Question

How is the shape of a monoprotic acid titration curve determined by the pk of the acidic proton?

Answer 1

Titration curves for strong acid vs. strong base

We'll take hydrochloric acid and sodium hydroxide as typical of a strong acid and a strong base.

Running acid into the alkali

You can see that the pH only falls a very small amount until quite near the equivalence point. Then there is a really steep plunge. If you calculate the values, the pH falls all the way from 11.3 when you have added 24.9 cm³ to 2.7 when you have added 25.1 cm

Running alkali into the acid

This is very similar to the previous curve except, of course, that the pH starts off low and increases as you add more sodium hydroxide solution.

Again, the pH doesn't change very much until you get close to the equivalence point. Then it surges upwards very steeply.

he equivalence point on the graph is where all of the starting solution (usually an acid) has been neutralized by the titrant (usually a base). It can be calculated precisely by finding the second derivative of the titration curve and computing the points of inflection (where the graph changes concavity); however, in most cases, simple visual inspection of the curve will suffice (in the curve given to the right, both equivalence points are visible, after roughly 15 and 30 mL of NaOH solution has been titrated into the oxalic acid solution. To calculate the acid dissociation constant (pK_a), one must find the volume at the half-equivalence point, that is where half the amount of titrant has been added to form the next compound (here, sodium hydrogen oxalate, then disodium oxalate). Halfway between each equivalence point, at 7.5 mL and 22.5 mL, the pH observed was about 1.5 and 4, giving the pK_a.

In monoprotic acids, the point halfway between the beginning of the curve (before any titrant has been added) and the equivalence point is significant: at that point, the concentrations of the two species (the acid and conjugate base) are equal. Therefore, the Henderson-Hasselbalch equation can be solved in this manner: