Piperazine, HN(C4H8)NH, is a diprotic weak base used as a corrosion inhibitor an
ID: 775521 • Letter: P
Question
Piperazine, HN(C4H8)NH, is a diprotic weak base used as a corrosion inhibitor and an insecticide and has the following properties:
pKb1=4.22pKb2=8.67
For writing the reactions of this base in water, it can be helpful to abbreviate the formula as Pip:
Pip+H2OPipH++H2O??PipH++OH?PipH22++OH?
The piperazine used commercially is a hexahydrate with the formula C4H10N2?6H2O. A 1.00-g sample of piperazine hexahydrate is dissolved in enough water to produce 100.0 mL of solution and is titrated with 0.500 M HCl. What is the pH at the first equivalence point?
Explanation / Answer
First calculate the starting molarity of the piperazine solution. The molar mass for piperazine hexahydrate = 4C (4 x 12.01) + 10H (10 x 1.008) + 2N (2 x 14.01) + 12H (12 x 1.008) + 6O (6 x 16.00) = 194.24 g/mole.
1.00 g Pip x (1 mole Pip / 194.24 g Pip) = 0.00515 moles Pip
Molarity Pip = moles Pip / L of solution = 0.00515 / 0.100 = 0.0515 M
pKb1 Pip = 4.22; Kb1 = 10^-4.22 = 6.0 x 10^-5
pKb2 Pip = 8.67; Kb2 = 10^-8.67 = 2.1 x 10^-9
(a). Set up an ICE chart.
Molarity . . . . . . . .Pip + H2O <==> PipH+ + OH-
Initial . . . . . . . . .0.0515 . . . . . . . .. . . 0 . . . . .0
Change . . . . . . . . .-x . . . . . . . . .. . . . x . . . . .x
Equilibrium . . . .0.0515-x . . . . . . . . . . x . . . . .x
Kb1 = [PipH+][OH-] / [Pip] = (x)(x) / (0.0515-x) = 6.0 x 10^-5
Because Kb1 is small, we can igniore the -x term after 0.0515 to simplify the math.
x^2 / 0.0515 = 6.0 x 10^-5
x^2 = 3.1 x 10^-6
x = 1.8 x 10^-3 M = [OH-]
pOH = -log [OH-] = -log (1.8 x 10^-3) = 2.75
pH = 14.00 - pOH = 14.00 - 2.75 = 11.25