Question

Calcutate the percent ionization of nitrous acid in a solution that is 0.219M in nitrous acid. The acid dissociation constant of nitrous acid is 4.50*10^4

Answer 1

By definition, in equilibrium:

Ka = [H+][A-]/[HA]

Then

Assume that [H+] = [A-] = x … due to stoichiometry (i.e. 1 mol of H+ per each mol of A-)

Then, in equilibrium [HA] = M – x (i.e. the original concentration minus the acid in equilibrium)

Substitute

Ka = (x*x)/(M-x)

4.5*10^-5 = x*x/(0.219-x)

solve for x (quadratic equation)

x = [H+] = 0.003116 M

[H+] = [A-] = x = 0.003116

The ionization:

% ionization = [H+] / M * 100% = 0.003116 /(0.219) * 100% = 1.422 %