Find the percent ionization of a 0.250 m solution of hc2h3o2. express your answer numerically to two significant figures.

From this reaction:
and by using ICE table:
                CH3COOH ↔ H + +  CH3COO-
initial       0.25                   0              0
change    - X                    +X             +X
Equ        (0.25 - X)            X                X

when we have Ka for acetic acid = 1.8x10^-5 

and Ka = [H+][CH3COO-] / [CH3COOH]
1.8x10^-5 = X*X / (0.25-X)
X = 0.00211 m
∴[H+] = 0.00211 m
∴percent ionization = [H+] / [initial acid] * 100
                                  = 0.00211 / 0.25 *100
                                  = 0.844 % 

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