..........HA --> H^+ + A^-
Initial.0.125....0......0
change....-x.....x......x
equil..0.125-x...x......x
Ka = (H^+)(A^-)/(HA)
Plug into the Ka expression and solve for x = (H^+).
Then % ion = [((H^+)/(0.125)]*100 = ?
The Ka of a monoprotic weak acid is 3.39 × 10-3. What is the percent ionization of a 0.125 M solution of this acid?
I just do not know where to even start...
2 answers
Oh, ok! I was over thinking the problem!!! Plus a fever may not help with my problem solving abilities... Thank you!
3 answers