What is the percent ionization of a monoprotic weak acid solution that is 0.188 M? The acid-dissociation (or ionization) constant, Ka, of this acid is 2.43 × 10 − 12 .

The percent ionization of a monoprotic weak acid solution that is 0.188 M is

Explanation:

Dissociation of weak acid is represented as:

 cM              0             0

So dissociation constant will be:

Give c= 0.188 M and = ?

Putting in the values we get:

Thus percent ionization of a monoprotic weak acid solution that is 0.188 M is