IDNLearn.com provides a collaborative platform for sharing and gaining knowledge. Ask your questions and receive accurate, in-depth answers from our knowledgeable community members.
Sagot :
1.4406 × 10⁻⁶ is the Ka(dissociation constant) for the acid.
The equilibrium constant for the reaction of an acid with water is the acid dissociation constant, where the acid, HA separates into H⁺ and A⁻ ions.
The acid dissociation constant is represented by (Ka).
So let's first imagine that the given monoprotic acid is HA.
HA will dissociate into H⁺ and A⁻ ions.
HA ⇒ H⁺ and A⁻
The formula used for Ka is
Ka = [H⁺] [A⁻] / [HA]
Given
pH = 2.83
[HA] = 1.50 M
From the given pH, we can calculate [H⁺] and [A⁻]
[H⁺] = [A⁻] = 1 × [tex]10^{-2.83}[/tex] = 1.47 × 10⁻³ M
Ka = 1.47 × 10⁻³ × 1.47 × 10⁻³ / 1.50
Ka = 1.4406 × 10⁻⁶
Hence, 1.4406 × 10⁻⁶ is the Ka for the acid.
Learn more about dissociation constant here https://brainly.com/question/3006391
#SPJ4
We appreciate your contributions to this forum. Don't forget to check back for the latest answers. Keep asking, answering, and sharing useful information. Your questions find clarity at IDNLearn.com. Thanks for stopping by, and come back for more dependable solutions.
