Question
Consider titrating 15.0ml of a .245M solution of hydazine, H2NNH2, with a .453M HCl solution. The Kb value for hydrazine is 1.3x10^-6 b) what volume of HCl needs to be added to the solution in oder to have a bffer system with the maximum capacity for added acid or base? c) what is the pH of the solution before any HCl has been added? d) what is the pH of the solution after the addition of 3.0 ml of HCl? e) What is the pH of the solution at the equivalence point? f) what is the pH of the solution when 5.0ml more of the HCl solution is added after the equivalence point has ben reached?
Explanation / Answer
What is the initial pH of the solution, before any titrant is added? given: Piperazine, HN(C4H8)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 + H20 PipH^+ + OH- PipH+ + H20 PipH2^2+ + OH- The piperazine used commercially is a hexahydrate with the formula: C4H10N2*6H2O. 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) 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)Calculate the pH at the 25%, 50%, and 75% neutralization points of the second neutralization, respectively. 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