Question

Calculate the percent ionization of nitrous acid in a solution that is 0.249 M in nitrous acid. The acid dissociation constant of nitrous acid is

Answer 1

4.26 %

Step-by-step explanation:

There is some info missing. I think this is the original question.

Calculate the percent ionization of nitrous acid in a solution that is 0.249 M in nitrous acid. The acid dissociation constant of nitrous acid is 4.50 × 10 ⁻⁴.

Step 1: Given data

Initial concentration of the acid (Ca): 0.249 M

Acid dissociation constant (Ka): 4.50 × 10 ⁻⁴

Step 2: Write the ionization reaction for nitrous acid

HNO₂(aq) ⇒ H⁺(aq) + NO₂⁻(aq)

Step 3: Calculate the concentration of nitrite in the equilibrium ([A⁻])

We will use the following expression.

`[A^(-) ] = √(Ca * Ka ) = \sqrt{0.249 * 4.50 * 10^(-4) } = 0.0106 M`

Step 4: Calculate the percent ionization of nitrous acid

We will use the following expression.

`\alpha = ([A^(-) ])/([HA]) * 100\% = (0.0106M)/(0.249) * 100\% = 4.26\%`