Problem: A 0.010 M solution of an acid is 8.0% ionized. What is Ka for this acid?A. 3.2 x 10-6B. 2.8 x 10-5C. 6.4 x 10-6D. 7.0 x 10-5E. 1.6 x 10-4

FREE Expert Solution

We’re being asked to calculate the equilibrium constant of an acid if a 0.010 M solution of an acid is 8.0% 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 the general acid HA is as follows:


HA(aq) + H2O(l)  H3O+(aq) + A(aq)


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Problem Details

A 0.010 M solution of an acid is 8.0% ionized. What is Ka for this acid?

A. 3.2 x 10-6

B. 2.8 x 10-5

C. 6.4 x 10-6

D. 7.0 x 10-5

E. 1.6 x 10-4

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