Help would be much appreciated! Here's the problem:
An unknown compound, X, is though 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 0.1M NaOH is added to 100 mL of a 0.1M solution of X at pH 2.0, the pH increases to 6.72. Calculate the pKa of the second ionizable group of X.
OK, so I'm not sure the pKa of the carboxyl group is relevant here, since at the initial pH, it will be 50% dissociated. As it goes up to pH 6.72, we will have most of it be dissociated. Thus, I need to find the A-/HA ratio at the new pH to use the Henderson-Hasselbalch equation (pH = pKa + log(A-/HA)). I tried it this way but I'm pretty sure there's something wrong: converted Molar of X solution to moles using the given volume and subtracted the moles added of NaOH (.1M x .075L) to get a pKa of 5.87. Another way I did this that sounds more reasonable is since the compound X is very acidic (initial pH 2) and the NaOH is obviously our base, I set HA = moles of X = (.1M)(.1L) and A- = moles of NaOH = (.1M)(.075L). Using the final pH of 6.72 and plugging into the H-H equation, I got a pKa of 6.84. Again, help would be much appreciated cuz I might be over-thinking this too much!