Question

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.

A.) What is the initial pH of the solution, before any titrant is added?

B.) The percent neutralization points for a titration indicate the percent of the reagent that has reacted. For example, at the 25% neutralization point for piperazine, 75% (0.75×[Pip]) of the base remains, and 25% is neutralized to form PipH+.

Calculate the pH at the 25%, 50%, and 75% neutralization points of the first neutralization, respectively.

C.) What is the pH at the first equivalence point?

D.) The percent neutralization points for a titration indicate the percent of the reagent that has reacted. For example, at the 25% neutralization point for PipH+, 75% (0.75×[PipH+]) of the base remains, and 25% is neutralized to form PipH22+.

Calculate the pH at the 25%, 50%, and 75% neutralization points of the second neutralization, respectively.

Please show all work!

Answer 1

Molar mass for piperazine=194.24 g/mole.

1.00 g Pip =1/194.24= 0.00515 moles Pip

Molarity = (0.00515 /100) * 1000= 0.0515 M

pK_b1 Pip = 4.22; K_b1 = 6.0 x 10^-5
pK_b2 Pip = 8.67; K_b2 = 2.1 x 10^-9

a.

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
As Kb1 is small, ignore 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


(b). During the first neutralization, Pip will react with HCl as follows:

Pip + H+ ==> PipH+

So Pip will be changed into PipH+. Above we calculated that the starting moles of Pip = 0.00515. When we finish the first neutralization, then moles PipH+ = 0.00515. At the 25% mark, then 25% of the 0.00515 moles of Pip have been changed into PipH+, or (0.25)(0.00515) = 0.00129 moles PipH+. That leaves 0.00515 - 0.00129 = 0.00386 moles of Pip remaining.

This mixture of Pip and PipH+ constitutes a buffer system where Pip is the weak base and PipH+ is the conjugate weak acid. Buffer problems are readily solved using the Henderson equation:

pOH = pK_b - log [ base/salt]

We need pKb1.

pOH= 4.22 - log[0.00386/0.00129] = 3.74

pH = 14 - 3.74 = 10.25

At the 50% point in the first neutralization,i.e., at the halfway point of titration, pOH = pKb

Thus, pOH= 4.22

pH= 14-4.22= 9.78

At the 75% point in the first neutralization, moles PipH+ = (0.75)(0.00515) = 0.00386 and moles Pip = 0.00129. This is the opposite of what we had at the 25% point.

pOH = 4.22 - log (moles Pip / moles PipH+) = 4.22 - log (0.00129 / 0.00386) = 4.69

pH= 14 -4.69= 9.31

(c) At the first equivalence point there is nothing left but PipH+, 0.00515 moles of it. But adding HCl during the titration changed the solution volume, so the molarity of PipH+ is different. Since HCl reacted with Pip in a 1:1 mole ratio, then moles HCl added = initial moles Pip = 0.00515.

moles HCl = M HCl x L HCl
0.00515 = (0.500)(L HCl)
L HCl = 0.0103 = 10.3 mL

Since we started with 100.0 mL of Pip solution, then the new volume = 100.0 + 10.3 = 110.3 mL = 0.1103 L.

M PipH+ = moles PipH+ / L of solution = 0.00515 / 0.1103 = 0.0467 M

Solve this one just like before. This time you're looking at the reaction of PipH+ with water to form PipH2 2+ and OH- so you'll have to use Kb2 (2.1 x 10^-9).

PipH+ + H2O <==> PiPH2 2+ + OH-

You'll find that [OH-] = 9.9 x 10^-6 and that pH = 9.00.


(d). This is exactly the same as what we did in (b), except the titration reaction is now
PipH+ + H+ ==> PipH2 2+. Use pKb this time. Answers will be
pH at 25% = 5.81
pH at 50% = pKa2 = 5.33
pH at 75% = 4.85