Consider the titration of a 60.0 mL of 0.365 M weak acid HA (Ka = 4.2 x 10⁻⁶) with 0.400 M KOH.

Question

After 75.0 mL of KOH have been added, what would the pH of the solution be?

Answer 1

HA + KOH ==> KA + H2O

moles HA = 60.0 ml x 1 L/1000 ml x 0.365 mol/L = 0.0219 moles HA

moles KOH = 75.0 ml x 1 L / 1000 ml x 0.400 mol/L = 0.0300 moles KOH

Since there are more moles of KOH than HA, all the HA will be neutralized, and there will be xs KOH

Moles of KOH in excess = 0.0300 - 0.0219 = 0.0081

Final volume = 60.0 ml + 75.0 ml = 135 ml = 0.135 L

[KOH] = 0.0081 mol/0.135 L = 0.06 M

pOH = -log 0.06 = 1.22

pH = 14 - pOH = 14 - 1.22

pH = 12.8

(NOTE: you could also include the hydrolysis of A- formed from the neutralization but the contribution to the pH would be small so I didn't include it). If you want to include it, see below:

moles A- formed = 0.0219 moles

[A-] = 0.0219 mol/0.135 L = 0.162 M

A- + H2O =>HA + OH-

Kb = 10^-14/4.2x10^-6 = 2.4x10^-9

2.4x10^-9 = (x)²/0.162

x² = 3.9x10^-10

x = [OH-] = ~2x10^-5 M add this to the 0.06 M found above from the xs KOH and you'll see the pH is about the same as calculated above.