Question

The percentage of pyridine (c5h5n) that forms pyridinium ion (c5h5n +h) in a 0.10 m aqueous pyridine solution (kb for c5h5n = 1.7 × 10–9) is

Answer 1

Ka x Kb = Kw

pKa + pKb = pKw= 14

where,

Ka = dissociation constant of acid

Kb = dissociation constant of base

Kw = ionic product of water

Ka = Kw/ Ke  x Ka = Kw/Ke =10^-14/ 1.7^-9 = 5.8^-6

C5H5N + H2O------------ C5H5NH + C5H5NH(+)  +   OH(-)

1-x                                                           x

5.8^-6 = x2

x = 7.6 ^-4

% of pyridine = 7.6^-4 x 100/ 0.1 = 0.76

Answer 2

Answer:

Explanation:

C5H5N + H2O ----> C5H5N^+ H + OH-

We know that,

α = √(Kb/M)

where

α --- degree of dissociation or association

Kb --- boiling point elevation constant

M --- Molarity

α = √(1.7×10^-9)/0.1

= √1.7×10^-8

= 1.3 × 10^-4

Percentage :

α % = 1.3 × 10^-4 × 100

= 0.013%