Question

1. The pKa's of the amino acid methionine are 2.28 and 9.21. The zwitterion of methionine is given below (I) S H3C O NH3 (a) Explain why structure (I) is an intermediate ion. Write balanced chemical reactions to support your reasoning. (b) Draw the conjugate acid AND conjugate base form of the amino acid shown above (c) A 15-mL aliquot solution consisted 0.25M of the zwitterion of methionine and 0.30M of its conjugate acid was titrated with 0.15M N2OH, calculate the equilibrium pH after the addition of 15mL and 65mL of N2OH NOTE: The 15-mL sample contains both 0.25M of the zwitterion of methionine and 0.30M of its conjugate acid of methionine. In other words, the sample total volume is only 15mL NOT 30mL

Answer 1

(a) Structure (I) is an intermediate ion because the basic amine group exists in NH₃⁺ form as a base accepts a proton and the acidic carboxylic acid group exists in COO^- form as an acid donates proton.

NH₂$\rightleftharpoons$ NH₃⁺

COOH $\rightleftharpoons$ COO^-

(b) The conjugate acid can be drawn as follows.

H₃C-S-CH₂-CH₂-CH(NH₃⁺)-COOH

The conjugate base can be drawn as follows.

H₃C-S-CH₂-CH₂-CH(NH₂)-COO^-

(c) pKa of carboxy terminal (COOH) = 2.28

15 mL of NaOH:

The resulting [Zwitter ion] = (15×0.25 + 15×0.15) mmol/(15+15) mL = 6 mmol/30 mL = 0.2 M

The resulting [Conjugate acid] = (15×0.3 - 15×0.15)/(15+15) mL = 0.075 M

According to the Henderson-Hasselbulch equation:

pH = pKa + Log([Zwitter ion]/[Conjugate acid])

i.e. pH = 2.28 + Log(0.2/0.075) = 2.71

65 mL of NaOH:

pKa of amino terminal (NH₂) = 9.21

The resulting [Conjugate base] = {65×0.15 - (15×0.25 + 15×0.3)} mmol/(15+65) mL = 1.5 mmol/80 mL = 0.01875 M

The resulting [Zwitter ion] = {(15×0.25 + 15×0.3) -1.5} mmol/(15+65) mL = 0.084375M

According to the Henderson-Hasselbulch equation:

pH = pKa + Log([Conjugate base]/[Zwitter ion])

i.e. pH = 9.21+ Log(0.01875/0.084375) = 8.56