Question 1

What is a polyprotic acid and how does it differ from a monoprotic acid?

Accepted Answer

A polyprotic acid contains multiple ionizable protons that dissociate in distinct steps, each with its own acid dissociation constant (Ka). Unlike monoprotic acids with one ionizable proton, polyprotic acids release protons sequentially, with each subsequent Ka being weaker than the previous one. For example, sulfurous acid has two ionizable protons with Ka1 = 1.6 × 10−2 and Ka2 = 6.4 × 10−8.

Question 2

Why does a polyprotic acid produce multiple equivalence points during titration?

Accepted Answer

During titration of a polyprotic acid with a strong base, each ionizable proton is neutralized in a separate step, generating a distinct equivalence point for each proton. The number of equivalence points equals the number of ionizable protons, provided the Ka values differ by more than ten thousand fold. A diprotic acid produces two equivalence points, while a triprotic acid produces three.

Question 3

How much base is required to completely neutralize a diprotic acid?

Accepted Answer

Two moles of base are needed to neutralize one mole of a diprotic acid completely. The first mole of base removes the first ionizable proton, generating hydrogen sulfite ions. The second mole neutralizes the second ionizable proton from the hydrogen sulfite ion, since its concentration equals the initial diprotic acid concentration.

Question 4

What is the pH at the half-equivalence point for each step of a polyprotic acid titration?

Accepted Answer

At the half-equivalence point of each titration step, the pH equals the pKa for that step. For the first neutralization step, pH = pKa1, and for the second step, pH = pKa2. This relationship holds because at half-equivalence, the concentrations of the acid and its conjugate base are equal, making the Henderson-Hasselbalch equation simplify to pH = pKa.

Question 5

How does the titration curve of sulfurous acid differ between the first and second neutralization steps?

Accepted Answer

Both steps produce similar titration curve features: an equivalence point and a half-equivalence point where pH equals the respective pKa. However, the first step resembles a weak monoprotic acid titration, while the second step occurs at higher pH. The second equivalence point lies in the basic region because the conjugate base of the second proton is a stronger base than the first.

Question 6

Why is the third equivalence point of phosphoric acid not easily visible on its titration curve?

Accepted Answer

Phosphoric acid is a triprotic acid that produces three equivalence points when titrated with a strong base like potassium hydroxide. However, the third equivalence point is not easily discernible because HPO42− is a very weak acid with an extremely small Ka3, making the third ionization barely detectable on the titration curve.

Question 7

What condition must be met for a polyprotic acid to show distinct equivalence points in its titration curve?

Accepted Answer

The acid dissociation constants of the ionizable protons must differ by more than ten thousand fold for distinct equivalence points to appear. When this condition is satisfied, each ionizable proton produces a separate, recognizable equivalence point on the titration curve. If Ka values are too similar, the equivalence points may overlap and become indistinguishable.

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