[cvc121]

In my chemistry textbook, there is an amino acid titration problem as follows: Calculate the pH when 16 mL of 0.2 M NaOH is added to 25 mL of 0.1 M glycine, whose initial pH = 2.0. The relevant pKa values for the ionizable groups of glycine are 2.2 and 9.8.

The model solution provided assumes that all of the glycine is in its protonated form, that is, HA = 2.5 x 10^-3 moles and A^- = 0 moles.

My question is, why is this assumption made? An initial pH of 2.0 is still very much in the buffering region since the pKa value for the carboxyl group of glycine is 2.2. Thus, wouldn't there be A^- present in solution as well at pH = 2.0, albeit, slightly less than HA? My initial instinct was to assign 2.5 x 10^-3 - x to HA and x to A^- to solve for the relative amounts of each by plugging the values into the H-H equation as such and solving for x: 2.0 = 2.2 + log x / 2.5 x 10^-3 - x

All clarification is very much appreciated. Thanks.

[Babcock_Hall]

I would infer that the model solution is designed for a simpler situation that you are encountering in this problem.  For example, glycine has two pK_a values, but the model system has only one.  I did not work this problem in detail, but your approach looks reasonable in terms of finding the starting amounts of each form.

[Borek]

My initial instinct was to assign 2.5 x 10^-3 - x to HA and x to A^- to solve for the relative amounts of each by plugging the values into the H-H equation as such and solving for x: 2.0 = 2.2 + log x / 2.5 x 10^-3 - x

That would be my approach as well. 10^0.2=1.6, HA and A^- concentrations are comparable.