Question

The following questions will be based on a solution made with aminobenzoic acid or one of its bases.

Pka1= 2.08 Pka2= 4.96

A. What is the pH of a 0.142 M aminobenzoic acid hydrochloride solution? (Assume activity coefficients are 1.0)

B. List all the assumptions that were made to simplify the pH calculation. Explain/justify why each assumption was made.

C. 10.00 mL of a 0.100 M NaOH was mixed with 25.00 mL of the 0.142 M aminobenzoic acid hydrochloride solution. What is the pH of the resulting solution? (Assume activity coefficients are 1.0).

Answer 1

Parts A and B.

pH = 7 - 1/2 (pKa2 + Log[aminobenzoic acid hydrochloride])

i.e. pH = 7 - 1/2 (4.96 + Log0.142)

i.e. pH = 7 - 1/2 (4.96 - 0.85)

i.e. pH = 7 - 1/2 (4.11)

i.e. pH = 7 - 2.06

i.e. pH = 4.94

Assumption: Since, NH2.HCl will have high pKa value, the pKa2 value is taken.

Part C.

The no. of millimoles of aminobenzoic acid (n_{aminobezoic acid}) = 25 mL * 0.142 mmol/mL = 3.55 mmol

The no. of millimoles of NaOH (n_NaOH) = 10 mL * 0.1 mmol/mL = 1 mmol

According to Henderson-Hasselbulch equation:

pH = pKa1 + Log{n_NaOH/(n_{aminobezoic acid} -n_NaOH)}

i.e. pH = 2.08 + Log{1/(3.55 - 1)}

i.e. pH = 1.67