Answer:
The pH of the solution at the half‑equivalence point is 4,14
Step-by-step 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) =( [0,52/2] [H^+])/([0,52/2])](https://img.qammunity.org/2020/formulas/chemistry/college/dco7q108rbvhe04g0fbk299xhrwll4qnx1.png)
7,3x10⁻⁵ = [H⁺]
As pH = -log [H⁺]
pH = 4,14
I hope it helps!