Question

Enough of a monoprotic acid is dissolved in water to produce a 1.35 M solution. The pH of the resulting solution is 2.93. Calculate the Ka for the acid.

Answer 1

The acid dissociation constant is defined as Ka = [H+][A-]/[HA] where [H+], [A-] and [HA] are the concentrations of protons, conjugate base, and acid in solution respectively. Assuming this is a weak acid as the pH is quite high for a 1.35 M solution, we can assume that the change in [HA] is negligible and therefore [HA] = 1.35 M.
To calculate [H+] we can use the relationship pH = -log[H+], rearranging to give: [H+] = 10^(-pH) = 10^(-2.93) = 1.17 x 10^(-3).
Since the acid is relatively concentrated we can assume therefore that       [H+] = [A-] as for each proton dissociated, a conjugate base is formed.
Therefore, we can calculate Ka as:
Ka = [H+]^2/[HA] = (1.17 x 10^-3 M)^2/1.35 = 1.01 x 10^-6 M