Question

Calculate the [H3O+] of 0.10 M HNO2 at equilibrium. Given: Ka = 4.46 × 10-4.

Answer 1

0.0067 mol/dm³

Step-by-step explanation:

The [H3O+] of HNO2 can be found by using the formula

`[H_3O^( + ) ] = √(Kac)`

Ka is the acid dissociation constant

c is the concentration of the acid

From the question we have

`[H3O^( + ) ] = \sqrt{4.46 * {10}^( - 4) * 0.1} \\ = 0.006678...`

We have the final answer as

0.0067 mol/dm³

Hope this helps you