 1

8 80 Ionization reaction in water and ICE table HGH 2 O + H2O - Hote, + Catholicae, T 0.10 C - x E (0.10-x) give: pH=1.96 [H2O+J = 10 = 10-'o = 0.01096478 M at equilibrium: [H20+) = OC = 0.01096470 Ka = [H₂O + ] [ C₂ H₂O₂C1-] [HC₂ H₂O₂ CI] (0.01096478) 2 (o. 10-x) (0.10 - 0.01096478) + Ka = 1.3503*10-3 1.4x10-3 Ka = 0.0014 | till 2 sig fig. 2 . 2