Originally Posted by craig97
question is with attachment on this thread
options for this are:
please provide explanation for your answer
I think it is +1 but I can not be too sure.
The way I do it is:
pKa for the carboxyl group is 2.2, meaning that at pH 2.2 half of the COOH will be in COO- and half will be in COOH form. As we increase the pH to 7.4 (physiologic), there are less protons around so more COOH will deprotonate. At 7.4 most of the carboxyl will be in COO- form and that is -1 in our net charge calculation.
pKa for both ammonium groups is well above 7.4. Half of the ammonium groups are in NH2 and half are in NH3+ at basic pH values. At a less basic pH of 7.4, there are more protons around so more NH2 will be protonated to form NH3+. Since there are two ammonium groups with pKa's of 9.2 and 10.7 they will both be protonated. That is +2 in our net charge calculation.
This might all be wrong and I will look terribly stupid but thats my take on it.