Part A Calculate the pH after 2.575 mL of HCl have been added. Enter the pH to t

Question

Part A

Calculate the pH after 2.575 mL of HCl have been added.

Enter the pH to two decimal places.

Part B

Calculate the pH when 12.875 mL of HCl has been added.

Enter the pH to two decimal places.

Piperazine, HN(C4Hs)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 The piperazine used commercially is a hexahydrate with the formula C4 H10N2-6H20.

Explanation / Answer

Find out the initial molar concentration of Pip from the data provided. We start with 0.00515 mole Pip in 100.0 mL solution so that the molar concentration is (0.00515 mole)/[(100 mL)*(1 L/1000 mL)] = 0.0515 mol/L = 0.0515 M.

The dissociation equations are given:

Pip + H2O <=====> PipH+ + OH-

Kb = [PipH+][OH-]/[Pip] = (x).(x)/(0.0515 – x) ……(1)

We have assumed [PipH+] = [OH-] = x M

Given pKb = 4.22, Kb = antilog (-pKb) = 10^(-.4.22) = 6.0256*10-5

Plug this value of Kb in expression (1).

6.0256*10-5 = x2/(0.0515 – x)

Make an assumption. Since Kb is small (of the order of 10-5), we can ignore x in the denominator and write

6.0256*10-5 = x2/0.0515

====> x2 = 3.103*10-6

====> x = 1.761*10-3

Therefore, [PipH+] = 1.761*10-3 M and moles of PipH+ = (100 mL)*(1 L/1000 mL)*(1.761*10-3 mol/L) = 1.761*10-4 mole.

Part A

We add 2.575 mL of 0.500 M HCl to the solution of Pip base. The moles of HCl added is (2.575 mL)*(1 L/1000 mL)*(0.500 mol/L) = 1.2875*10-3 mole.

The reaction taking place is

Pip + HCl --------> PipH+ + Cl-

Moles of HCl added = moles of PipH+ formed = moles of Pip consumed = 1.2875*10-3 mole.

Moles of PipH+ at equilibrium = (1.761*10-4 + 1.2875*10-3) mole = 1.4636*10-3 mole.

Moles of Pip at equilibrium = (0.00515 – 1.2875*10-3) mole = 3.8625*10-3 mole.

The volume of the solution remains the same [= (100 + 2.575) mL = 102.575 mL] and hence we can write the Henderson-Hasslebach equation in terms of number of moles of Pip and PipH+.

Therefore,

pOH = pKb + log [PipH+]/[Pip] = 4.22 + log (1.4636*10-3 mole)/(3.8625*10-3 mole) = 4.22 + log (0.3789) = 4.22 + (-0.4214) = 3.789 3.79.

Therefore, pH = 14 – pOH = 14 – 3.79 = 10.21 (since pH + pOH = 14) (ans).

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