...........HA ==> H^+ + A^-
...........1M......0....0
C.......-0.01...0.01..0.01
E........0.99....0.01..0.01
Ka = (H^+)(A^-)/(HA)
Substitute the above into Ka expression and solve for Ka. I think you should obtain about 1E-4 but that isn't exact.
Then.........HA ==> H^+ + A^-
I............5......0.....0
C............-x.....x.....x
E...........5-x.....x....x
Ka = (H^+)(A^-)/(HA)
about 1E-4 = (x)(x)/(5-x)
Solve for x and I get about 0.02 (you need to do it more accurately).
% ionization = (0.02/5)*100 =?
A weak acid HA is 1.0% dissociated in a 1.0 M solution. What would its percent dissociation be
in a 5.0 M solution?
1 answer
1 answer