My calculator is broken. You calculate H^+ from pH = -log(H^+). I will call that value, whatever it is, x
.....................HA ==> H^+ + A^-
Initial..........0.06..........0........0
change..........-x............x..........x
equilibriium..0.06-x.......x..........x
Ka = (H^+)(A^-)/(HA)
Substitute the E line into Ka expression and solve for Ka.
Post your work if you get stuck.
The pH of a 0.060 M weak monoprotic acid, HA is 3.44. Calculate the Ka of the acid.
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