Question

draw a titration curve for oxalic acid from a potentiometric titration of 20.0 mL of 0.10 M oxalic acid with 0.10 M NaOH.

Ka1 = 5.9 x 10-2

Ka2 = 6.4 x 10-5

Identify the following in the titration curve:

A) pK1 and pK2

B) equivalence points

C) buffering regions

i will give thumbs up

Answer 1

The titration curve would be the following, where I've marked the important volumes of NaOH:

Titration Curve (oxalic acid/NaOH) Hd 10 20 30 40 Volume of NaOH added (ml)

Since the concentaration of acid and base are the same, it is easy to realize that 20 and 40 mL of NaOH indicate the equivalence points, which occur when the number of moles of NaOH are equal to:
1) The number of moles of oxalic acid (this means that the first deprotonation of the acid has occured completely)
2) Double the number of moles of oxalic acid (the second deprotonation is total).

pKa 1 is the pH corresponding to the volume halfway to the first equivalence point, which is given by 10 mL (pH value at this point is approximately 1.2). This can be deduced from the fact that halfway to the first equivalence point, we will have titrated half of the first dissociation H⁺ of the acid. This means that the concentration of deprotonated acid is the same as the concentration of non-deprotonated acid. If we use the Henderson-Hasselbach equation to calculate the pH at this point:

$pH=pKa_{1}+log(\frac{[C_{2}O_{4}H^{-}]}{[C_{2}O_{4}H_{2}]})$

The reation between the concentration would be 1, and since log(1) = 0, we get that pH = pKa₁.

In an analogous way we can derive the fact that pKa₂ is equal to the pH at the volume halfway between the first and second equivalence points. This is the pH corresponding to a volume of 30 mL of NaOH, which is approximately 4.

c) The buffering regions are those in which there aren't big changes in pH with the addition of titrant. In this case, they are marked with red circles. They are the regions around the pKa values, where the mixtures buffer the most.