Asked by Mahnoor
                A 1.0 x 10-2 M solution of a weak acid is found to be 3.7% dissociated.  Calculate Ka of this acid.
I found the amount dissociated by (amount dissociated/initial conc)*100=3.7
Where do I go from here?
            
        I found the amount dissociated by (amount dissociated/initial conc)*100=3.7
Where do I go from here?
Answers
                    Answered by
            DrBob222
            
    I think you need to back up before going forward.
(HA) = 0.01
(H^+) = (A^-) = 0.01 x 0.037 = 3.7E-4
............HA ==> H^+ + A^-
I.........0.01.....0.....0
C........-3.7E-4..3.7E-4..3.7E-4
E.......you finish
Substitute the E line into the Ka expression and solve for Ka. Your answer should be approx 1E-5 if I plugged in the right numbers.
    
(HA) = 0.01
(H^+) = (A^-) = 0.01 x 0.037 = 3.7E-4
............HA ==> H^+ + A^-
I.........0.01.....0.....0
C........-3.7E-4..3.7E-4..3.7E-4
E.......you finish
Substitute the E line into the Ka expression and solve for Ka. Your answer should be approx 1E-5 if I plugged in the right numbers.
                                                    There are no AI answers yet. The ability to request AI answers is coming soon!
                                            
                Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.