[salleebrowne]

How can I tell what the most abundant species would be at a certain pH; say of 9.8 or 12.0? Does it become more basic as it protonates?

[Smrt guy]

For each of the given pKa values, where pH = pKa, [HA] = [A-] where HA represents the protonated form and A- represents the deprotonated.  So as you increase pH from pH = pKa to pH > pKa the base form will predominate.  As a good rule of thumb, any time pH - pKa > 2, only the base form will exist in solution (for all intents and purposes).  The opposite is true for acids as well (pKa - pH > 2 is all acid form). 

[salleebrowne]

Thanks for your response. I understand the differences in pH and pKa. So does that mean that the Arginie ion labeled "C" in the diagram is most abundant at a pH of 9.8 and "D" is most abundant at pH=12.0?

[orgopete]

C will be most abundant at pH 9.8, but it will also be the most abundant at pH 12.0 also. If the pH were 12.5, it would equal the pKa and an equal amount of C and D would be present. At pH 11.5, the ratio would be 91:9 C/D and at pH 13.5 the ratio would be 99:1 C/D. Since the pH is less than 12.5, it will still favor C.

[salleebrowne]

Thank You! This is starting to make sense. How did you arrive at those ratios?

[orgopete]

This was well explained in Paula Bruice's organic chemistry book. It goes like this. From the Henderson-Hasselbalch equation,

pH = pKa + log (A-/HA)

If [A-] = [HA], then pH = pKa.

If [A-] = 10 x [HA], then log (A-/HA) = log (10) = 1.
pH = pKa + 1

If 100 x [A-] = [HA] , log (A-/HA) = log (1/100) = -2
pH = pKa -2

How can (A-/HA) = 10, 100, 0.1 or 0.01 (1/100)? If [A-] = 0.01M and [HA] = 1M.

[salleebrowne]

You're awesome!! Thanks