Question

Answer the following questions using the graph and labeling the equivalence point and half-way point.

Trial 1 Tri Mass of oxalic acid 0.2099 M: pH at equivalence point PH at vo Volume of titrant added at equivalence point Volume of titrant added half-way to equivalence point pH at half-way point ha ра pKa of oxalic acid Ka of oxalic acid(5 pts) Calculations

Answer 1

I'm not sure if this is the titration curve for oxalic acid, since oxalic acid is a dicarboxyllic acid, so there should be two equivalence points. Still, it's a good titration curve of a weak acid and the questions can be definitively answered.

The equivalence point is the point at which the curve is more steep. By visual apporximation, I would say this happens at a pH of around 8.5, where the volume of added titrant is approximately 32 mL.

Half-way to the equivalence point would be equivalent to a volume of 32/2 = 16 mL. The pH at this point is roughly 3.2. If we take into account that at this point the concentration of base added is equal to half the concentration of total acid, we have half the acid protonated and half deprotonated. If we now take into account the Henderson Hasselbach equation, which allows us to calculate the pH of this type of mixtures:

$pH=pKa+log(\frac{[A^{-}]}{[AH]})$

And [A^-] = [HA], the value inside the log would be 1, which means that log(1) = 0 and pH = pKa. So the pH halfway to the equivalence point is equal to the pKa of the acid; that is: pKa = 3.2.

Ka can be calculated as:

$Ka=10^{-pKa}=10^{-3.2}=6.3x10^{-4}$