Problem: The Ka of a monoprotic weak acid is 5.55x10-3. What is the percent ionization of a 0.127 M solution of this acid?

FREE Expert Solution

We’re being asked to calculate the percent ionization of a 0.127 M aqueous solution of a weak acid.


Recall that the percent ionization is given by:


% ionization=[H+][HA]initial×100


We know the initial concentration of HA, 0.127 M. We just need to find the concentration of H+ at equilibrium.


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 = 5.55 × 10–8


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Problem Details

The Ka of a monoprotic weak acid is 5.55x10-3. What is the percent ionization of a 0.127 M solution of this acid?

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