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1. Class, Friday, Dec 3, 2004 Announcements ______________________________________ • Changes from schedule in the syllabus: • pH and Spectrometer Labs due Friday, Dec 10. • Exam 3 on Wed, Dec 8 for Exam 3. You may start after 1:00 PM. • I plan to complete lecturing today, so you have Mon and Fri of next week off. • Final Exam, 10:00 AM, Wed., Dec 15.

2. Friday we will discuss the following problem; work on it on your own between now and then. ______________________________________ • The conjugate acid of the weak base cyclohexamine has a pKa = 10.57. 30.00 mL of a 0.06000 M solution of cyclohexamine is to be titrated with a 0.04500 M solution of standard HCl. Find points (pH values) at Vo, 0.25 Vept, 0.50 Vept 0.75 Vept , Vept , 1.1Vept and 1.2 Vept. Use these points to sketch a titration curve.

3. Weak Base Titration Curve Problem – the Solution ______________________________________ 1) First, find the volume of HCl required to reach the equivalence point. At the eq pt, the mmol of acid = mmol of base (30.00)(0.06000) = VHCl (0.04500) VHCl = 40.00 mL Then we wish to calculate the pH at the following VHCl: 0, 10, 20, 30, 40, 44, and 48 mL

4. Weak Base Titration Curve Problem – the Solution ______________________________________ • 2) If the pKa of the acidic form is 10.57, then the • pKb of the base = 14 – 10.57 = 3.43. • Kb = 103.43 =3.72 X 104

5. Weak Base Titration Curve Problem – the Solution ______________________________________ • 3) Let Cy represent cyclohexamine; the chemical equation then is • Cy+ H2O < == > HCy+ + OH and • Kb = [HCy+ ] [OH] / [Cy] = 3.72 X 104

6. Weak Base Titration Curve Problem – the Solution, at VHCl = 0 ______________________________________ • 4) At VHCl = 0, this is a weak base problem: • Kb = x2 / 0.600 = 3.72 X 104 • x2 = 2.23 X 105 • x = 4.72 x 103 = [OH] • pOH = 2.33 • pH = 14 – 2.33 = 11.67

7. Weak Base Titration Curve Problem – the Solution, the buffer region ______________________________________ • 5) Each of the volumes between 0 and the eq pt of 40 are buffer problems. For example, at 0.25 Vept or 10 mL, the [Cy]  (40 – 10), or 30 and the [HCy] 10. Using the Henderson-Hasselbalch equation, • pH = 10.57 + log10 (30/10) • pH = 10.57 + log10 (3) = 10.57 + 0.48 • pH = 11.05 • Likewise, the pH values at 20 and 30 mL are calculated; they ar 10.57 and 10.09 respectively.

8. Weak Base Titration Curve Problem – the Solution, the eq pt. ______________________________________ • At the eq pt, the acidic properties of the conjugate acid form establishes the pH. The chemical reaction is • HCy+ + H2O < == > Cy + H3O+ • The pKa was given as 10.57, so the • Ka = 1010.57 = 2.69 x 1011 • Ka= [Cy][H3O+] / [HCy+] = x2 / [HCy+]

9. Weak Base Titration Curve Problem – the Solution, the eq pt. ______________________________________ • 8) After the eq pt, the pH is established by the amount of excess HCl added. For example at VHCl = 44.00 mL, there are 4.00 mL of HCl past the eq pt. The mmol of HCl = (4.00)(0.04500) = 0.1800mmol • The total volume = 30.00 + 44.00 = 74.00, so calculate the • [H3O+] = 0.1800 / 74.00 = 2.43 x 103 • pH = 2.61. • Likewise at VHCl = 48.00, the pH = 2.34

10. Weak Base Titration Curve Problem – the Solution ______________________________________ • In a table the results are

11. Weak Base Titration Curve Problem – the Plot ______________________________________

12. Weak Base Titration Curve Problem – the Plot ______________________________________