Question

A 0.10 M solution of ethylamine (C2H5NH2) has a measured pH of 11.87. Calculate the kb of this base.

Answer 1

Molarity---. C2H5NH2 + H2O <----> C2H5NH3+ + OH-
Initial Condition -------- .0.100 . . . . . . . . . . . . . . .0 . . . . . .0
Change------- -x . . . . . . . . . . . . . . . . x . . . . . .x
At Equilibrium 0.100- x . . . . . . . . . . . ..x . . . . . .x

Given pH = 11.87
So, pOH = 14.00 - pH = 14.00 - 11.87 = 2.13
[OH-] = 10^-pOH = 10^-2.13 = 0.0074 = x

Kb = [C2H5NH3+][OH-] / [C2H5NH2] = x^2 / (0.100 - x) = (0.0074)^2 / (0.100 - 0.0074) = 5.9 x 10^-4