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 0.435 M solution of pyridine? (Assume that the temperature is 25 ∘C.)

Answer 1

Answer:

pH of the solution = 9.45

Explanation:

pKb = 8.57

pKb = -logKb

Kb = antilog pkb

$K_b = 1.78 \times 10^{-9}$

ICE table for the ionization of pyrodine is:

$C_5H_5N+H_2O \leftrightharpoons C_5H_5NH^+ + OH^-$

initial  0.435                         0   0

equi.    0.435 - x                   x   x

so,

$k_b =\frac{ [C_5H_5NH^+][OH^-]}{[C_5H_5N]}\\1.78\times 10^{-9}=\frac{x^2}{0.453 - x}$

As kb is very less that means there is very less ionization, hence x can be ignored as compared to 0.435

$x^2 = 1.78\times 10{-9} \times 0.435\\x= 2.78\times 10^{-5}$

pOH = -log[OH-]

$pOH = -log[2.78\times 10^{-5}]\\=4.55$

pH = 14 - pOH

     = 14 - 4.55

     = 9.45