Question

What is equilibrium concentration of pyridine to 6 decimal places (C₅H₅N, K_b=1.7x10^-9) in a solution of 0.095 M C₅H₅NH⁺¹?

Answer 1

use:

Ka = (1.0*10^-14)/Kb

Ka = (1.0*10^-14)/1.7*10^-9

Ka = 5.882*10^-6

C5H5NH+ + H2O -----> C5H5N + H+

9.5*10^-2 0 0

9.5*10^-2-x x x

Ka = [H+][C5H5N]/[C5H5NH+]

Ka = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((5.882*10^-6)*9.5*10^-2) = 7.475*10^-4

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

so, x = 7.475*10^-4 M

[C5H5NH+] = x = 7.475*10^-4 M = 0.000748 M

Answer: 0.000748 M