35.0 mL of 0.13 M weak acid (Ka = 6.3 x 10^-5) is titrated with 0.15 M NaOH. What is the pH at the half-equivalence point?

Method 1 :

For the titration of a weak acid with a strong alkali at the half-equivalence point :
pH = pKa = -log(Ka) = -log(6.3 × 10⁻⁵) = 4.2

The answer: D : 4.2


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Method 2 :

Denote the weak acid as HA.

At the half-equivalence point, a half of HA is converted to NaA, i.e. [HA] = [NaA]
NaA completely dissociates in the solution give A⁻ ions. Due to the common ions in the presence of A⁻ ions and HA is a weak acid, the dissociation of HA is to a very small extent.
Hence, at the half-equivalence point, [HA] = [A⁻]
Therefore, [A⁻]/[HA] = 1

Consider the dissociation of HA :
HA(aq) ⇌ H⁺(aq) + A⁻(aq)

pH = pKa + log([A⁻]/[HA]
pH = -log(6.3 × 10⁻⁵) + log(1)
pH = 4.2

The answer: D : 4.2

pH = pKa = 4.2
thus
D.