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.

Show the titration curve by plotting the following points:

• the initial pH of the solution. (pH=11.23)

• the pH at the 25% neutralization point of the first neutralization (pH=10.26)

• the pH at the 50%neutralization point of the first neutralization. (pH=9.78)

• the pH at the 75% neutralization point of the first neutralization. (pH=9.30)

• the pH at the first equivalence point (pH=7.56)

• the pH at the 25% -neutralization point of the second neutralization. (pH=5.81)

• the pH at the 50% -neutralization point of the second neutralization. (pH=5.33)

• the pH at the 75% -neutralization point of the second neutralization. (pH=4.85)

• the pH at the second equivalence point.(pH=3.35)

Be sure to plot all nine points. (Find the HCl volumes of the pH's in mL. Also plot points on graph.)

Answer 1

Given the pH of the weak base = 11.23

=> pOH = 14 - 11.23 = 2.77

=> [OH-] = 10^-2.77 = 1.70x10^-3

Since Kb2 is very small the OH- concentration is mainly due to first dissociation

For weak base, [OH-] =  1.70x10^-3 = underroot(Kb1 x C)

=> 6.03x10^-5 xC = 2.89x10^-6  

=> C = 0.050 M

Hence the concentration of Piperazine, HN(C4H8)NH in the initial solution = 0.050 M

Since the volume of the base and the concentration of the acid taken is not given in the question, let's consider it from ourself.

Let's consider 50 mL of 0.050 M Piperazine solution and 0.050 M HCl solution

Since Piperazine is a diprotic acid, we need 100 mL of 0.050 M HCl to neutralize completely.

at the 25% neutralization point of the first neutralization, volume of 0.050M HCl added = 100 / 8 = 12.5 mL

at the 50% neutralization point of the first neutralization, volume of 0.050M HCl added = 2x100 / 8 = 25 mL

at the 75% neutralization point of the first neutralization, volume of 0.050M HCl added = 3x100 / 8 = 37.5 mL

at the first equivalence point, volume of 0.050M HCl added = 4x100 / 8 = 50.0 mL

at the 25% neutralization point of the second neutralization, volume of 0.050M HCl added = 5x100 / 8 = 62.5 mL

at the 50% neutralization point of the second neutralization, volume of 0.050M HCl added = 6x100 / 8 = 75.0 mL

at the 75% neutralization point of the second neutralization, volume of 0.050M HCl added = 7x100 / 8 = 87.5 mL

at the second equivalence point, volume of 0.050M HCl added = 8x100 / 8 = 100.0 mL

Now the graph can be plotted as