Quote (cialda @ Oct 29 2013 07:42pm)
Originally/initially, it should be 0 for both CH3COO- and H3O+. then both of them change by +x each, and you know that H3O+ = 0.0019
in this case, it could be easy to do without an ICE table, but im sure they are just using basic ones for now. when you get to more complicated problems, the ICE table will come in handy
both are originally "0" and the reaction proceeds in the direction of producing H3O+ and CH3COO-. when it proceeds, it produces 1 mol of H3O+ / 1 mol of CH3COO- and consumes 1 mol of CH3COOH / 1 mol of H3O & 1 mol of CH3COO- produced
You already know that [CH3OO-] = [H+] you just don't know that you know it. That's why you write +x and +x in the "C" row. Since they have the same molar ratio (given in the initial equation) no coefficients are needed. 1:1:1
Like ^ said, this first one doesn't really need a table since they give you [H+] in the beginning but it's a good habit to get into when problems get harder.