Question

The base dissociation equilibrium constant (Kb) for C5H5N is 1.40×10-9 (eq. 1).

(eq. 1): C5H5N(aq) + H2O(l) = OH-(aq) + C5H5NH+(aq)

Calculate the value of the equilibrium constant (Ka) for the hydrolysis of C5H5NH+ as shown in the reaction (eq. 2).

(eq. 2): C5H5NH+(aq) + H2O(l) = H3O+(aq) + C5H5N(aq)

sorry for so many questions
Ka = (Kw/Kb).
Just remember that KaKb = Kw. The Ka and Kb are interchangeable with that formula.

Related Questions

Consider a solution that contains both C5H5N and C5H5NHNO3. (Kb for pyridine, C5H5N, is 4.0 x 10^-4)... Calculate the equilibrium constant for the weak base C5H5N, if a solution of the base with an initia... Calculate the pH of a buffer system composed of 0.880 M C5H5N and 0.660 M C5H5NHNO3. The Kb for C5H5... 'What is the pH of 0.083 M C5H5N if the Kb value for C5H5N is 2.9 x 10^-9?' I wrote the equation...