Piperazine, HN(C_4H_8)NH is a diprotic weak base used as a corrosion inhibitor a
ID: 485502 • Letter: P
Question
Piperazine, HN(C_4H_8)NH is a diprotic weak base used as a corrosion inhibitor and an insecticide and has the following properties pK_b1 = 4.22 pK_b2 = 8.67 For writing the reactions of this base in water, it can be helpful to abbreviate the formula as Pip: Pip + H_2O PipH^+ + OH^- PipH^+ - H_2O PipH_2^2+ + OH^- The piperazine used commercially is a hexahydrate with the formula C_4H_10N_2 middot 6H_2O A 0.00515-mol sample of piperazine hexahydrate is dissolved in enough water to produce 100 0 111L of solution and is titrated with 0.500 A/HC1 Calculate the pH after 2 575 mL of HCI have been added. Enter the pH to two decimal places. Calculate the pH when 12 875 mL of HCI has been added Enter the pH to two decimal places.Explanation / Answer
Moles of HCl added = volume of HCl in L * Molarity = 0.002575 L * 0.5 M = 1.29*10^-3 mol
Moles of piparazine initially present = 0.00515 mol
HCl will react with piparazine to convert it to its protonated form (conjugate acid).
Moles of PipH+ formed = moles of acid added = 1.29*10^-3 mol
moles of Pip present after adding HCl = 0.00515 - 1.29*10^-3 mol = 3.86*10^-3 mol
The base and the conjugate acid together foems a buffer system. Ph of a buffer can be calculated by using Hinderson Hasselbalch equation.
pOH = pKb + log [PipH+/Pip]
= 4.22 + log[1.29*10^-3 mol/3.86*10^-3 mol]
= 3.74
pH =10.26
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Moles of Hcl added = 0.012875 L * 0.5 M = 6.44*10^-4 mol
Moles of PipH+ present = moles of HCl added = 6.44*10^-4 mol
moles of Pip present = 0.00515-6.44*10^-4 mol= 4.5*10^-3 mol
pOH = pKb + log [PipH+/Pip]
= 4.2 + log[6.44*10^-4 mol/4.5*10^-3 mol]
= 3.37
pH of the solution = 14-3.37 = 10.62