Calculate the percent ionization of nitrous acid in a solution that is 0.139 M in nitrous acid. The acid dissociation constant of nitrous acid is 4.50 ⋅ 10-4.

Ok first, we have to create a balanced equation for the dissolution of nitrous acid.

HNO2 H (+) + NO2 (-)

Next, create an ICE table

HNO2 H + NO2-

[]i 0.139M 0M 0M

Δ[] - x + x + x

[]f 0.139-x x x

Then, using the concentration equation, you get

4.5x10^-4 = [H+][NO2-]/[HNO2]

4.5x10^-4 = x*x /.139 - x

However, because the Ka value for nitrous acid is lower than 10^-3, we can assume the amount it dissociates is negligable,

assume 0.139-x ≈ 0.139

4.5x10^-4 = x^2/0.139

Then, we solve for x by first multiplying both sides by 0.139 and then taking the square root of both sides.

We get the final concentrations of [H+] and [NO2-] to be x, which equals 0.007M.

Then to find percent dissociation, you do final concentration/initial concentration.

0.007M/0.139M =.0503 or

≈5.03% dissociation.