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):
[tex]7,3x10^{-5} =\frac{ [0,52/2] [H^+]}{[0,52/2]}[/tex]
7,3x10⁻⁵ = [H⁺]
As pH = -log [H⁺]
pH = 4,14
I hope it helps!
