A 25.0 mL sample of 0.124 M lactic acid (HC3H5O3, pKa = 3.86) is titrated with 0.124 M NaOH solution.

Question

Lactic acid is a common by-product of cellular respiration and is often said to cause the "burn" associated with strenuous activity. A 25.0 mL sample of 0.124 M lactic acid (HC₃H₅O₃, pK_a = 3.86)

 is titrated with 0.124 M NaOH solution. (Assume that all solutions are at 25°C.)

Calculate the pH after the addition of the following amounts of NaOH.

Vol. NaOH

0 mL

4.0 mL

8.0 mL

12.5 mL

20.0 mL

24.0 mL

24.5 mL

24.9 mL

25.0 mL

25.1 mL

26.0 mL

28.0 mL

30.0 mL

I do not really know how to approach this problem?? Please show specific steps I am really confused. :((

Answer 1

You should be getting the hang of it by now. What exactly are you confused about? Again, I will get you started, but there are a lot of additions, and I don't want to do them all. The first thing you must realize is that you are titrating a weak acid (lactic acid) with a strong base (NaOH), so you will form a BUFFER. This is very similar to a previous question using benzoic acid and NaOH. You should also note that both the acid and the base are 0.124 M.

(a) no NaOH added:

pKa = 3.86 therefore Ka = 1x10^-3.86 = 1.38x10^-4

HA ==> H⁺ + A^- and Ka = [H⁺][A-]/[HA] = (x)(x)/0.124) = 1.38x10^-4

x² = 1.71x10^-5

x = 4.14x10^-3 M = [H⁺]

pH = 2.38

(b) 4.0 ml of 0.124 M NaOH added: (HA will represent lactic acid)

moles HA = 25 ml x 1 L/1000 ml x 0.124 mol/L = 0.0031 moles HA

moles OH- = 4 ml x 1 L/1000 ml x 0.124 mol/L = 0.000496 moles OH-

HA + OH- ===> A- + H2O

0.0031...0.000496.....0..............Initial

-.000496...-0.000496....+0.000496....Change

0.002604.....0.................0.000496....Equilibrium

Final volume = 29 ml = 0.029 L

[HA] = 0.002604/0.029 = 0.0898 M

[A-] = 0.000496/0.029 = 0.0171 M

pH = pKa + log [A-]/[HA] = 3.86 + log (0.0171/0.0898) = 3.86 + (-0.720)

pH = 3.14

Do all the others the same way. When you get to 12.5 mls, you should end up with a pH = 3.86

Then let's look at addition of 25.1 ml of NaOH.

moles HA = 0.0031

moles OH- = 25.1 ml x 1 L /1000 ml x 0.124 mol/L = 0.00311 moles

excess OH- = 0.00311 - 0.00310 = 0.00001 moles OH in excess

Final volume = 50.1 ml = 0.0501 L

Final [OH-] = 0.00001 mol/0.0501 L = 0.0001996 M

pOH = 3.699

pH = 10.3

Do the others that have excess OH- the same way. Find moles OH added, subtract that from 0.0031 moles HA, divide by new volume, find pOH, and then pH.