# chem

A weak acid, HA, has an ionization constant of 4.34 x 10-8. If we prepare a solution that is 0.273 M in HA what percent of the acid will be ionized

1. ...........HA --> H^+ + A^-
I.......0.273.....0......0
C.........-x......x......x
E.......0.273-x...x......x

Ka = (H^+)(A^-)/(HA)
Substitute the E line and solve for x = (H^+). Then %ion = [(H^+)/0.273]*100 = ?

posted by DrBob222

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