Question

The pH of an aqueous solution of 0.593 M quinoline (a weak base with the formula C9H7N) is

Answer 1

kb of quinoline = 6.3*10^-10

C9H7N dissociates as:

C9H7N +H2O -----> C9H7NH+ + OH-

0.593 0 0

0.593-x x x

Kb = [C9H7NH+][OH-]/[C9H7N]

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 ((6.3*10^-10)*0.593) = 1.933*10^-5

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

so, x = 1.933*10^-5 M

so.[OH-] = x = 1.933*10^-5 M

use:

pOH = -log [OH-]

= -log (1.933*10^-5)

= 4.7138

use:

PH = 14 - pOH

= 14 - 4.7138

= 9.29

Answer: 9.29