Question

Calculate the molar concentration of OH^- ions in a 8.1 x 10^-2 M solution of ethylamine (K_b= 6.4 x 10^-4).

Calculate the pH of this solution.

Answer 1

Molarity . . . . . .C2H5NH2 + H2O <==> C2H5NH3+ + OH-
Initial . . . . . . . . . .0.81 . . . . . . . . . . . . . . .0 . . . . . .0
Change . . . . . . . . . .-x . . . . . . . . . . . . . . . .x . . . . . .x
At Equil. . . . . . 0.81 - x . . . . . . . . . . . . . . .x . . . . . .x

pH = 11.87
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 / (0.81 - x) = 6.4 x 10^-4

x^2 = 5.184 x 10^-4 - 6.4 x 10^-4 * x

x = 0.0224

so [OH-] = 0.0224 mol/L or (M)

pOH = -log( [OH-]) = 1.65

pH = 14 - pOH = 14 - 1.65 = 12.35 Answer