Consider the titration of a 40.0 mL of 0.229 M weak acid HA (Ka = 2.7 x 10⁻⁸) with 0.100 M LiOH.

Question

1. What is the pH of the solution before any base has been added?
2. What would be the pH of the solution after the addition of 20.0 mL of LiOH?
3. How many mL of the LiOH would be required to reach the halfway point of the titration?
4. What is the pH of the solution at the equivalence point?
5. What would be the pH of the solution after that addition of 100.0 mL of LiOH?

Answer 1

HA + OH^- ==> A^- + H₂O This creates a buffer consisting of the weak acid HA and the conjugate base A^-

(1). Ka = [H⁺][A^-]/[HA] = 2.7x10^-8 = (x)(x) / 0.229

x² = 6.18x10^-9

x = [H⁺] = 7.9x10^-5 M

pH = -log [H⁺] = 4.10

(2). HA + OH^- ==> A^- + H₂O Initial HA = 0.040L x 0.229 mol/L = 0.00916 mol HA. OH- = 0.02Lx0.1mol/L=0.002

....0.00916..0.002......0............Initial

-0.002........-0.002....+0.002.....Change

0.00716.....0............0.002.......Equilibrium

Henderson Hasselbalch: pH = pKa + log [conj.base]/[acid]

[conj.base] = 0.002 mol/0.06 L = 0.0333 M

[acid] = 0.00716 mol/0.06 L = 0.119 M

pKa = -log Ka = -log 2.7x10-8 = 7.57

pH = 7.57 + log (0.0333/0.119) = 7.57 - 0.756

pH = 6.81

(3). Calculate volume of 0.1 M LiOH needed to produce 1/2 x 0.00916 moles. Since there are originally 0.00916 mols HA, it will take 1/2 of that (0.00458 mols) LiOH to reach half the equivalence. Finding the volume, we have

0.00458 mols LiOH x 1 L / 0. 1mols = 0.0458 L = 45.8 mls LiOH

(4). At equivalence, all the HA is now in the form of A- and that would be 0.00916 moles A- in a volume of 40.0 ml + 45.8 mls or a total volume of 85.8 ml (0.0858 L). So, the [A-] is 0.00916 mol / 0.0858 L = 0.2 M.

A- + H2O =>HA + OH- and solve for OH- using Kb = 1x10^-14 / 2.7x10^-8.

2.7x10^-8 = [HA][OH^-] / 0.2 = x²/0.2 and then x² = 5.4x10^-9

x = [OH^-] = 7.35x10^-5 and pOH = -log 7.35x10^-5 = 4.13

pH = 14 - 4.13

pH = 9.87

(5). 100 ml LiOH x 1 L/1000 ml x 0.1 mol/L = 0.01 mols LiOH added

excess LiOH = 0.01 mols - 0.00916 mols = 0.00084 mols LiOH left over

Final volume = 100 ml LiOH + 40 ml HA = 140 mls = 0.140 L

Final [OH-] = 0.00084 mol OH- / 0.140 L = 0.006 M

pOH = -log 0.006 = 2.22

pH = 14 - 2.22 = 11.78