Question

The pH of a 0.145 mM solution of the alkaloid arecoline is 8.77. Determine Kb for arecoline from these data. (Assume Kw = 1.01 ✕ 10−14.) C8H13NO2(aq) + H2O(l) equilibrium reaction arrow HC8H13NO2+(aq) + OH −(aq)

Answer 1

we have below equation to be used:

pH = -log [H+]

8.77 = -log [H+]

log [H+] = -8.77

[H+] = 10^(-8.77)

[H+] = 1.698*10^-9 M

we have below equation to be used:

[OH-] = Kw/[H+]

Kw is dissociation constant of water whose value is 1.01*10^-14 at 25 oC

[OH-] = (1.01*10^-14)/[H+]

[OH-] = (1.01*10^-14)/(1.698*10^-9)

[OH-] = 5.947*10^-6 M

Lets write the dissociation equation of C8H13NO2

C8H13NO2 +H2O -----> HC8H13NO2+ + OH-

1.45*10^-4 0 0

1.45*10^-4-x x x

Kb = [HC8H13NO2+][OH-]/[C8H13NO2]

Kb = x*x/(c-x)

Kb = 5.947*10^-6*5.947*10^-6/(1.45E-4-5.947*10^-6)

Kb = 2.54*10^-7

Answer: 2.54*10^-7