Question

a) Calculate [OH^-] of a 3.20×10^-1 M aqueous solution of triethylamine ((C₂H₅)₃N, K_b = 4.0×10^-4).

____mol/L

b) Calculate the pH of the above solution.

____pH

Answer 1

a) Dissociation equillibrium of (C2H5)3N is

(C2H5)3N(aq) + H2O(l) <-------> (C2H5)3NH⁺(aq)   + OH^-(aq)

Kb = [(C2H5)3NH+][OH^-]/[(C2H5)3N]

Initial concentration

[(C2H5)3N]= 0.320

[(C2H5)3NH+]=0

[OH^-]= 0

change in concentration

[(C2H5)3N] = -x

[OH^- ] = + x

[(C2H5)3NH+] = +x

equillibrium concentration

[(C2H5)3N] = 0.320 -x

[(C2H5)3NH+]= x

[OH^- ] = x

Therefore

x²/(0.320 - x) = 4.0×10^-4

solving for x

x = 0.0111

Therefore,

[OH^-] = 0.0111M

b) pOH = -log[OH^- ]

pOH = - log (0.0111)

pOH = 1.95

pH + pOH = 14

pH = 14 - pOH

pH = 14 - 1.95

pH = 12.05