Calculate the percent ionization of nitrous acid in a solution that is 0.311 M in nitrous acid (HNO2) and 0.189 M in potassium nitrite (KNO2). The acid dissociation constant of nitrous acid is 4.50 × 10-4.

Answer 1

........HNO2 ==> H^+ + NO2^-
I.....0.311......0.......0
C......-x.........x........x
E.....0.311-x......x......x

......KNO2 ==>K^+ + NO2^-
I.....0.189...0......0
C....-0.189..0.189...0.189
E......0.....0.189..0.189

Ka = (H^+)(NO2^-)/(HNO2)
Substitue for Ka,
For (H^+) = x
For (NO2^-) = x(from HNO2) + 0.189(from KNO2)
(HNO2) = 0.311-x
Solve for x = (H^+), then
%ion = [(H^+)/(0.311)]*100 = ?

Answer 2

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