The Ka of the 0.50 M monoprotic acid with a pH of 5.25 is approximately 6.30 x 10^-11. The [H3O+] from the autoionization of water in this solution is negligible compared to the concentration from the acid ionization.
The student asked two questions regarding a 0.50 M monoprotic acid solution with a pH of 5.25:
a. What is the Ka of the acid?
b. What is the [H3O+] that is produced in this solution by the autoionization of water?
a. To find the Ka (acid dissociation constant) of the acid, first, we calculate the concentration of H3O+ using the pH value:
[H3O+] = 10-pH = 10-5.25
The concentration of H3O+ is thus approximately 5.62 x 10-6 M.
Since the acid is monoprotic, each molecule of acid that ionizes will produce one H3O+. The equilibrium expression for the ionization of the acid is Ka = [H3O+][A-]/[HA].
At equilibrium, the concentration of H3O+ is equal to the concentration of A-, and [HA] is approximately the initial concentration of the acid because only a small fraction ionizes. Therefore, Ka can be approximated as:
Ka = [H3O+]2 /[HA] = (5.62 x 10-6)2/0.50 = 6.30 x 10-11.
b. The concentration of [H3O+] produced by the autoionization of water is negligible compared to the concentration produced by the acid, especially at a pH of 5.25. For pure water at 25°C, the concentration of H3O+ is 1 x 10-7 M. However, due to the significantly higher concentration of H3O+ from the acid, the autoionization H3O+ contribution can be considered negligible in this solution.