Concentration of Ionic Species

Question

What are the concentrations of the various ionic species in a 0.1M solution of lysine at pH 4,

7, and 10?

Studying for an exam and this is a study question. I need help making a step by step so I can use it on every problem like this.

The answers are

pH 4: [Lys+1] = 0.099 M and [Lys+2] = 1.0 x 10-3 M

pH 7: [Lys0] = 1.0 x 10-3 M and [Lys+1] =0.099 M

pH 10: [Lys+1] = 7.1 x 10-3 M, [Lys0] = 0.071 M, [Lys-1] = 0.022 M

Answer 1

I'm pretty sure that we'd need the pI for lysine, or at the very least, the pKa values for the ionizable groups.

These values will vary depending on the source one uses, but here are values from one source:

pKa COOH = 2.2

pKa NH₃ = 9

pKa NH₃ side chain = 10.5

pI given = 9.8

pI calculated = (9 + 10.5) / 2 = 9.75

At pH = pKa1= 2.2, there would be an equal amount of dicationic (COOH, NH₃⁺, NH₃⁺) and monocationic form (COO-, NH₃⁺, NH₃⁺).

As the pH increases above 2.2, there will be more monocationic (COO-, NH₃⁺, NH₃⁺) compared to the dicationic (COOH, NH₃⁺, NH₃⁺) form .

At pH =pKa2 = 9.0, there will be an equal amount of the monocationic (COO-, NH₃⁺, NH₃⁺) form and zwitterion (COO-, NH₃⁺, NH₂)

At pH above 9.0, there will be more COO-, NH₂, NH₂

Calculation of the molarity of the different forms requires using the Henderson Hasselbalch equation.

pKa = pH + log [HA] / [A-]