An unknown compound, X is thought to have a carboxyl group with a pKa of 2.0 and another ionizable group with a pKa between 5 and 8. When 75 mL of 0.1 M NaOH was added to 100mL of a 0.1 M solution of X at pH 2.0, the pH increased to 6.72. Calculate the pKa of the second group of X. ok, here's another problem. I got the answer of 7.2 while the answer in the back says 7.3. I just found out that i was actually wrong in my other post of "Dissociation of glycine". This book is pretty accurate as far as the answers go and when we're dealing with logs, .1 is still quite a big difference. Here's how i solved it: i knew that since pKa = pH at the beginning there must be equal amounts of X carboxyl group and it's ionized form. Since there were .01 moles in total there must be .005 moles of X carboxyl group. When .0075 moles of OH(-) were added the carboxyl group completely dissociated and there were still .0025 moles of OH(-) left in the solution. there were .01 moles of X in the beginning, and I'm assuming that none of the second group has been dissociated. The remaining OH(-) will react with the .01 moles X-second group, producing .0025 moles of the ionized form of X-second group, while only 0.0075 moles of X-second group remain. Thus we go back the henderson hasselbach equation of 6.72 = pKa + log(.0025/.0075) and solve for pKa i get 7.2. However, I'm thinking that the book could be wrong, in that it assumed that the .01 moles of X-second group did not react, and thus 6.72 = pKa + log(0.0025/.01) will be roughly 7.3. Can some one explain this. Thanks.