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.

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