Question

A 3.452 M solution of triethylamine ((C2H5)3N, Kb= 4e-4) has a pH of 12.57. What is the percent ionization of this solution?

Answer 1

Answer –

We are given, [(C₂H₅)₃N] = 3.452 M , pH = 12.57

First we need to calculate [OH^-] from the given pH, because trimethylamine is weak base and we need to calculate how much concentration of OH^- ,

We know formula

pH + pOH = 14

pOH = 14-pH

         = 14-12.57

         = 1.43

Now from the pOH calculate [OH^-]

pOH = -log [OH^-]

[OH^-] = 10^-pOH

         = 10^-1.43

        = 0.0372 M

This is the concentration at equilibrium , x = 0.0372 M

We know formula for calculating percent dissociation

Percent dissociation = x / initial concertation * 100 %

                                  = 0.0372 M / 3.452 M * 100 %

                                   = 1.08 %