Problem: When a 0.0538 M solution of an unknown weak acid, HA, is made, only 3.57% acid ionized. What is the Ka of the unknown acid?

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FREE Expert Solution

We’re being asked to calculate the equilibrium constant of HA if a 0.0538 M solution is 3.57% ionized.


Recall that the percent ionization is given by:


% ionization=[H3O+][HA]initial×100


Remember that weak acids partially dissociate in water and that acids donate H+ to the base (water in this case). 


The dissociation of HNO2 is as follows:

HA(aq) + H2O(l)  H3O+(aq) + A(aq)


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When a 0.0538 M solution of an unknown weak acid, HA, is made, only 3.57% acid ionized. What is the Ka of the unknown acid?

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