HNO2(aq) + H2O(l) → ← H3O+(aq) + NO2–(aq)
Ka = [H3O+]e[NO2–]e [HNO2]e
Initial0.0100 0 0
Change–x +x +x
Equilibrium0.0100 – x x x
We can use the per cent ionization to find the value of x:
x = [H3O+]e
α = [H3O+]e [HNO2]init ×100%
or
[H3O+]e = α×[HNO2]init/100% = 19%×(0.0100)/100% = 0.0019 M
[NO2–]e = [H3O+]e = 1.9×0–3 M
[HNO2]e = 0.0100 – x = 0.0100 – 0.0019 = 0.0081 M