For a diprotic acid, why does the autoionization of water have a greater impact on the pH at the second equivalence point?

For a diprotic acid, the autoionization of water has a greater impact on the pH at the second equivalence point primarily because the concentration of the second disassociated proton is much lower compared to the first one, making the system more sensitive to changes from other sources, such as water autoionization.

A diprotic acid is capable of donating two protons (H⁺ ions) sequentially. The dissociation of a diprotic acid can be represented by two distinct equilibrium reactions:

1. The first dissociation: HA₂ ⇌ H⁺ + A⁻₁

2. The second dissociation: A⁻₁ ⇌ H⁺ + A⁻₂

At the first equivalence point, we have fully titrated the first proton, and the solution contains largely A⁻₁. The concentration of protons (H⁺) in solution is relatively high because the first dissociation constant (Kₐ1) for most diprotic acids is greater than Kₐ2 (the second dissociation constant).

When we reach the second equivalence point, we have also titrated the second proton, and the predominant species is A⁻₂. Because the second dissociation constant (Kₐ2) is typically much smaller than Kₐ1, the concentration of free protons in solution at the second equivalence point is lower than at the first equivalence point.

The autoionization of water, which can be represented as:

H₂O ⇌ H⁺ + OH⁻

is a very weak process resulting in a very low concentration of hydronium (H₃O⁺, represented as H⁺) and hydroxide ions (OH⁻) in pure water (at 25°C, this concentration is about 10⁻⁷ M for each). However, because the concentration of protons from the second dissociation of the diprotic acid is so low, the autoionization of water now represents a significant source of protons compared to the second dissociation. As a result, the pH of the solution at the second equivalence point is more strongly influenced by the ionization of water than the first dissociation of the acid.