1. The problem statement, all variables and given/known data prepare 1L of a .1M acetate buffer with a pH of 5.22 2. Relevant equations pK of acetate is 4.70 3. The attempt at a solution this is what my professor did: ph = pK + log(base/acid) 5.22 = 4.70 + log(base/acid) antilog(.52) = (base/acid) 3 = (acetate/acetic acid) = 3.3/1 you therefore require 3 moles of acetate for every mole of acetic acid. -------------------- I don't understand why the .3 was just ignored, it dosen't seem like she should have rounded down....any suggestions? I understand that eventually we will have to use .1M, which will reduce the sig figs to one, but that is not until the next step, and I thought you were not supposed to drop figures early.