Question

Pyridine is an organic base that reacts with water as follows:

C5H5N(aq) + H2O(l) <---> C5H5NH+(aq) +OH-(aq)

Kb=1.58*10^-9 for pyridine. What is the concentration of the C5H5NH+ ion in a solution initially 1.0M in pyridine? What is the pH of the solution? Clearly state and check any approximations you make.

Answer 1

1)

C5H5N dissociates as:

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

1.0 0 0

1.0-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 ((1.58*10^-9)*1) = 3.975*10^-5

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

so, x = 3.975*10^-5 M

[C5H5NH+] = x = 3.975*10^-5 M

Answer: 3.975*10^-5 M

2)

[OH-] = x = 3.975*10^-5 M

use:

pOH = -log [OH-]

= -log (3.975*10^-5)

= 4.40

use:

PH = 14 - pOH

= 14 - 4.40

= 9.60

Answer: 9.60