Post a New Question

Chemistry

posted by .

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...

  • Chemistry -

    ..........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 = ?

  • Chemistry -

    Oh, ok! I was over thinking the problem!!! Plus a fever may not help with my problem solving abilities... Thank you!

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

More Related Questions

Post a New Question