Problem: The Ka of a monoprotic weak acid is  2.08 × 10−3. What is the percent ionization of a .194 M solution of this acid?

FREE Expert Solution

We’re being asked to calculate the percent ionization of a 0.194 M aqueous solution of a monoprotic acid HA.


Since HA has a low Ka value, it’s a weak acid. Remember that weak acids partially dissociate in water and that acids donate H+ to the base (water in this case). The dissociation of HA is as follows:


HA(aq) + H2O(l)  H3O+(aq) + A(aq); Ka = 2.08 × 10−3


From this, we can construct an ICE table. Remember that liquids are ignored in the ICE table.

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Problem Details

The Ka of a monoprotic weak acid is  2.08 × 10−3. What is the percent ionization of a .194 M solution of this acid?

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