Question

Dozens of pharmaceuticals ranging from Cyclizine for motion sickness to Viagra for impotence are derived from the organic compound piperazine. Solutions of piperazine are basic (Kb1 = 5.38 x 10^-5, Kb2 = 2.15 x 10^-9). What is the pH of a 0.0130 M solution of piperazine? Please explain!

Answer 1

Piperazine is a base with 2 N atoms; each can accept a proton. If we call piperazine "B" then the first proton accepted results in BH+ and the 2nd proton accepted = BH2 2+. The Ka values that they gave you are: Ka1 for BH2 2+ ==> BH+ + H- and Ka2 for BH+ ==> B + H+. But we need Kb values to do this problem. Note that the first reaction is

B + H2O <++> BH+ + OH- . . .BH+ is the conjugate acid of B. Hence,

Kb1 x Ka2 = 1 x 10^-14
Kb1 = (1 x 10^-14) / (1.86 x 10^-10) = 5.38 x 10^-5

In the second step,

BH+ + H2O <==> BH2 2+ + OH- . . .BH2 2+ is the conjugate acid of BH+. Hence,

Kb2 = (1 x 10^-14) / (Ka1) = (1 x 10^-14) / (4.65 x 10^-6) = 2.15 x 10^-9

Make an ICE chart for the first reaction.

Molarity . . . . . .B + H2O <==> BH+ + OH-
Initial . . . . . . .0.207 . . . . . . .. . . . .0 . . . . .0
Change . . . . . .-x . . . . . . . . . . . . .x . . . . . x
Equilibrium . . 0.207-x . . . . . . . . . .x . . . . . x

Kb1 = [BH+][OH-] / [B] = (x)(x) / (0.207-x) = 5.38 x 10^-5

Because Kb1 is fairly small, we can neglect the -x term to simplify the math.

x^2 / 0.207 = 5.38 x 10^-5
x^2 = 1.11 x 10^-5
x = 3.33 x 10^-3 M = [BH+] = [OH-]
[B] = 0.207-x = 0.207 - 0.00333 = 0.204 M

pOH = -log[OH-] = -log (3.33 x 10^-3) = 2.48
pH = 14.00 - pOH = 14.00 - 2.48 = 11.52

Note that Kb2 = 2.15 x 10^-9. The additional [OH-] created from this reaction (and the amt. of BH+ consumed) is negligible.

If you want to know [BH2+] you must draw up an ICE chart for that Kb2 reaction.

Molarity . . . . . . .BH+ + H2O <==> BH2 2+ + OH-
Initial . . . . . . . 0.00333 . . . . . . . . . . . .0 . . . . .0.00333
Change . . . . . . . .-x . . . . . . . . . . . . . .x . . . . . . . .x
Equilibrium . .0.00333-x . . . . . . . . . . . .x . . . .0.00333+x

Kb2 = [BH2 2+][OH-] / [BH+] = (x)(0.00333+x) / (0.00333-x) = 2.15 x 10^-9

Because Kb2 is very small, we can neglect the -x and +x terms to simplify the math.

x = 2.15 x 10^-9 M = Kb2 = [BH2 2+]