1. Analytical chemistry SCT6660E Spring 2016 Instructor: Tom Brenner Room 3-113, Phone no. 03-3238-4691 (office), brenner@sophia.ac.jp, physicalchemistrytom@gmail.com 1

2. Analytical chemistry SCT6660E Spring 2016 1. Introduction, Acids & bases, buffers & several equilibria 2. Several equilibria 3. K sp – the solubility product 4. Quiz, solubility product 5. Chelation equilibria 6. Chelation equilibria 7. Mid-term 8. Debye-Huckel theory 9. Redox I 10. Redox II 11. Redox III 12. Quiz, partitition coefficients & ion-exhcange 13. Preparation for exam 14. Exam 2

3. 3 HW assignments HW assignments will always be posted on http://pweb.cc.sophia.ac.jp/tom_brenner/Analyti cal_chemistry Hand in during class or before class in the A4 box by my office; attendance is not mandatory, but you will be graded on two quizzes and the mid- term! Homework is not compulsory!

4. 4 Evaluation 1.Quizzes 7.5 x 2 = 15% 2.Midterm 25 % 3.Final exam 60% The mid-term and quizzes are only used in the grading if you score higher on them than on the final exam!

5. Start from the scope of this course: qualitative and quantitative chemistry 1.Qualitative: How do we test whether chloride anions are present? (Hopefully students know the answer) 2. Quantitative: A. How do we quantify chloride ions? (see you next semester......) B. How do we quantify dissolved chlorine? Chlorine demand: Cl 2 + 2I - → I 2 + 2Cl -, back titration of I 2 I 2 + 2 S 2 O 3 2- → S 4 O 6 2- + 2 I - How do we do this? 1. Add an EXCESS of KI; Cl 2 + 2I - → I 2 + 2Cl - (actually, under slightly basic conditions, we have H 2 O + Cl 2 ⇋ OCl - + 2H + + Cl - ; we are titrating both OCl - and Cl 2, usually under slightly acidic conditions). Now titrate with K 2 S 2 O 3 until the blue color of the starch indicator is gone! The chlorine demand is the DECREASE in amount of I - needed to react all the Cl 2. Anything that can be oxidized (i.e., can lose electron) can remove chlorine ions from the solutions.

6. The quantification of dissolved chlorine is based on a redox reaction The qualitative identification of chloride anions is based on solubility In this course, we cover: 1.Solubility (The solubility product K sp ) 2.Complex formation in solution 3.Acid-base equilibria revisited - buffering capacity and pH calculations using quadratic and cubic equations 4.Partition coefficients and ion-exchange 5.Redox reactions: reduction potentials, Nernst equation & use in chemistry

7. 2H 2 O H 3 O + + OH - Definition of the pH in H 2 O Water molecules dissociate to give hydronium (H 3 O + ) hydroxide (OH - ) and ions: In neutral water at 25 °C, [H 3 O + ] = [OH - ] = 10 -7 M. The P operator indicates -log 10 : pH = -log 10 ([H 3 O + ]) = 7 in neutral water Addition of acid to water leads to pH < 7, ACIDIC Addition of base to water leads to pH > 7, BASIC/ ALKALINE (autoprotolysis or self-ionization)

8. Importance and applications Some examples: pH of human blood ≈7.35-7.40; pathology indicator Acids and bases are very important in industrial processes: As reactants: Ammonium nitrate production As catalysts: Formation of carbocations in oil refining Control of pH and understanding of acid-base chemistry is important in virtually all scientific research structure design & maintenance, safety…..

9. Lewis acids & bases Do not necessarily comprise a different group! ….A different feature is emphasized: Lewis base: electron-pair donor; Lewis acid: electron- pair acceptor Example:

10. Typical acids & bases Acids: Hydrogen halides (HF, HCl, HBr, HI) Oxyacids (H 2 CO 3, H 3 PO 4, HNO 3, HNO 2 …) (Lewis acids) High-charge (transition) metals (Al 3+, Cr 3+, Fe 3+ ) Bases Alkali and alkaline earth hydroxides (NaOH, KOH, Ca(OH) 2… ) Alkali and alkaline earth oxides (K 2 O, CaO, MgO…) Metal hydrides (LiH, NaH, KH….)

11. The strength of acids and bases is described by their equilibrium constants Acid dissociation constant K a ; pK a = -log(K a )

12. HA + B A - + BH + Acid Base Conjugate base Conjugate acid Reactions always proceed in the direction of the weaker acid and base! pK a + pK b = pK w ; The conjugate base of a strong acid is a very weak base (non-reactive); The conjugate base of a weak acid is a weak base Acid-base conjugate pairs HA/A - and B/BH + are both acid/base conjugates:

13. pH calculation in aqueous solutions Strong acids (pK a < 0) dissociate completely: [H 3 O + ] = C HA, pH = -log(C HA ) Weak acids dissociate negligibly, so [HA] ≈ C HA

14. What happens when C HA is ≈ K a ?

15. Let’s start calculating… Acetic acid, K a = 1.75·10 -5 ; 1.C HA = 0.05 M 2.C HA = 0.005 M 3.C HA = 0.001 M Chloroacetic acid, K a = 1.36·10 -3 ; 1.C HA = 0.05 M 2.C HA = 0.005 M 3.C HA = 0.001 M (what is the K a of trichloroacetic acid??)

16. Can you identify the last approximation???

17. Henderson Hasselbalch: mix acid HA with its conjugate base A -

18. Lastly: amphiprotic compounds

19. H 2 PO4 - + H + ⇋ H 3 PO 4 K b1 = 1.41·10 -12 H 2 PO4 - ⇋ HPO 4 2- + H + K a2 = 6.32·10 -8

20. Henderson-Hasselbalch

22. Nernst equation 22

23. Nernst equation 23 A half cell containing 0.5 M AgCl solution is connected to to a half cell containing 0.25 M RbCl. 1.In which direction do the electrons flow? 2.What is the voltage difference in the cell? Assume T = 298 K

24. Nernst equation 24 A half cell containing 0.5 M AgCl solution is connected to to a half cell containing 0.25 M ZnCl 2. 1.In which direction do the electrons flow? 2.What is the voltage difference in the cell? Assume T = 298 K

25. Alkaline batteries 25 In an alkaline battery, the cathode and anode are contained in a single cell. As a result, water and hydroxide ions are not used up by the reaction(s): Zn 2+ (aq) + H 2 O (l) + 2e ⇋ Zn (s) + 2OH - (aq) E° = -1.28 V 2MnO 2 (s) + H 2 O (l) + 2e ⇋ Mn 2 O 3 (s) + 2OH - (aq) E° = 0.15 V 1. What is the initial concentration of Zn 2+ (aq) if the voltage of the battery is 1.5 V? 2. What is the concentration of Zn2+ when is the battery is dead (E = 0.8 V)? 3. If the mass of the cathode changes by -0.15 g, what is the change in mass of the anode? (M Zn = 65.39 Da, M Mn = 54.94 Da)

26. How do we quantify dissolved chlorine? Chlorine demand: Cl 2 + 2I - → I 2 + 2Cl -, back titration of I 2 I 2 + 2 S 2 O 3 2- → S 4 O 6 2- + 2 I - How do we do this? 1. Add an EXCESS of KI; Cl 2 + 2I - → I 2 + 2Cl - (actually, under slightly basic conditions, we have H 2 O + Cl 2 ⇋ OCl - + 2H + + Cl - ; we are titrating both OCl - and Cl 2, usually under slightly acidic conditions). Now titrate with K 2 S 2 O 3 until the blue color of the starch indicator is gone! The chlorine demand is the DECREASE in amount of I - needed to react all the Cl 2. Anything that can be oxidized (i.e., can lose electron) can remove chlorine ions from the solutions.

27. How do we quantify the „oxygen demand“? The oxygen demand corresponds to the presence of organic material that oxidizes dissolved oxygen (O 2 (aq) ) in water. To quantify the oxygen demand, we measure how much „oxidizable“ (by dichormate) material has disappeared: 14H + + Cr 2 O 7 2- + 6e → 2Cr 3+ + 7H 2 O Or 4H + + O 2 + 4e → 2H 2 O Then again use a back-titration (usually with peroxydisulfate, S 2 O 8 2- + 2e → 2SO 4 2- ) Diphenylamine sulfonic acid (oxidized) ⇋ Diphenylamine sulfonic acid (reduced) (Weak) orange Green ColorlessViolet

28. Example 15.0 mg of Cl 2 are added to a 5.00 L water solution. Excess KI (150.0 mg) is added to the sample. A K 2 S 2 O 3 titrant is prepared by dissolving 68.5 mg in a 250.00 mL volumetric flask. Following are the titration results: What is the chlorine demand after 20 min and after 2 hours? SolutionVolume titrant to end-point 250 mL blank (no Cl 2 added) 31.22 mL 250 mL after 15 min30.80 mL 250 mL after 2 hours30.16 mL

29. Example Excess K 2 Cr 2 O 7 is added to a 500 mL water solution. S 2 O 8 2- titrant (1.33×10 -3 M) is used to titrate the solution to a violet end-point using diphenylamine sulfonic acid as the indicator. If 40.80 mL of the titrant are required to reach the end-point, what is the oxygen demand of the solution?