Question

Pyridine is a weak base that is used in the manufacture of pesticides and plastic resins. It also is 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 .1M Solution of Pyridine?

Answer 1

Given pKb = 8.75

=> -logKb = 8.75

=>Kb = 10^-8.75 = 1.78*10^-9

The dissociation reaction of pyridine is

C5H5N+H2O <=> C5H5NH+ + OH-

Init conc(M): 0.1 0 0

eqm conc(M): (0.1 - x) x x

Kb = [C5H5NH+]*[OH-] / [C5H5N]

=> 1.78*10^-9 = x² / (0.1 - x)

Since x<<0.1, (0.1-x) is nearly equals to 0.1.

=> 1.78*10^-9 = x² / 0.1

=> x = underroot(0.1*1.78*10^-9 ) = 1.334x10^-5 M

Hence [salt] = [C5H5NH+] = x = 1.334x10^-5 M

and [base] = [C5H5N] = nearly equals to 0.1

Now applying Hendersen equation

pOH = pKb + log[salt]/[base] = 8.75 + log(1.334x10^-5 M / 0.1M) = 4.88

=>pH = 14 - pOH = 14 - 4.88 = 9.12

Hence pH of 0.1M Solution of Pyridine is 9.12