The Ka of a monoprotic weak acid is 8.11 x 10 ^ -3. What is the percent ionization of a 0.125M solution of this acid?

Answer 1

........HA ==>H^+ + A^-
I....0.125M...0......0
C......-x.....x......x
E....0.125-x...x......x

Ka = ..... Sutstitute the numbers in to Ka expression and solve for x. (It looks like you MAY--check it out---be required to use the quadratic equation.)
Then %ion = [(H^+)/0.125]*100 = ?

Answer 2

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Answer 3

You just have to use the quadratic formula to find x then divide by the initial concentration