1. ______________________________________ Class, Wednesday, Oct 27, 2004 Calcium Determination is due this Friday at class time. Exam 2 on Wed, Nov 3. Covers Chapters 7 – Gravimetric and Combustion Analysis 8 – Acids and Bases 9 – Buffers 12 – Chemical Equililbrium 13 – EDTA Titrations Volumetric Chloride Lab Calcium EDTA Lab Exam starts at 1:00 PM on Nov 3; you may work as needed. Study sheet should be ready this PM during lab

2. ______________________________________ Homework Assignment, Ch 16 The following Problems are assigned from chapter 16: 13, 14, 15, 17, 20 They are due at class time on Friday, Nov 5

3. ______________________________________ Redox Problems 1 – Brass is an alloy of copper and zinc. A sample whose mass was 8.5679 g was dissolved in nitric acid and the solution transferred quantitatively to a 500.0 mL volumetric flask. A 25.00 mL aliquot of the solution was treated with bromine water to convert all copper to the Cu+2 state as you did in the copper-iodine lab. After the pH was adjusted, excess KI was added and 21.32 mL of a 0.09855 M solution of sodium thiosulfate was required to reach the starch endpoint. What is the % copper in the brass sample? Chemical Reactions: 2Cu +2 (aq) + 4 I  (aq)  I 2 (aq) + 2CuI (s) I 2 (aq) + 2S 2 O 3  2 (aq)  2 I  (aq) + S 4 O 6  2 (aq)

4. ______________________________________ Redox Problem 1 Chemical Reactions: 2Cu +2 (aq) + 4 I  (aq)  I 2 (aq) + 2CuI (s) I 2 (aq) + 2S 2 O 3  2 (aq)  2 I  (aq) + S 4 O 6  2 (aq) From the chemical reactions, # mmol of I 2 = ½ x mmole of Cu +2, and the mmol of S 2 O 3  2 = 2 x mmol of I 2. Therefore, the mmol of Cu +2 = mmol S 2 O 3 -2. The mmol of S 2 O 3 -2 = 21.32 x 0.09855 = 2.101 = mmol Cu +2 The mass (mg) of Cu +2 = 2.101 mmol x 63.546 mg/mmol = 133.51 mg The aliquot represents (25/500) of the total sample of brass weighed. The total copper in the sample = (500/25) x 133.51 mg = 2670.3 mg = 2.6703g Finally, the % Cu in the brass = 2.6703 x 100 / 8.5679 = 31.166% = 31.17%

5. ______________________________________ Redox Problem 2 2 – A 0.3017 g sample of uranium ore was dissolved in strong mineral acids and treated with a reducing agent to convert all of the uranium to the +4 state. The solution required 28.37 mL of 0.01374 M KMnO 4 to oxidize the U +4 to the uranyl ion UO 2 +2 in an acidic solution. Calculate the % uranium in the sample. Chemical Reactions: U +4 (aq) + MnO 4  (aq)  UO 2 +2 (aq) + Mn +2 (aq) (unbalanced) Note: Omit H 2 O on your work sheet. Water may appear on either side in the net balanced equation.

6. ______________________________________ Ended, Wednesday, Oct 27, 2004 Continue on Friday

7. ______________________________________ Redox Problem 3 3 – 10 mL of an intravenous solution of glucose weighed 12.42 g. The 10 mL were diluted to 250.0 mL with water. A 10.00 mL aliquot was treated with 25.00 mL of 0.1537 M Ce +4 and heated for 10 minutes which oxidized the glucose to formic acid. The excess Ce +4 was back titrated with 21.06 mL of a 0.1146 M Fe +2 solution. What is the percent glucose in the original IV solution? Chemical Reactions: C 6 H 12 O 6 (aq) + Ce +4 (aq) → HCO 2 H(aq) + Ce +3 (aq) (unbalanced) Ce +4 (aq) + Fe +2 (aq) → Ce +3 (aq) + Fe +3 (aq)

8. ______________________________________ Redox Problem 4 4 – Pool treatment chemicals contain Ca(OCl) 2 (calcium hypochlorite) as the active ingredient. A 0.2500 g sample was dissolved in 0.5000 M sulfuric acid and an excess of KI added. The liberated iodine required 38.17 mL of 0.1187 M sodium thiosulfate to reach the starch endpoint. What is the percent calcium hypochlorite in the sample? Chemical Reactions: Ca(OCl) 2 + 2H + (aq)  Ca +2 (aq) + 2HOCl (aq) OCl  (aq) + 2H + (aq) + 2 I  (aq)  Cl  (aq) + I 2 (aq) + H 2 O

9. ______________________________________ Redox Problem 5 5 – A 1.407 g sample of paint from an old building was dissolved in organic solvents and treated with appropriate reagents to remove potential interferences. The solution was treated with hydrogen sulfide to precipitate the lead as PbS. The precipitate was collected, washed to remove the excess H 2 S, and dissolved in 25.00 mL of 3 M HCl that was also 0.1200 M in I 2. The H 2 S freed from the solution of the lead sulfide was oxidized by the iodine to elemental sulfur. The remaining iodine required 28.41 mL of 0.04136 M Na 2 S 2 O 3 for titration to the starch endpoint. What is the % lead in the paint sample? Chemical Reactions: PbS(s) + 2H + (aq)  H 2 S (aq) + Pb +2 (aq) H 2 S (aq) + I 2 (aq)  2 I  (aq) + S (s) I 2 (aq) + 2S 2 O 3  2 (aq)  2 I  (aq) + S 4 O 6  2 (aq)

10. ______________________________________ Redox Problem 6 6 - The sulfur dioxide in 25.00 Liters of air was collected by drawing the sample through 200.0 mL of 0.02017 M I 2 at a pH of 1.0. The iodine remaining after reaction with the sulfur dioxide was titrated with 42.07 mL of 0.1041 M Na 2 S 2 O 3. Calculate the concentration of the sulfur dioxide in units of mg of SO 2 per Liter of air. Chemical Reactions: SO 2 (g) + I 2 (aq) + 2H 2 O  2 I  (aq) + SO4  2 (aq) + 2H + (aq) I 2 (aq) + 2S 2 O 3  2 (aq)  2 I  (aq) + S 4 O 6  2 (aq)