[OhGosh]

Hey guys

Encountered this little problem and I've actually solved it, but I don't think my answer is correct.

Given glycine which is NH2CH2COOH. I was asked to calculate [+NH3CH2COO-]/[+NH3CH2COOH] at pH 5.5

I was given the Ka of COOH group which is 4.47 x 10^-3 as well as Kb of NH2 group which is 6.03 x 10-5 .



What I've done was first calculate the [H+] : -lg[H+] = 5.5  --> [H+] = 10^(-5.5) M

After which : I did  ([+NH3CH2COO-][H+])/[+NH3CH2COOH] = 4.47 x 10^-3
                         
By subbing in the [H+] value found and using the Ka value provided for the -COOH group, i found the answer (the value of [+NH3CH2COO-]/[+NH3CH2COOH]) .

However by doing this I'm assuming the Ka given to be for the reaction :  +NH3CH2COOH --> H+ + +NH3CH2COO-

Am I right to assume this? Or is the Ka(for COOH group) given for the reaction : NH2CH2COOH --> NH2CH2COO-  +  H+


 I did not use the Kb value given at all, therefore I think I might not be correct.


Any tips on this?


Thanks in advanced for the help.

[rabolisk]

You are correct. The method you used is fine, and you could've also used the Henderson-Hasselbach equation. The given Ka is indeed for +NH₃CH₂COOH   +NH₃CH₂COO^- + H⁺. What you were supposed to do is realize that there are two acidic groups, the COOH group and the NH₃⁺ group. You can find the Ka of the amino group (Ka2) using the Kb given. You can see that Ka2 << Ka1, or that pKa2 >> pKa1. Then this amino acid is like a diprotic acid (e.g. acetic acid), where you have two protons that can be dissociated. The first proton (COOH) has to come off before the second proton (NH₃⁺), which means that the reaction has to be the way you assumed it would be. The only thing is that this assumption isn't really an assumption, it can be shown by the Ka and Kb values given.