Question
In a 0.05M solution of a weak monoprotic acid,[H+]=.0018.What is its Ka?
Answers
DrBob222
............HA ==> H^+ + A^-
initial....0.05....0......0
change..-0.0018...0.0018..0.0018
equil..0.05-0.0018..0.0018..0.0018
Ka = (H^+)(A^-)/(HA)
Substitute and solve for Ka.
initial....0.05....0......0
change..-0.0018...0.0018..0.0018
equil..0.05-0.0018..0.0018..0.0018
Ka = (H^+)(A^-)/(HA)
Substitute and solve for Ka.