[Immunoglobulin]

Hi guys, I was wondering if any of you can help me with a homework question that I'm trying to figure out. I'm taking a biochem course in university right now, but it's been awhile since I've done chemistry in general since I've taken some years off. Needless to say it's been a bit of a struggle to refresh some of this knowledge.

Anyway, the question goes like this:
-There is an unspecified amino acid (37.238g of its dried zwitterion form) and you dissolve it in 1.00L of 0.100M HCl.
-You then take a 50mL sample and titrate it with 0.8812M NaOH.
-On your titration curve, you find that the first equivalence point is at 22.70 mL and the second equivalence point is at 36.86 mL of NaOH added.
-What is the molecular weight of the amino acid?

Thanks in advance for your help

[Babcock_Hall]

Welcome to the forum.  It is a rule that you must show an attempt before we can help you.  However, this problem is constructed in an unusual way, so it might take a little bit more thought than a typical titration problem.  Maybe you can give an outline of how you think you should proceed, or some more general thoughts.

[Immunoglobulin]

Welcome to the forum.  It is a rule that you must show an attempt before we can help you.  However, this problem is constructed in an unusual way, so it might take a little bit more thought than a typical titration problem.  Maybe you can give an outline of how you think you should proceed, or some more general thoughts.

Thanks!

Yeah I attempted the question last night and I did get an answer that I think is reasonable - not sure if it's correct though.

So, assuming the 37.238g of amino acid dissolved evenly throughout the 1L solution of 0.1M HCl, a 50mL sample of this solution would contain 1.862g of the amino acid:
37.238g / 1L = x / 0.050L, x = (37.238g x 0.050L) / 1L = 1.862g in the 50 mL sample

NaOH required to reach the 1st equivalence point: nNaOH =  8812M x 0.0227L = 0.02mol
NaOH required to reach the 2nd equivalence point: nNaOH = 0.8812M x 0.03686L = 0.0325mol
0.0325 – 0.02 = 0.0125mol of NaOH is required to deprotonate 1 equivalent of the amino acid, meaning that there are 0.0125mol of the amino acid in the solution.
 
MWAmino Acid = (1.862g) / (0.0125mol) = 148.96g/mol ≈ 149g/mol
Therefore, the molecular weight of the unknown amino acid is 149g/mol.

[Babcock_Hall]

That looks reasonable.  One might hazard a guess that this is methionine.