1. The problem statement, all variables and given/known data A 0.0200 mol/L acid solution has a pH of 2.347. Calculate the percent ionization of this acid. Calculate the Ka of this acid 3. The attempt at a solution I first calculated the hydronium ion concentration: 10^-pH = 10^-2.347 = 4.5*10^-3 I then used this concentration to calculate the percent ionization: 0.0200 / 4.5*10^-3 = 22.5% Can anyone confirm this answer. Now the second part of the question asks for the Ka. My prof said that if you got 1.3*10^-3 as the Ka value it is wrong. I keep getting this value and it makes no sense why it would be anything other than this value: Since the ratio in the balanced chemical equation is 1:1 , the equilibrium expression is: Ka = (4.5*10^-3)(4.5*10^-3) / (0.0200 - 4.5*10^-3) solving for x I get 1.3*10^-3, however this isn't suppose to be the right answer.. can anyone tell me where I've gone wrong or maybe my prof is mistaken? Thanks.