When weak neutral acids and bases are put in water, they form ions. For example, if a weak acid like HF is place in water, it will form both the hydronium ion, H3O+ as well as its conjugate base the fluoride ion, F-.
\[{ \rm HF(aq) + H_2O(l) \rightleftharpoons H_3O^+(aq) + F^-(aq)}\]
When the HF is place into water it is decomposing into H+ and F-. This dissociation can also be referred to as "ionization" as the compound is forming ions. The extent to which this happens can be determined from the equilibrium constant Ka, but it can also be quantified by the percent ionization. This is the percentage of the compound that has ionized (dissociated). Strong acids (bases) ionize completely so their percent ionization is 100%. The percent ionization for a weak acid (base) needs to be calculated. It can be more intuitive when thinking about solution to think about the percent ionized versus the concentrations or the equilibrium constant. It is essentially the answer to the question "what fraction actually reacted when it was put into the water?"
The percent ionized is the fraction of the original compound that has ionized. Thus we compare the concentration of the ion in solution to the original concentration of the neutral species. For the example of HF in water, the percent ionization would be
\[ {\rm percent \; ionized \;\; = {{[F^-] \over C_{HF}} \times 100\% }} \]
where [F-] is the concentration of F- at equilibrium and CHF is the concentration of HF in the original solution. The Ka for HF is 6.8 x 10-4. If we calculate the concentration of F- in a 0.1M solution of HF, we find that [F-] = 8.3 x 10-3. The percent ionization for this solution is then.
\[ {\rm percent \; ionized \;\; = {{[F^-] \over C_{HF}} \times 100\%} = {{(8.3x10^{-3}) \over (0.1)}\times 100\%} = 8.3\% } \]
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