...........HOAc + H2O ==> H3O^+ + OAc^-
I..........0.5M............0.......0
C..........-x..............x.......x
E.........0.5-x............x.......x
Substitute the E line into Ka expression and solve for x = (H3O^+) then convert to pH.
If you know Ka (and you do) then KaKb = Kw = 1E-14. You always know Kw; if you know either Ka or Kb the other can be calculated.
If a solution of 0.5 M HOAc dissociates as follows: HOAc + H2O „® H3O+ + OAc-, what is the final [H3O+] in the solution? Ka for HOAc = 1.8„ª10-5. What is the pH of the above solution? What is the Kb for HOAc? not sure of the initial set up please help
1 answer
2 answers