Question

4. Calculate the percentage of pyridine (C5H5N) that forms pyridinium ion, C3H5NH', in a 0.10 Maqueous solution of pyridine (Kb = 1.7 * 10-9). A) 0.013% B) 0.0060% C) 0.77% D) 0.060% E) 1.6%

Answer 1

C5H5N dissociates as:

C5H5N +H2O -----> C5H5NH+ + OH-

0.1 0 0

0.1-x x x

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

Kb = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((1.7*10^-9)*0.1) = 1.304*10^-5

since c is much greater than x, our assumption is correct

so, x = 1.304*10^-5 M

So, [OH-] = x = 1.304*10^-5 M

% C5H5NH+ = [C5H5NH+]*100 / initial [C5H5N]

= x*100 / 0.1

= (1.304*10^-5)*100/0.1

= 0.013 %

Answer: A