What is the initial pH of the solution, before any titrant is added? Info: Piper

Question

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

Info:

Piperazine,HN(C4H8)NH , a diprotic weak base has the following properties: pKb1 = 4.22, pKb2= 8.67

For writing the reactions of this base in water,abbreviate the formula as Pip:

Pip+H2O-->/<--PipH+OH-, PipH+ + H2O-->/<-- PipH2+ + OH-

The piperazine used commercially is a hexahydrate with the formula C4H10N2 * 6H20.

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 .

Explanation / Answer

First calculate the starting molarity of the piperazine solution. The molar mass for piperazine hexahydrate = 4C (4 x 12.01) + 10H (10 x 1.008) + 2N (2 x 14.01) + 12H (12 x 1.008) + 6O (6 x 16.00) = 194.24 g/mole. 1.00 g Pip x (1 mole Pip / 194.24 g Pip) = 0.00515 moles Pip Molarity Pip = moles Pip / L of solution = 0.00515 / 0.100 = 0.0515 M pKb1 Pip = 4.22; Kb1 = 10^-4.22 = 6.0 x 10^-5 pKb2 Pip = 8.67; Kb2 = 10^-8.67 = 2.1 x 10^-9 (a). Set up an ICE chart. 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 Because Kb1 is small, we can igniore 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-Hasselbalch equation: pH = pKa + log (moles weak base / moles weak acid) We need pKa1 for PipH+. Since pKa1 + pKb1 = 14.00 (always), then pKa1 = 14.00 - pKb = 14.00 - 4.22 = 9.78. pH = 9.78 + log (moles Pip / moles PipH+) = 9.78 + log (0.00386 / 0.00129) = 9.78 + 0.48 = 10.26 At the 50% point in the first neutralization, moles Pip = moles PipH+ = (0.50)(0.00515) = 0.002575. pH = 9.78 + log (moles Pip / moles PipH+) = 9.78 + log (0.002575 / 0.002575) = 9.78 + 0 = 9.78 Note: at the halfway point of any titration, pH = pKa. 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. pH = 9.78 + log (moles Pip / moles PipH+) = 9.78 + log (0.00129 / 0.00386) = 9.30 (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 I did in (a) using an ICE chart. 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+. Be sure to use pKa2 = 14.00 - pKb2 = 14.00 - 8.67 = 5.33. The moles of PipH+ and PipH2 2+ are exactly as they were in (b). You should get these results: pH at 25% = 5.81 pH at 50% = pKa2 = 5.33 pH at 75% = 4.85

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