Choose chlorous acid, hclo2, from the list above and press the equilibrate button. when the 0.10 m solution of hclo2 is allowed to come to equilibrium, what percent of the acid molecules are ionized?

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)

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%

Learn more about acid dissociation constant:https://brainly.com/question/1301963