c. You have 0.05M x 50 mL = 2.5 mmols of the amino acid. When you have titrated with 25 mL of 0.1 M NaOH, you have neutralized all of the COOH part but none of the amine part and you have the zwitter ion. That is the first equivalence point (or the isoelectric point and pH = (pka1 + pka2)/2.
b. You start with 2.5 mmols alanine. When all has been converted to the zwitter ion by the addition of the 25 mL of 0.1 M NaOH, the concn of that material then is 25 mmols/75 mL = ?M.
So this is a three part question. I solved the first part correctly but I need help with the second and third.
Q: The dissociation equilibrium constants for the protonated form of of alanine (a diprotic amino acid, H2X+)are Ka1 = 4.6 x 10^-3 and Ka2 = 2.0 x 10^-10. Assume that you start with 50.0 mL of 0.050 M solution of alanine.
a) What is the pH of this initial solution? My answer: pH = 1.88
b) What is the concentration of HX after 25.0 mL of 0.100 NaOH has been added to the solution?
c) What is the pH of the solution after 25.0 mL of 0.100 NaOH has been added to the solution?
I need more help understanding how to do the actual problem on a) and b) than the actual answer. If someone can set me on the right path I can do all the math to solve it. Thanks.
2 answers
Thank you so much DrBob222. I really appreciate your help!
1 answer