I'd really appreciate help :) 1. The problem statement, all variables and given/known data A 0.05 M aqueous solution of sodium hydrogen sulfate, NaHSO4, has a pH of 1.73. Calculate Ka2 for sulfuric acid. Sulfuric acid is a strong electroylyte, so you can ignore the hydrolysis of the HSO4- ion. 2. Relevant equations If I knew that I'd know the answer ;) 3. The attempt at a solution A 0.05 M aqueous solution of sodium hydrogen sulfate, NaHSO4, has a pH of 1.73. Calculate Ka2 for sulfuric acid. Sulfuric acid is a strong electrolyte, so you can ignore the hydrolysis of the HSO4- ion. Sulfuric acid is H2SO4. The Ka1 equation of sulfuric acid would be H2SO4 + H2O <-> H3O + HSO4 I would of thought Ka2 would be HSO4- + H2O <--> SO4(2-) + H3O+, but it says to ignore the hydrolysis. In saying to ignore it because of being a strong electrolyte, I assume it means that H+ and SO4(2-) are the products of HSO4-. That looks the same to me as the hydrolysis, assuming the H+ eventually bonds with a hydronium ion. So I guess the Ka2 must be HSO4- --> H+ + SO4(2-) giving [H][SO4(2-)]/[HSO4] = Ka2... but if it is a strong electrolyte it will completely go to products, so it would have no Ka, since dividing by zero would be infinity. Being stuck there I try looking at the sodium hydrogen sulfate. The first thing that would happen would be for the Na and HSO4- to disassociate in water. It says the pH is 1.73, so the [H3O+] = 10^-1.73, and the concentration of SO4(2-) would be the same. The Ka of the HSO4- would be (10^-1.73)^2/0.05 = 0.0693. The equilibrium of HSO4- is the same thing as that as the Ka2 of sulfuric acid, but my answer is not matching the back of the book, which was 0.011. I had one other question which is really simple and has nothing to solve. I was reading a example given about titrations where you are titrating NH3 with HCl and it is at the equivalence point. The part of the solution I did not follow was using this formula: NH3 + HCl --> NH4+ + CL- What I don't understand is why the NH3 did not already partially react with water, since it was aqueous. Shouldn't it have reacted like NH3 + H2O <--> NH4+ + OH-? I thought that since all acids and bases are in water that they have already reacted with water before you use them for anything unless it is a weak acid or base... like in a container of .1M NH3, it contains mostly water... shouldn't it react before anything is added?