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Alchen [17]
3 years ago
15

At a certain temperature, the percent dissociation (ionization) of chlorous acid, HClO2, in a 1.43 M solution water is 8.0%. Cal

culate the value of Ka for chlorous acid at this temperature.
Chemistry
1 answer:
lakkis [162]3 years ago
5 0

Answer:

Kₐ = 9.9 × 10⁻³

Explanation:

HClO₂ + H₂O ⇌ H₃O⁺ + ClO₂⁻

0.080 × 1.43 = 0.1144

So, at equilibrium,

[H₃O⁺] = [ClO₂⁻] = 0.1144

[HClO₂] = 1.43 – 0.1144 = 1.316

Kₐ = {[H₃O⁺][ClO₂⁻]}/[HClO₂];        Substitute values

Kₐ = (0.1144 × 0.1144)/1.316           Do the operations

Kₐ = 9.9 × 10⁻³

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