In the laboratory, a general chemistry student measured the pH of a
0.507 M aqueous solution of acetic
acid to be 2.504.
Use the information she obtained to determine the Ka for
this acid.
Concentration of acetic acid = 0.507 M
pH = 2.504
[H+] = 10-pH
[H+] = 10-2.504
[H+] = 3.133 x 10-3 M
ICE table | HAc (aq) | H+ (aq) | Ac- (aq) | |
Initial conc. | 0.507 M | 0 | 0 | |
Change | -x | +x | +x | |
Equilibrium conc. | 0.507 M - x | +x | +x |
Ka = [H+]eq[Ac-]eq / [HAc]eq
Ka = [(x) * (x)] / (0.507 M - x)
Ka = (x2) / (0.507 M - x)
x = [H+] = 3.133 x 10-3 M
Ka = (3.133 x 10-3 M)2 / (0.507 M - 3.133 x 10-3 M)
Ka = 1.95 x 10-5 M
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