Question

The half‑equivalence point of a titration occurs half way to the equivalence point, where half of the analyte has reacted to form its conjugate, and the other half still remains unreacted. If 0.480 moles of a monoprotic weak acid (Ka=3.0×10−5) is titrated with NaOH, what is the pH of the solution at the half‑equivalence point?

Answer 1

pH = 4.52

Step-by-step explanation:

A monoprotic acid, HA, reacts with NaOH as follows:

HA+ NaOH → A⁻ + H₂O

When the weak acid HA, is in solution with its conjugate base, A⁻, a buffer is produced. That means in the titration of the weak acid with NaOH you are producing a buffer.

The pH of a buffer can be found using H-H equation:

pH = pKa + log [A⁻] / [HA]

Where pKa is -logKa = 4.52 and [] can be understood as the moles of A⁻ and HA.

pH = 4.52 + log [A⁻] / [HA]

In the half-equivalence point, the half of HA was converted in A⁻ and the other half still remains as HA.

That means moles of A⁻ and HA are: 0.480/2 = 0.24 moles of both A⁻ and HA

Replacing in H-H equation:

pH = 4.52 + log [A⁻] / [HA]

pH = 4.52 + log [0.24] / [0.24]

pH = 4.52

-In the half-equivalence point of a titration of a weak acid, pH = pKa-