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 0.520 moles of a monoprotic weak acid (Ka=7.3×10−5) is titrated with NaOH, what is the pH of the solution at the half‑equivalence point?

Answer 1

Answer:

The pH of the solution at the half‑equivalence point is 4,14

Explanation:

The acid dissociation of this monoprotic acid is:

HA ⇄ H⁺ + A⁻; ka = 7,3x10⁻⁵

ka is defined as:

ka = [A⁻] [H⁺] / [HA] (1)

When HA is titrated with NaOH the reaction is:

HA + NaOH → A⁻ + Na⁺ + H₂O

At half-equivalence point, 0,52/2 moles of HA remains while 0,52/2 A⁻ moles are produced.

Replacing this values in (1):

$7,3x10^{-5} =\frac{ [0,52/2] [H^+]}{[0,52/2]}$

7,3x10⁻⁵ = [H⁺]

As pH = -log [H⁺]

pH = 4,14

I hope it helps!