Question

Starting with 120 mL of a 0.45 M solution of phenylalanine, shown above at pH = 7.0, how many mL of a 0.400 M sodium hydroxide solution must be added so that phenylalanine has a net charge of -1? Your answer should be in mL.

Answer 1

Phenylalanine has two ionizable groups, the α-carboxyl group having pK_a = 1.83 and the α-amino group having pK_a 9.13.

When pH < pK_a1, the α-carboxyl group is protonated (carries charge 0) while when pH > pK_a1, the α-carboxyl group is deprotonated (carries charge 1-). Similarly, when pH < pK_a2, the α-amino group is protonated (carries charge 1+) while when pH > pK_a2, the α-amino group is deprotonated (carries charge 0).

At pH 9.13, the α-carboxyl group and the α-amino groups are completely deprotonated and the charge is 1-. The volume of NaOH required will be equal to the volume of NaOH required to completely neutralize phenylalanine.

Considering phenylalanine as a diprotic acid (H₂A), the dissociation reaction is

H₂A (aq) + 2 NaOH (aq) --------> Na₂A (aq) + H₂O (l)

As per the stoichiometric equation,

1 mole H₂A = 2 moles NaOH.

Millimoles of H₂A added = (120 mL)*(0.45 M) = 54 mmole.

Millimoles of NaOH required to completely neutralize H₂A = (54 mmole)*(2 moles NaOH/1 mole H₂A) = 108 mmole.

Volume of NaOH required = (108 mmole)/(0.400 M) = 270 mL (ans).