Question
The Ka of a monoprotic weak acid is 2.04 × 10-3. What is the percent ionization of a 0.181 M solution of this acid?
Answers
DrBob222
............HA ==> H^+ + A^-
initial..0.181.....0......0
change......-x.....x......x
equil.....0.181-x..x.......x
Ka = (H^+)(A^-)/(HA)
Substitute ICE chart into Ka expressiion and solve for H^+; then
%ion = [(H^+)/(0.181)]*100 = ?
initial..0.181.....0......0
change......-x.....x......x
equil.....0.181-x..x.......x
Ka = (H^+)(A^-)/(HA)
Substitute ICE chart into Ka expressiion and solve for H^+; then
%ion = [(H^+)/(0.181)]*100 = ?