c. Example 3. Determine the pH of a 0.0100 mol/L HC2H3O2 solution. The Ka of
the acid is 1.75 X 105.
Solution. In determining the pH of a solution of weak electrolyte, you must
first calculate the hydrogen ion concentration. Acetic acid ionizes as follows:
+
HC2H3O2 <====> H + C2H3O2
+
[H ][C2H3O2 ]
Ka = 
[HC2H3O2]
Substitute the given information, and solve for the hydrogen ion
concentration.
+
[H ] [C2H3O2]
1.75 X 105 mol/L = 
0.100 mol/L
+
Substitute the variable x for H and C2H3O2, since one hydrogen ion and
one acetate ion are produced for every dissociated molecule of HC2H3O2.
+

[H ] [C2H3O2 ] = (1.75 X 105) (0.0100 mol/L)
x2 = (1.75 X 105 mol/L) (0.0100 mol/L)
x2 = 1.75 X 107 mol2/L2
+
x = [H ] = 4.18 X 104 mol/L
Using the calculated value for the hydrogen ion concentration, determine the
pH of the solution.
+
pH = log [H ]
pH = log [4.18 X 104 mol/L] = (.6212  4)
pH = (3.38)
pH = 3.38
MD0837
910