So this drug has two pKa values, the tertiary amine with a value of 9.54 and the sulfonamide of a pka of 11.24.
The definition of a pka is the pH at which half of the molecule population is charged and the other half is neutral, i.e., the pka for ammonia is 9.25. So, if the water is set to a pH of 9.25 and you have 100 ammonia molecules you will have 50 NH3 and 50 NH4+ molecules.
So back to your question, if you dissolve your drug into a buffered solution of pH 7 then both the tertiary amine and indole will be protonated (the tertiary amine will have a positive charge and the indole will be protonated). If you set the pH of your solution at a value of 10 most of the population of tertiary amines will be uncharged (deprotonated) but the sulfonamide will still be charged (protonated).
So the protonation state of the drug is dependent on the pH of the solution.
Having had a look at the structure of the Sumatriptan on Wikipedia (http://en.wikipedia.org/wiki/Sumatriptan), I assume these two values are referring to the two Ka's of the molecule when fully protonated: (C14H23N3O2S)++. The two resulting tertiary ammonium cations have different acidities.
when the solvent and the temperature of a solute change, the values of Ka of acids change, too. I guess these values are given for the molecule in water at 25 °C.
again, from what I see from Wikipedia, I see Sumatriptan is a diprotic base so you take the pKb1 (= 14 - 11.24) and pKb2 (= 14 - 9.54) at 25 °C in water..
So this drug has two pKa values, the tertiary amine with a value of 9.54 and the sulfonamide of a pka of 11.24.
The definition of a pka is the pH at which half of the molecule population is charged and the other half is neutral, i.e., the pka for ammonia is 9.25. So, if the water is set to a pH of 9.25 and you have 100 ammonia molecules you will have 50 NH3 and 50 NH4+ molecules.
So back to your question, if you dissolve your drug into a buffered solution of pH 7 then both the tertiary amine and indole will be protonated (the tertiary amine will have a positive charge and the indole will be protonated). If you set the pH of your solution at a value of 10 most of the population of tertiary amines will be uncharged (deprotonated) but the sulfonamide will still be charged (protonated).
So the protonation state of the drug is dependent on the pH of the solution.
Firstly, a zwitter ion is a molecule that contains both negative and positive charge such that the net charge of the molecule is zero.
Back to your question. In a simplistic view (constant temperature and pressure) the pKa value of a molecule does not change. So the pKa of the tertiary amines of all the molecules you are interested in are essentially the same (pKa = 9.5). The other pKa values you are noting are of the indole unit and are really high and that would require a strong base to deprotonate.
So if you were to have a solution with a pH of 5 and 6 all of your molecules of interested with have a protonated tertiary amine (the amine will have an added hydrogen and will have a positive charge). At pH of 9 you will have a percentage of charged molecules.
I would suggest you read a little about the Henderson-Hasselbalch equation, this equation describes what a pKa is. You can also use this equation to predict what percentage of the molecule population will be charged or uncharge.
Regarding to your question, the strongest acidic pKa refers to N-H dissociation of the indol part, while the strongest basic pKa refers to the dissociation of the protonated tertiary base.
If you are interested in the solvent composition, search for marvinsketch and you can see the distribution of the various ionized states at different pH values. As a starting point that is enough.
From GSK, sumatriptan molecule has 2 pKas (tertiary amine, 9.63) and (sulphonamide, >12). The indole nitrogen doest not involve in either protonation or deprotonation (within pH 1 - 14).