Question

A nonstandard amino acid has three ionizable groups, with pKa values of 2.6, 7.4, and 10.13. A 0.1 M solution of this amino acid is made, 100 ml total, at pH 2.6. To this solution is added 10 ml of 1 M NaOH. What is the final pH of this solution after the addition?

A) 3.2
B) 8.0
C) 7.4
D) 6.8
E) 10.13

Answer 1

Recall that the amino acid will be:

For pKa = 2.6M we are talking about the acidic proton, COOH

pH = pKa + log(COO-/COOH)

2.6 = 2.6 + log(COO-/COOH)

(COO-/COOH) = 10^0 = 1

COO-/COOH = 1

total species --> mol = MV = 0.1*100 = 10 mmol

COO-/COOH = 1 --> COO- = COOH

COO- + COOH = 10

COO- = 5; COOH = 5

after addition of NaOH:

mmol of NaOH = MV = 10*1 = 10 mmol

this will neutralize:

COOH = 5-5 = 0 mmol of COOH; still 5 mmol of OH- left

COO- formed = 5 + 5 = 10 mmol fo COO-

now...

2nd point is pK = 7.4; most likely another COOH group so

pH = pKa2 + log(COO-/COOH)

mmol of OH- left:

mmol of COOH(2nd) = 10

mmol of COO-(2nd) = 0

after 5 mmol neutralize:

mmol of COOH(2nd) = 10 - 5 = 5

mmol of COO-(2nd) = 0 + 5

Substitut ein pH:

pH = pKa2 + log(5/5)

pH = 7.4 + log(1)

pH = 7.4

since this is now in the SECOND half equivalence point