1. Analytical chemistry SCT6660E
Spring 2018
Instructor: Tom Brenner
Room 9-756A, Phone no (office),

2. Analytical chemistry SCT6660E
Spring 2018
1. Introduction, Acids & bases, buffers & several equilibria (chapters 9, 15)
2. Several equilibria (chapter 9, 10, 11)
3. Ksp – the solubility product (chapter 10, 11)
4. Quiz, solubility product
5. Chelation equilibria (chapter 17)
6. Chelation equilibria (chapter 17)
7. Mid-term
8. Debye-Huckel theory
9. Redox I (chapter 18, 19)
10. Redox II (chapter 18, 19)
11. Redox III (chapter 18, 19)
12. Quiz, ion-exhcange
13. Preparation for exam
14. Exam

3. HW assignments HW assignments will always be posted on
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. White board

5. Quizzes 7.5 x 2 = 15% Midterm 25 % Final exam 60%
Evaluation
Quizzes 7.5 x 2 = 15%
Midterm 25 %
Final exam 60%
Quizzes 7.5 x 2 = 15%
Final exam 85%
OLD! NEW!
The mid-term and quizzes are only used in the grading if you score higher than on the final exam!

6. Start from the scope of this course: qualitative and quantitative chemistry
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: Cl2 + 2I- → I2 + 2Cl-, back titration of I2
I2 + 2 S2O32- → S4O I-
How do we do this?
1. Add an EXCESS of KI; Cl2 + 2I- → I2 + 2Cl-
(actually, under slightly basic conditions, we have H2O + Cl2 ⇋ OCl- + 2H+ + Cl-; we are titrating both OCl- and Cl2, usually under slightly acidic conditions).
Now titrate with K2S2O3 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 Cl2.
Anything that can be oxidized (i.e., can lose electron) can remove chlorine ions from the solutions.

7. 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:
Solubility (The solubility product Ksp)
Complex formation in solution
Acid-base equilibria revisited - buffering capacity and pH calculations using quadratic and cubic equations
Partition coefficients and ion-exchange
Redox reactions: reduction potentials, Nernst equation & use in chemistry

8. Definition of the pH in H2O
Water molecules dissociate to give hydronium (H3O+) hydroxide (OH-) and ions:
(autoprotolysis or self-ionization)
2H2O H3O+ + OH-
In neutral water at 25 °C, [H3O+] = [OH-] = 10-7 M. The P operator indicates -log10:
pH = -log10([H3O+]) = 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

9. Importance and applications
Some examples:
pH of human blood ≈ ; 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…..

10. 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:

11. Typical acids & bases Acids: Hydrogen halides (HF, HCl, HBr, HI)
Oxyacids (H2CO3, H3PO4, HNO3, HNO2…)
(Lewis acids) High-charge (transition) metals (Al3+ , Cr3+, Fe3+)
Bases
Alkali and alkaline earth hydroxides (NaOH, KOH, Ca(OH)2…)
Alkali and alkaline earth oxides (K2O, CaO, MgO…)
Metal hydrides (LiH, NaH, KH….)

12. Acid-base conjugate pairs
HA/A- and B/BH+ are both acid/base conjugates:
HA B A BH Acid Base Conjugate base Conjugate acid Reactions always proceed in the direction of the weaker acid and base! pKa + pKb = pKw ; 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

13. Acid-base conjugate pairs

14. pH calculation in aqueous solutions
Strong acids (pKa < 0) dissociate completely:
[H3O+] = CHA, pH = -log(CHA)
Weak acids dissociate negligibly, so [HA] ≈ CHA
𝐾 𝑎 = H3O+ [A−] HA
[H3O+] = [A-] = X
X2 = Ka[Ha] ≈ KaCHA
X = [H3O+] = 𝐾 𝑎 𝐶 𝐻𝐴
pH = −log 𝐾 𝑎 𝐶 𝐻𝐴 = 1 2 (𝐾 𝑎 −𝑙𝑜𝑔 (𝐶 𝐻𝐴 ))

15. What happens when CHA is ≈ Ka?
𝐾 𝑎 = H3O+ [A−] HA
[H3O+] = [A-] = X
X2 = Ka[Ha] ≈ KaCHA - XKa  X2 + XKa - KaCHA = 0  𝐾 𝑎 = H3O+ [A−] 𝐶 𝐻𝐴 − 𝐻 +
X = [H3O+] = −Ka± 𝐾 𝑎 𝐾 𝑎 𝐶 𝐻𝐴 2
We may use the approximation X = [H3O+] = 𝐾 𝑎 𝐶 𝐻𝐴 only when CHA >> Ka. For the error to be smaller than 5%, CHA must be at least 110 times bigger than Ka.
Notes
How to identify the validity of an approximation.
What the physically meaningful solution is
How to identify the last inherent assumption: [HA] = CHA - [H3O+]

16. Let’s start calculating…
Acetic acid, Ka = 1.75·10-5;
CHA = 0.05 M
CHA = M
CHA = M
Chloroacetic acid, Ka = 1.36·10-3;
(what is the Ka of trichloroacetic acid??)

17. Can you identify the last approximation???
𝐾 𝑎 = H3O+ [A−] HA
[H3O+] = [A-] = X
𝐾 𝑎 = H3O+ [A−] 𝐶 𝐻𝐴 − 𝐻 +
How to deal with cases when this approximation is not correct?

18. Henderson Hasselbalch: mix acid HA with its conjugate base A-
HA = 𝐶 𝐻𝐴 − 𝐻 + +[OH-]
A− = 𝐶 𝑁𝑎𝐴 + 𝐻 + -[OH-]
𝐻 + ≈ 𝐾 𝑎 𝐶 𝐻𝐴 𝐶 𝑁𝑎𝐴
So, how do we choose buffers??

19. Lastly: amphiprotic compounds

20. Lastly: amphiprotic compounds
H2PO4- + H+ ⇋ H3PO4 Kb1 = 1.41·10-12
H2PO4- ⇋ HPO42- + H+ Ka2 = 6.32·10-8

21. Henderson-Hasselbalch
For a large range of pH (say 2-12) and buffer concentrations not much lower than 0.1 M, we always have:
𝐶 𝐻𝐴 𝐶 𝐴− 𝐾 𝑎 = 𝐻 +

23. Reduction potentials I2 (s)+ 2e  2I- (aq) E° = +0.536 V
I2 (aq)+ 2e  2I- (aq) E° = V
K < Ca < Na < Mg < Al < Fe < Ni < Sn < Pb < (H) < Cu < Hg < Ag < Pt < Au

24. Acid dissociation constant Ka; pKa = -log(Ka)
The strength of acids and bases is described by their equilibrium constants
General acid HA: HA+ H2O H3O+ + A- Equilibrium constant K: 𝐾= 𝑎 H3O+ 𝑎[A−] 𝑎 HA 𝑎[H2O] ≈ H3O+ A− HA H2O …But the concentration of water is almost constant: 𝐾 𝑎 =K H2O = H3O+ [A−] HA
Acid dissociation constant Ka; pKa = -log(Ka)

25. Nernst equation E = 𝑬 𝟎 − 𝑹𝑻 𝒏𝑭 𝒍𝒏𝑸 ∆ 𝐺 0 =−𝑛𝐹 𝐸 0
∆𝐺=−𝑛𝐹𝐸=∆ 𝐺 0 +𝑅𝑇𝑙𝑛𝑄;
𝐸= −∆ 𝐺 0 𝑛𝐹 − 𝑅𝑇𝑙𝑛𝑄 𝑛𝐹 = 𝐸 0 − 𝑅𝑇 𝑛𝐹 𝑙𝑛𝑄
E = 𝑬 𝟎 − 𝑹𝑻 𝒏𝑭 𝒍𝒏𝑸
K > Ca > Na > Mg > Al > Fe > Ni > Sn > Pb > (H) > Cu > Hg > Ag > Pt > Au

26. Nernst equation: Who‘s reacting with whom? Why?
K > Ca > Na > Mg > Al > Zn> Fe > Ni > Sn > Pb > (H) > Cu > Hg > Ag > Pt > Au

27. Nernst equation: Who‘s reacting with whom? Why?
K > Ca > Na > Mg > Al > Zn> Fe > Ni > Sn > Pb > (H) > Cu > Hg > Ag > Pt > Au

28. Nernst equation: Who‘s reacting with whom? Why?
K > Ca > Na > Mg > Al > Zn> Fe > Ni > Sn > Pb > (H) > Cu > Hg > Ag > Pt > Au

29. Nernst equation E = 𝑬 𝟎 − 𝑹𝑻 𝒏𝑭 𝒍𝒏𝑸
A half cell containing 0.5 M AgCl solution is connected to a half cell containing 0.25 M RbCl.
In which direction do the electrons flow?
What is the voltage difference in the cell? Assume T = 298 K

30. Nernst equation E = 𝑬 𝟎 − 𝑹𝑻 𝒏𝑭 𝒍𝒏𝑸
A half cell containing 0.5 M AgCl solution is connected to to a half cell containing 0.25 M ZnCl2.
In which direction do the electrons flow?
What is the voltage difference in the cell? Assume T = 298 K

31. Alkaline batteries
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):
ZnO (s) + H2O(l) + 2e ⇋ Zn(s) + 2OH-(aq) E° = V
2MnO2 (s) + H2O(l) + 2e ⇋ Mn2O3 (s) + 2OH-(aq) E° = 0.15 V
1. What is the reaction quotient if the initial cell potential is 1.5 V?
2. What is the reaction quotient when the battery is dead (0.8 V)?
3. If the mass of the cathode changes by g, what is the change in mass of the anode? (MZn = Da, MMn = Da)

32. How do we quantify dissolved chlorine?
Chlorine demand: Cl2 + 2I- → I2 + 2Cl-, back titration of I2
I2 + 2 S2O32- → S4O I-
(thiosulfate) (tetrathionate)
How do we do this?
1. Add an EXCESS of KI; Cl2 + 2I- → I2 + 2Cl-
(actually, under slightly basic conditions, we have H2O + Cl2 ⇋ OCl- + 2H+ + Cl-; we are titrating both OCl- and Cl2, usually under slightly basic conditions).
Now titrate with K2S2O3 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 Cl2.
Anything that can be oxidized (i.e., can lose electron) can remove aqueous chlorine from the solutions.

33. Violet Colorless How do we quantify the „oxygen demand“?
The oxygen demand corresponds to the presence of organic material that is oxidized by dissolved oxygen (O2 (aq)) in water. To quantify the oxygen demand, we measure how much „oxidizable“ (by dichormate) material has disappeared:
14H+ + Cr2O e- → 2Cr3+ + 7H2O Or
4H+ + O2(aq) + 4e- → 2H2O
Then again use a back-titration (usually with peroxydisulfate, S2O e → 2SO42-)
Indicator:
Diphenylamine sulfonic acid (oxidized) ⇋ Diphenylamine sulfonic acid (reduced) + H+
(Weak) orange
Green
Violet
Colorless

34. What is the chlorine demand after 20 min and after 2 hours?
Example
15.0 mg of Cl2 are added to a 5.00 L water solution. Excess KI (150.0 mg) is added to the sample. A K2S2O3 titrant is prepared by dissolving 68.5 mg in a mL volumetric flask. Following are the titration results:
What is the chlorine demand after 20 min and after 2 hours?
Solution
Volume titrant to end-point
250 mL blank (no sample added)
31.22 mL
250 mL after 15 min
30.80 mL
250 mL after 2 hours
30.16 mL

35. Example
Excess K2Cr2O7 is added to a 500 mL water solution. (Following excessive reflux) S2O82- 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 mL of the titrant are required to reach the end-point, what is the oxygen demand of the solution?
(Easiest route to solving):
L × 1.33 mmol/L = mmol S2O82-
Each S2O82- is going to react with 2e-, which it does by oxidizing the Cr3+ cation. Note we are interested in how much dissolved oxygen, a strong oxidant, could have achieved the same effect as the amount of Cr2O7 that reacted. So we have directly that /2 mmol O2 (aq) had reacted, as each O2 molecule reacts with 4e-. Therefore:
mmol O 2 aq 2∗500 mL solution = mmol O 2 aq L 𝑜𝑟 1.74 𝑚𝑔 O 2 aq L
Is the oxygen demand