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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
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