Question

At a certain temperature, the percent dissociation (ionization) of chlorous acid, HClO2, in a 1.43 M solution water is 8.0%. Calculate the value of Ka for chlorous acid at this temperature.

Answer 1

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

Answer 2

The Ka for chlorous acid at this temperature 9.7 × 10^-3.

The Ka refers to the equilibrium constant for the dissociation of an acid. It shows the extent to which an acid dissociates in solution and is also a measure of the strength of an acid.

The equation of the reaction is;

          HClO2(aq) ⇄         H^+(aq)  + ClO2^-(aq)

I           1.43                         0                0

C        -0.08                       0.08         0.08

E         1.43 - 0.08           1.43(0.08)   1.43(0.08)

The equilibrium dissociation constant Ka is obtained from;

Ka = [ 1.43(0.08)] [ 1.43(0.08)]/  1.43 - 0.08

Ka = 9.7 × 10^-3

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