The half-equivalence point of a titration occurs half way to the end point, where half of the analyte has reacted to form its conjugate, and the other half still remains unreacted. If 0.220 moles of a monoprotic weak acid (Ka = 5.7 × 10-5) is titrated with NaOH, what is the pH of the solution at the half-equivalence point?

pKa=-log(Ka)

Then to find pH use, pKa+log(0.220/0.220)
Thus, pH=pKa

To calculate the pH at the half-equivalence point, we need to determine the concentration of the acid and its conjugate base at this point.

First, let's calculate the initial concentration of the weak acid, HA. We know that 0.220 moles (or 220 mmol) of the weak acid was used, but we need to determine its initial volume. This information is not provided in the question, so let's assume it is a 100 mL solution.

The concentration (C) of the weak acid is given by the formula C = n/V, where n is the number of moles and V is the volume in liters. Therefore, C = 0.220 moles / 0.100 L = 2.2 M.

At the half-equivalence point, half of the weak acid has reacted to form its conjugate base, A-. This means that the concentration of the weak acid, [HA], and the conjugate base, [A-], will be equal.

Since the weak acid is monoprotic, the reaction can be represented as follows:
HA + OH- --> A- + H2O

The equilibrium constant expression for this reaction is given by Ka = [A-][H2O] / [HA][OH-], but in this case, [OH-] is equal to the concentration of NaOH added at the half-equivalence point. Let's represent the concentration of NaOH as [NaOH] and [OH-].

Since [OH-] = [NaOH], the concentration of NaOH is equal to the number of moles of NaOH divided by the volume of the solution. We'll assume that 0.110 moles (or 110 mmol) of NaOH is needed to reach the half-equivalence point (as half of 0.220 moles of weak acid), and that the total volume of the solution is still 100 mL.

[NaOH] = 0.110 moles / 0.100 L = 1.1 M.

Now, let's rearrange the equilibrium constant expression to solve for the concentrations:

Ka = [A-][H2O] / [HA][OH-]
[HA] = [A-] = x (since they are equal at the half-equivalence point)
[H2O] = constant (approximately 55.5 M)

Ka = (x)(55.5) / (x)(1.1)
Ka = 5.7 × 10^(-5)

Simplifying the equation, we get:
5.7 × 10^(-5) = 55.5 / 1.1

Solving for x:
x^2 = (55.5 / 1.1) / 5.7 × 10^(-5)
x^2 = 431.82
x ≈ 20.8

Therefore, the equilibrium concentrations of HA and A- at the half-equivalence point are approximately 20.8 mM.

Now, let's calculate the pH at this point. Since we know the concentration of the weak acid, we can use the Henderson-Hasselbalch equation:

pH = pKa + log([A-] / [HA])

The pKa is given in the question as 5.7 × 10^(-5), and the concentrations are both approximately 20.8 mM.

pH = -log(5.7 × 10^(-5)) + log(20.8 / 20.8)
pH = 4.24

Therefore, at the half-equivalence point, the pH of the solution is approximately 4.24.