Amphoteric compounds-effect of pH

[mesna]

Hello friends!

I have a query for which i have been trying to find an answer for a long time but in vain!

Can anybody tell me how do we find the effect of pH on amphoteric compounds?I am currently working on Amphotericin B which happens to have both the acidic and the amino groups. I need to know how such compounds would behave in solutions with different pH values, especially physiological pH. (The -COOH pka is 5.7 and the -NH2 pka is 10.0).

I would be really really grateful if some1 could answer this question!

Thanks a ton!
Mesna

 

[Hootenanny]

Hello friends!

I have a query for which i have been trying to find an answer for a long time but in vain!

Can anybody tell me how do we find the effect of pH on amphoteric compounds?I am currently working on Amphotericin B which happens to have both the acidic and the amino groups. I need to know how such compounds would behave in solutions with different pH values, especially physiological pH. (The -COOH pka is 5.7 and the -NH2 pka is 10.0).

I would be really really grateful if some1 could answer this question!

Thanks a ton!
Mesna

The pH of a solution of amino acids depends on the conditions. The amino acid acts as a buffer and the carboxyl group will deproponate in alkaline conditions and the amino group will proponate in acidic solutions. If you are looking for the pH of a solution where the amino acids will be predominantly zwitterions (neutral charge) then you need to find the isoelectric point (pI) of the amino acids. The isoelectric point can be found by averaging the pK_a values of each of the ionisable groups on your amino acid (this will give you the pH).

As amino acids act as buffers you can use the Henderson-Hasselbach equation to determine the ratio of charged amino groups to non-charged and charged carboxyl groups to non-charged. For example, the ratio of carboxylic acid to carboxyl ions at physiological pH (7.4) is;

$$pH = pK_{a} + \log\left( \frac{\left[COO^{-}_{(aq)} \right]}{\left[COOH_{(aq)} \right]}\right)$$

$$7.4 = 5.7 + log\left( \frac{\left[COO^{-}_{(aq)} \right]}{\left[COOH_{(aq)} \right]}\right)$$

$$\frac{\left[COO^{-}_{(aq)} \right]}{\left[COOH_{(aq)} \right]} = 10^{7.4-5.7}$$

$$\frac{\left[COO^{-}_{(aq)} \right]}{\left[COOH_{(aq)} \right]} = 50.1$$

So at pH 7.4 there are fifty times more carboxyl ions than carboxyl groups. Hope this was helpful.

 

[GCT]

If you ever to a titration of such amphoteric compounds, you'll find that, for the most part, you'll need to account for the individual acid and base components themselves with respect to pH. That is, as long as the pKa of the groups are adequately separated, you'll completely be concerned with the titration of one group, before the start of the other. Use the H-H equations that Hootenany gave you for each of the acid components for the amino acid.


For example, the pKa for the carboxylic acid group is said to be 5.2 in for your problem, when the pH=5.2 you'll have half of all of the carboxylic acid groups in the acid and conjugate base form and none of the amine groups will be affected, that is all of them will still be protonated. At the pH of 10.0, all of the carboxylic acid groups will be deprotonated (in their conjugate base form) and half and half acid/conjugate acid form for the amine group.

 

[mesna]

Thank you guys, thanks alot. It really helped.

cheers
Mesna