Question

What is the pH of a 0.60 M pyridine solution that has K_b = 1.9 × 10^-9? The equation for the dissociation of pyridine is

C₅H₅N(aq) + H₂O(l) ⇌ C₅H₅NH⁺(aq) + OH^-(aq).

Answer 1

Lets write the dissociation equation of C5H5N

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

0.6 0 0

0.6-x x x

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

Kb = x*x/(c-x)

Assuming small x approximation, that is lets assume that x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((1.9*10^-9)*0.6) = 3.376*10^-5

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

so, x = 3.376*10^-5 M

so.[OH-] = x = 3.376*10^-5 M

we have below equation to be used:

pOH = -log [OH-]

= -log (3.376*10^-5)

= 4.47

we have below equation to be used:

PH = 14 - pOH

= 14 - 4.47

= 9.53

Answer: 9.53