Question

Pyridine is a weak base that is used in the manufacture of pesticides and plastic resins. It is also a component of cigarette smoke. Pyridine ionizes in water as follows:

C5H5N+H2O⇌C5H5NH++OH−

The pKb of pyridine is 8.75. What is the pH of a 0.450 M solution of pyridine?

Answer 1

C5H5N+H2O⇌C5H5NH++OH−

This is a base in water so, let the base be "B" and HB+ the protonated base "HB+"

there are free OH- ions so, expect a basic pH

B + H2O <-> HB+ + OH-

The equilibirum Kb:

Kb = [HB+][OH-]/[B]

let "x" be OH- in solution

in equilibrium due to sotichiometry:

[HB+]= x= [OH-]

Account for the dissolved base in solution vs. not in solution:

[B] = 0.45-x

Substitute in Kb

Kb = [HB+][OH-]/[B]

10^-8.75= x*x/(0.45-x)

Solve for x, using quadratic equation

x = OH- = 2.828*10^-5

pOH = -log(OH-) = -log(2.828*10^-5) = 4.5485

pH = 14-pOH = 14-4.5485 = 9.45

pH = 9.45

pH = 8.96426