Respuesta :
Chlorous Acid Ionizes as,
HClO₂ ⇆ H⁺ + ClO₂⁻
Ka = [H⁺] [ClO₂⁻] / [HClO₂]
Ka of Chlorous Acid = 1.1 × 10⁻²
The concentration of H⁺ ions at equilibrium are calculated as,
Initial 0.1 M ⇆ 0 0
At Equilibrium 0.1-X ⇆ X X
So,
Ka = [X] [X] / [0.1-X]
Putting value of Ka,
1.1 × 10⁻² = X² / 0.1-X
Solving for X,
(www.cymath.com)
X = 0.028 M
Hence at equilibrium concentration of H⁺ and ClO₂⁻ is 0.028 M.
Percentage Ionization is calculated as,
= [H⁺] / [HA] × 100
= (0.028 / 0.1) × 100
= 0.28 × 100
= 28 % (Percentage Ionization)
HClO₂ ⇆ H⁺ + ClO₂⁻
Ka = [H⁺] [ClO₂⁻] / [HClO₂]
Ka of Chlorous Acid = 1.1 × 10⁻²
The concentration of H⁺ ions at equilibrium are calculated as,
Initial 0.1 M ⇆ 0 0
At Equilibrium 0.1-X ⇆ X X
So,
Ka = [X] [X] / [0.1-X]
Putting value of Ka,
1.1 × 10⁻² = X² / 0.1-X
Solving for X,
(www.cymath.com)
X = 0.028 M
Hence at equilibrium concentration of H⁺ and ClO₂⁻ is 0.028 M.
Percentage Ionization is calculated as,
= [H⁺] / [HA] × 100
= (0.028 / 0.1) × 100
= 0.28 × 100
= 28 % (Percentage Ionization)
The percent dissociation of the chlorous acid is 0.33 or 33%.
Given that we have the concentration of the acid from the question as 0.1M, we also know from the standard tables that the Ka of chlorous acid is 1.1 × 10⁻².
Now, given the formula;
α = √Ka/C
Ka = acid dissociation constant
C = concentration
Therefore;
α = √1.1 × 10⁻²/0.1
α = 0.33 or 33%
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