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SAMPLE EXERCISE 16.10 continued
We can do a little accounting to determine the concentrations of the species involved in the equilibrium. We
imagine that the solution is initially 0.10 M in HCHO2 molecules. We then consider the ionization of the acid
into H+ and CHO2–. For each HCHO2 molecule that ionizes, one H+ ion and one CHO2– ion are produced in
solution. Because the pH measurement indicates that [H+] = 4.2 10–3 M at equilibrium, we can construct the
following table:
Notice that we have neglected the very small concentration of H+(aq) that is due to the autoionization of H2O.
Notice also that the amount of HCHO2 that ionizes is very small compared with the initial concentration of the
acid. To the number of significant figures we are using, the subtraction yields 0.10 M:
We can now insert the equilibrium concentrations into the expression for Ka :
Check: The magnitude of our answer is reasonable because Ka for a weak acid is usually between 10–3 and Acids
10–10.
and
Bases