Question

A 35.00mL sample of aqueous lactic acid(HC3H5CO3) is titrated to the equivalence point with 65.00mL of 0.15 M KOH(aq). What is the pH of the resulting solution? The Ka for lactic acid is 8.3x10^-4. The correct answer is 8.03 but I am unsure about the steps in getting there.

Answer 1

First, let's calculate the concentration of the lactic acid:
Ma = 0.15 * 65/35 = 0.2786 M

Now, at the equivalence point where the moles are the same for base and acid, we see that in equilibrium we have this:

HC₃H₅CO₃ + OH^- <----------> C₃H₅CO₃^- + H₂O

At this point, all the acid was consumed and the base will begin to be in excess, so the reaction in solution when this happens is the following:
C₃H₅CO₃^- + H₂O <---------> HC₃H₅CO₃ + OH^-  Kn = 1x10^-14/8.3x10^-4 = 1.2x10^-11

Now, the moles and further concentration of the conjugate base of lactic acid are:
moles = 0.15 * 0.065 = 0.00975 moles
Total volume = 65+35 = 100 mL
[C₃H₅CO₃^-] = 0.00975/0.1 = 0.0975 M

Now, the ICE for this reaction:
r: C₃H₅CO₃^- + H₂O <---------> HC₃H₅CO₃ + OH^-  Kb = 1.2x10^-11
i: 0.0975 0 0
e: 0.0975-y y y

1.2x10^-11 = y²/0.0975-y --> Kb is small so we can neglect 0.0975-y to 0.0975 only:
1.2x10^-11 * 0.0975 = y²
y = [OH^-] = 1.08x10^-6 M
pOH = -log(1.08x10^-6) = 5.97
pH = 14-5.97
pH = 8.03

Hope this helps