acetic acid is HAc.
.........HAc ==> H^+ + Ac^-
I..l....0.31.....0......0
C........-x......x......x
E......0.31-x....x......x
Substitute the E line into the Ka expression and solve for x.
Then % ion = [(x)/0.31]*100 = ?
Calculate the percent ionization of a .31 molar solution of acetic acid. The ionization constant of acetic acid is 1.8 x 10^-5
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