5. Using Equation 1, calculate the percent ionization of acetic acid for each concentration you used. Show the work for one of the calculations below. Report the information in Table 2.
6. Be sure to provide a satisfactory title for Table 2.
Table 2:
[Solution]
pH
[H3O+]
Ka
% ionization
0.10 M H
-
-
Average
-
7. Compare your average experimentally determined Ka with the accepted value at 25 °C (1.8 x 10^-5) and calculate the percent error.
8. Comment on whether the initial HC2H3O2 concentration has any effect on Ka. Briefly explain.
9. What effect does the initial HC2H3O2 concentration seem to have on the percent ionization?
10. Calculate the initial concentration of acetic acid in the vinegar solution (show your calculation).
11. In the past, many students have listed that the accidental addition of too much acetic acid contributed greatly to the difference between the experimental value and the accepted value. Suppose that Student A was supposed to make a 0.18 M solution by diluting 9.0 mL of 2.0 M acetic acid to 100.0 mL. The expected pH for this solution is 2.74. The Ka of acetic acid is 1.8 x 10^-5.
What would be the expected pH if Student A accidentally diluted 9.1 mL (instead of 9.0 mL) of the acid to 100 mL?
M1V1 = M2V2
M1 = 0.182
1.8 x 10^-5 = [H+]/0.182
[H+] = 1.81 x 10^-3
pH = -log(1.81 x 10^-3)
= 2.74
If Student A measured the above calculated pH, what would be the resultant Ka of acetic acid given that they expected the acid to have an initial concentration of 0.18 M?
[H+]^2/acetic acid
Ka = 3.2 x 10^-6/0.18 M
Ka = 1.82 x 10^-5