Consider a solution that contains both C5H5N and C5H5NHNO3. (Kb for pyridine, C5H5N, is 4.0 x 10^-4)

Question

(a) Calculate the ratio of [C5H5N]/[C5H5NH+] if the solution has a pH of 4.50.

(b) Calculate the ratio of [C5H5N]/[C5H5NH+] if the solution has a pH of 5.00.

Answer 1

C5H5N + HOH ==> C5H5NH^+ + OH^-
Ka = (Kw/Kb) = (C5H5NH^+)(OH^-)/(C5H5N).
You know Kw and Kb. From pH you can calculate (OH^-). That leaves the ratio of C5H5N/C5H5NH^+ to calculate. The reciprocal is what you are looking for. Part b is done the same way.