Question

Piperazine, HN(C4H8)NH, is a diprotic weak base used as a corrosion inhibitor and an insecticide and has the following properties: pKb1=4.22, pKb2=8.67 For writing the reactions of this base in water, it can be helpful to abbreviate the formula as Pip: Pip + H2O ⇌ PipH+ + OH− PipH+ + H2O ⇌ PipH22+ + OH− The piperazine used commercially is a hexahydrate with the formula C4H10N2⋅6H2O. A 0.00515-mol sample of piperazine hexahydrate is dissolved in enough water to produce 100.0 mL of solution and is titrated with 0.500 MHCl.

a. Calculate the pH after 2.575 mL of HCl have been added. (two decimal places)

b. Calculate the pH when 12.875 mL of HCl has been added. (two decimal places)

Answers: a) 10.26 b) 5.81 Need further explanation!

Answer 1

Piperazine titration with HCl

pKb1 = 4.22 ; pKb2 = 8.67

initial [piperazine] = 0.00515 mol/0.1 L = 0.0515 M

a. After 2.575 ml of 0.5 M HCl added

initial moles piperazine = 0.0515 mol x 100 ml = 5.15 mmol

added HCl = 0.5 M x 2.575 ml = 1.29 mmol

PipH+ formed = 1.29 mmol

Pip remained = 5.15 - 1.29 = 3.86 mmol

Using Hendersen-Hasselbalck equation,

pOH = pKb1 + log(PipH+/Pip)

        = 4.22 + log(1.29/3.86)

        = 3.74

pH of solution = 14 - pOH = 14 - 3.74 = 10.26

b. after 12.875 ml of 0.5 M HCl added

initial moles piperazine = 0.0515 mol x 100 ml = 5.15 mmol

added HCl = 0.5 M x 12.875 ml = 6.44 mmol

5.15 mmol PipH+ first forms which reacts with (6.44 - 5.15) mmol HCl to form PipH2^2+

PipH2^2+ formed = (6.44 - 5.15) = 1.29 mmol

PipH+ remained = 5.15 - 1.29 = 3.86 mmol

Using Hendersen-Hasselbalck equation,

pOH = pKb2 + log(PipH2^2+/PipH+)

        = 8.67 + log(1.29/3.86)

        = 8.19

pH of solution = 14 - pOH = 14 - 8.19 = 5.81