1. Lecture 28 11/09/05

2. At the equivalence point: clarification

3. Pre-reduction SnCl 2 –Sn 2+  Sn 4+ + 2e - –Destroy with HgCl 2 (Sn 2+ gets oxidized to Sn 4+ ) CrCl 2 –Excess Cr 2+ oxidized in air SO 2 –Boil off H 2 S –Boil off

4. Pre-reduction Jones Reductor –Zn coated with Zn amalgam (contains Hg) –Zn(Hg)(s)  Zn 2+ + Hg + 2e - –Can be put into a column –Not selective (Zn powerful reducing agent) –E° = -0.764V Walden reductor –Solid Ag and 1 M HCl –Ag(s) + Cl -  AgCl (s) + e - –More selective since not as powerful reducing agent –E°= 0.222V

5. Titrants: Oxidation with KMnO 4 Very acidic (pH ≤ 1) –MnO 4 - + 8H + + 5e - ⇄ Mn 2+ + 4H 2 O –E° = 1.507 V –Can be its own indicator (stop when you see pink) Neutral / alkaline solutions –MnO 4 - + 4H + + 3e - ⇄ MnO 2 (s) + 2H 2 O –E° = 1.692 V Strongly alkaline –MnO 4 - + e - ⇄ MnO 4 2- –E° = 0.56 V

6. Oxidation with Ce 4+ Ce 4+ + e - ⇄ Ce 3+ –E° = 1.70 V in 1 F HClO 4 –E° = 1.61 V in 1 F HNO 3 –E° = 1.47 V in 1 F HCl –E° = 1.44 V in 1 F H 2 SO4

7. K 2 Cr 2 O 7 Cr 2 O 7 2- (aq) + 14H + (aq) + 6e -  2Cr 3+ (aq) + 7H 2 O E° = 1.36 V –Potential is slightly lower in 1M acid –Potential is very low in basic solutions

8. Question 16-19 Nitrite (NO 2 - ) can be determined by oxidation with excess Ce 4+, followed by back titration of unreacted Ce 4+. A 4.030-g sample of solid containing only NaNO 2 (FM 68.995) and NaNO 3 was dissolved in 500.0-mL. A 25.00-mL sample of this solution was treated with 50.00-mL 0.1186M Ce 4+ in strong acid for 5 minutes and the excess Ce 4+ was back titrated with 31.13 mL of 0.04289 M ferrous ammonium sulfate. 2Ce 4+ + NO 2 - + H 2 O  2Ce 3+ + NO 3 - + 2H + Ce 4+ + Fe 2+  Ce 3+ + Fe 3+ Calculate the weight percentage of NaNO 2 in the solid.

9. Iodine (Iodimetry) Analyte is oxidized –So Iodine is the titrant Titration with I 3 - + 2e -  3I - –I 2 + I -  I 3 - (increases solubility) Starch is used since excess I 2 turns solution dark blue –Add starch at beginning of titration

10. Iodometry 3I -  I 3 - + 2e - Analyte is reduced: (H 2 O 2, O 2, etc) Back-titrate I 3 - –I 3 - (aq) + 2S 2 O 3 2-  3I - (aq) + S 4 O 6 2- –Add starch right before endpoint Color plate 11

11. Winkler method to measure O 2 2Mn 2+ (aq) + 4OH - (aq) + O 2 (aq)  2MnO 2 (s) + 2H 2 O MnO 2 (s) + 3I - (aq) +4H 3 O + (aq)  Mn 2+ (aq) + I 3 - (aq) + 6H 2 O I 3 - (aq) + 2S 2 O 3 2-  3I - (aq) + S 4 O 6 2-

12. Organic Analysis COD

13. 16-25 A 3.026-g portion of a copper(II) salt was dissolved in a 250-mL volumetric flask. A 50.0-mL aliquot was analyzed by adding 1 g of KI and titrating the liberated iodine with 23.33 mL of 0.04668 M Na 2 S 2 O 3. Wt% of copper in salt? Copper = 63.5 g/mol

14. Using the Winkler Method (equations below), find the mg/L of O 2 (FM=32g/mol), if 10 mL of 0.1 M Na 2 S 2 O 3 was needed to titrate a 1 L sample. 2Mn 2+ (aq) + 4OH - (aq) + O 2 (aq)  2MnO 2 (s) + 2H 2 O MnO 2 (s) + 3I - (aq) +4H 3 O + (aq)  Mn 2+ (aq) + I 3 - (aq) + 6H 2 O I 3 - (aq) + 2S 2 O 3 2-  3I - (aq) + S 4 O 6 2- Quiz 12

15. 2Cu 2+ + 5I -  2CuI(s) +I 3 - e - + Cu 2+ + I -  CuI(s) 3I -  I 3 - + 2e -