Question
a 0,25M solution of a week acid is 3% dissociated.calculate ka
Answers
DrBob222
0.25 x 0.003 = 7.5E-4
..............HA ==> H^+ + A^-
initial.......0.25.....0.....
change.....-7.5E-4...7.4E-4..7.5E-4
equl..........?........?......?
Ka = (H^+)A^-)/(HA)
Substitute and solve for Ka.
..............HA ==> H^+ + A^-
initial.......0.25.....0.....
change.....-7.5E-4...7.4E-4..7.5E-4
equl..........?........?......?
Ka = (H^+)A^-)/(HA)
Substitute and solve for Ka.