What is the pH of the solution after the addition of 1.0 x 10-3 moles of NaOH?

Question

100 ml of 0.0150 M solution of benzoic acid (C6H5COOH). Ka(C6H5COOH) = 6.4 x 10-5.

a) What is the pH of the benzoic acid solution?

b) What is the pH of the solution after the addition of 1.0 x 10-3 moles of NaOH? You may assume no volume change to the solution upon addition of the NaOH.

This question appeared on a test I just got back a few weeks ago but I got question B incorrect. I worked out that question A is pH=3.0 correctly. But I am lost on how to answer question B. My attempt gained the answer pH=2 but this was incorrect. What would the working be? I have exams coming up and I would like to brush up on my knowledge of questions like this. Any help is greatly appreciated! Thank you so much!!!

Answer 1

C₆H₅COOH ==> C₆H₅COO^- + H⁺

a) Ka = 6.4x10-5 = [C₆H₅COO^-][H⁺]/[C₆H₅COOH] = (x)(x)/0.150 - x

If we assume that x is small relative to 0.150 M, we can neglect it and avoid using the quadratic:

6.45x10^-5 = x²/0.150 and x² = 9.675x10^-6

x = [H⁺] = 3.11x10^-3 M (this is only 2% of 0.150 M so out assumption was correct)

pH = -log [H+] = -log 3.11x10^-3 = 2.51

Not sure how you arrived at a pH of 3.0 and got it correct. You should have included your workings. Also, since you are adding a base, the pH would have to go up, not down as you have concluded.

b) Addition of NaOH will react with the acid to reduce the concentration of C₆H₅COOH and increase the concentration of C₆H₅COO^- thus forming a buffer. C₆H₅COOH + OH^- ==> C₆H₅COO^- + H₂O

Initial moles acid = 0.1 L x 0.150 mol/L = 0.0150 moles

Final moles acid = 0.0150 moles - 0.001 moles = 0.014 moles

Final moles conjugate base (C₆H₅COO^-) = 0.001 moles

Since no volume change, final [acid] = 0.014 mol/0.1L = 0.14 M and final [conj.base] = 0.001mol/0.1L = 0.01M

Using the Henderson Hasselbalch equation pH = pKa + log [conj.base]/[acid] we can solve for pH.

pKa = -log Ka = -log 6.4x10^-5 = 4.19

pH = 4.19 + log (0.01/0.14) = 4.19 + log 0.0714

pH = 4.19 + (-1.15)

pH = 3.04