Intersection 13: Acid/Base Titration & Electrochemistry Introduction

1. Intersection 13:Acid/Base Titration & Electrochemistry Introduction 11/28/06 Reading: 5.4 p 185-188 19.1-19.2 p 909-917

2. Gateway Chemistry 130/125/126 Section 600 6 – 8 PM Thursday Nov. 30th A Gateway to Scientific Ethics • Scientists Employ the Scientific Method to Explore the Physical World. • What are the ethical considerations associated with being a scientist? • How do they relate to the use of the Scientific Method? • Are there ethical considerations beyond the Scientific Method? • We will employ a case-study approach to examining these questions including a prominent case of fraud from Lucent Technologies as well • as the classic Millikan-Ehrenhaft debate. Dinner Provided Please RSVP by email to Prof Banaszak Holl (mbanasza@umich.edu) Gateway evenings are optional and will not affect your course grade

3. Outline • Acid/Base Titration • Electrochemistry • Oxidation numbers

4. M Titration

5. M Titrations: A Closer Look • A titration is a method of determining the concentration of a dissolved substance by addition of a reagent of known concentration until a stoichiometric amount has been added indicated by a known effect (precipitation, color change, change in heat, etc.) • Titrations are popular because: • Absolute content of a sample can be determined. • Speed. • Versatility • Sample size: micrograms up to several grams • Accuracy and reproducibility • Price

6. M Common Uses of Titrations • TAN(Total Acid Number) and TBN(Total Base Number) determinations of motor oil • Various weak base drug assays by titration with perchloric acid in a glacial acetic acid media • Differentiation of nitric and sulfuric acids in nitrating acid used in the explosives industry. • Determination of salt in various foods and snacks. • Determination of chloride in drinking water • Simultaneous determination of chloride, bromide, and iodide in seawater. • Determination of 925 silver in the jewelry industry • Determination of silver in photographic developing solutions.

7. M Titration • Precipitation: Cl-(aq) + AgNO3(aq)→AgCl(s) • What was the purpose? • How did you find the end-point? • Acid-base (Oxiclean) • Redox (coming up…)

8. M Acid/Base Reactions Need to determine the concentration of a solution of HCl. How? 20 mL sample of HCl diluted to 50 mL 1.0 M NaOH used for titration Reaction: How will you know when you have reached the end-point?

9. M Determining End-point?

10. M Finding the Concentration Suppose it takes 15 mL of NaOH, what was the concentration of HCl? 20 mL sample of HCl diluted to 50 mL 1.0 M NaOH used for titration

11. M Problem 1 Strong Acid/Strong Base HCl(aq) + NaOH(aq) H2O(l) + NaCl(aq) Weak Acid/Strong Base CH3COOH (aq) + NaOH (aq) 25 mL 20 mL of 0.1 M What is the concentration CH3COOH?

12. A Question 1 Each of the solutions in the table has equal volume and the same concentration of 0.1 M Which solution requires the greatest volume of 0.1 M NaOH to reach the titration end-point (stoichiometric point)? Explain

13. A What happens when you add less than a stoichiometric amount of a strong base to a weak acid? 2 mols CH3COOH + 1 mol NaOH in 1 L of water? (Ka CH3COOH = 1.8 x10-5)

14. A Question 2 Each of the solutions in the table has equal volume and the same concentration of 0.1 M? Which will have the highest pH at the stoichiometric point?

15. M Problem 2 When 40.00 mL of a weak monoprotic acid solution is titrated with 0.100 M NaOH, the equivalence point is reached when 35.00 mL of base have been added. After 20.00 mL of NaOH solution has been added, the titration mixture has a pH of 5.75. Calculate Ka.

16. M Solution: Initial Concentration HA HA + NaOH  H2O + NaA 40 mL 35 mL 0.10 M 0.1 mol NaOH * (0.035L) * 1 mol HA = 0.0035 mol HA in 0.040 L L 1 mol NaOH = 0.0875 M HA

17. M Solution: Finding Initial Values for ICE Table HA + NaOH -> H2O + NaA 40 mL 20 mL 0.0875 M 0.1 M 0.0035 mol 0.002 mol limiting reagent ? -0.002 mol -0.002 mol + 0.002 mol + 0.002 mol 0.0015 mol 0 0.002 mol 0.002 mol

18. M Solution: ICE table to find Ka 0.025 0.033 1.78 x10-6 pH = 5.75 = -log[H3O+] 1.78x10-6 = [H3O+] = x Ka = [H3O+][A-] = 1.78x10-6*0.033 [HA] 0.025 = 2.35 x10-6

19. A Exam 3 • Tuesday, December 5th 8-10 pm CHEM 1400 • Focus on material through this slide • Please remember your significant figures

20. A Electrochemistry http://genchem.chem.wisc.edu/demonstrations/Gen_Chem_Pages/17electropage/potato_clock.htm

21. A Redox Reactions • Chemical reactions: acid/base or reduction/oxidation ("redox") • Redox reactions involve the transfer of electrons from one reactant to another. • The reaction energy can also be harnessed in the form of electricity.   • The branch of chemistry that deals with the relationship between electricity and chemical reactions is called electrochemistry.

22. A Oxidation States The concept of oxidation state (or oxidation number) was developed to help scientists keep track of the movement of electrons in reactions. 2 Fe(s) + 3 Cl2(g)  2 FeCl3(s)

23. A Oxidation Number “Rules” • For an atom in its elemental form, the oxidation state is always 0. • For any monatomic ion, the oxidation state equals the charge on the ion. 2 Fe(s) + 3 Cl2(g)  2 FeCl3(s)

24. A 3.Specific elements a) The oxidation state of oxygen is almost always -2.  The only major exception is in peroxides (H-O-O-H), where oxygen has an oxidation state of -1. b) The oxidation state of hydrogen is +1 when bonded to non-metals and -1 when bonded to metals. c) The oxidation state of fluorine is always -1.  The other halogens will usually have an oxidation state of -1 in binary compounds.  In some instances the halogens other than fluorine may have different oxidation states.  For example, in oxyanions the halogens have positive oxidation states. H2O HNO3 ZnH2 CF4 NaClO4

25. A • The sum of the oxidation states of all atoms in a neutral compound must equal 0.  The sum of the oxidation states of all atoms in a polyatomic ion equals the charge of the ion. Usually, the most electronegative atom in the molecule will have a negative oxidation state. 2 Fe(s) + 3 Cl2(g)  2 FeCl3(s)