Problem: A 0.142 M solution of a monoprotic acid has a percent dissociation of 1.60%. Determine the acid ionization constant (Ka) for the acid.

FREE Expert Solution

We’re being asked to calculate the acid ionization constant (Ka) for a monoprotic acid if a 0.142 M solution has a percent dissociation of 1.60%.


Recall that the percent ionization is given by: 

(Let HA → monoprotic acid)


% 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 HA is as follows:

HA(aq) + H2O(l)  H3O+(aq) + A(aq)


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Problem Details

A 0.142 M solution of a monoprotic acid has a percent dissociation of 1.60%. Determine the acid ionization constant (Ka) for the acid.

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