1. aqueous solution reactions1 Aqueous-solution Reactions Classify a reaction by Homogeneous chemical reactions: gas phase solutions aqueous-solution (common occurrence) non-aqueous-solution Heterogeneous (more than one phase) chemical reactions: gas / liquid gas / solid liquid / solid Know the meaning of terms

2. aqueous solution reactions2 Nature of Aqueous Solutions Nature of compounds molecular substances (polar, non-polar, H-bonding) non-electrolytes ionic substances (acids, bases, salts) strong electrolytes (completely ionized in solution) week electrolytes (not completely ionized in solution) Know your terms and species (in the solution)

3. aqueous solution reactions3 Dissolving a Strong Electrolyte See them in your imagination

4. aqueous solution reactions4

5. 5 Strong Electrolytes Strong acids: HNO 3, H 2 SO 4, HCl, HClO 4 Strong bases: MOH (M = Na, K, Cs, Rb etc) Salts: All salts dissolving in water are completely ionized. The ions may react with water (to be discussed in Chem 123) Stoichiometry & concentration relationship NaCl (s)  Na + (aq) + Cl – (aq) Ca(OH) 2 (s)  Ca + (aq) + 2 OH – (aq) AlCl 3 (s)  Al 3+ (aq) + 3 Cl – (aq) (NH 4 ) 2 SO 4 (s)  2 NH 4 + (aq) + SO 4 2– (aq)

6. aqueous solution reactions6 Concentration of Ions A bottle labeled as 0.100 M Al 2 (SO 4 ) 3. [Al 3+ ] = _____ M (mol / L) [SO 4 2– ] = _____ M Assume sea water is 0.438 M NaCl, 0.0512 M MgCl 2, and 0.001 M CaCl 2 [Na + ] = _____ M [Mg 2+ ] = _____ M [Ca 2+ ] = _____ M [Cl – ] = _____ M Know how to calculate your quantities

7. aqueous solution reactions7 Precipitations When ions form a solid that is not very soluble, a solid is formed. Such a phenomenon is called precipitation. The formation of a precipitation is also an equilibrium phenomenon (a subject to be covered in Chem123) Ag + (aq) + Cl – (aq)  AgCl(s) or AgCl(s)  Ag + (aq) + Cl – (aq) K sp = [Ag + ][Cl – ] is a constant  the solubility product

8. aqueous solution reactions8 Precipitation Reactions Ag + (aq) + NO 3 – (aq) + Cs + (aq) + I – (aq)  AgI (s) + NO 3 – (aq) + Cs + (aq) Ag + (aq) + I – (aq)  AgI (s) (net reaction) or Ag + + I –  AgI (s) Heterogeneous Reactions Spectator ions or bystander ions Soluble ions Alkali metals, NH 4 + nitrates, ClO 4 -, acetate Mostly soluble ions Halides, sulfates Mostly insoluble Silver halides Metal sulfides, hydroxides carbonates, phosphates

9. aqueous solution reactions9 Acid-base Reactions HCl (g)  H + (aq) + Cl – (aq) NaOH (s)  Na + (aq) + OH – (aq) neutralization reaction: H + (aq) + OH – (aq)  H 2 O (l) Explain these reactions Mg(OH) 2 (s) + 2 H +  Mg 2+ (aq) + 2 H 2 O (l) CaCO 3 (s) + 2 H +  Ca 2+ (aq) + H 2 O (l) + CO 2 (g) Mg(OH) 2 (s) + 2 HC 2 H 3 O 2  Mg 2+ (aq) + 2 H 2 O (l) + 2 C 2 H 3 O 2 – (aq) acetic acid

10. aqueous solution reactions10 Oxidation-reduction reactions Oxidation reaction must be accompanied by reductions – redox reactions Increasing oxidation state is oxidation (loss e –, LEO) Decreasing oxidation state is reduction (gain e –, GER) What elements are oxidized and reduced in each reaction? Work out the oxidation state changes for them as well! 2 KClO 3  2 KCl + 3 O 2 Fe 2 O 3 + 3 CO  2 Fe + 3 CO 2 MnO 2 + 4 H + + 2 Cl –  Mn 2+ + 2 H 2 O + Cl 2

11. aqueous solution reactions11 Review Oxidation States Oxidation states is an assigned number. Formal charge concept may be used to assign oxidation states. Work out the oxidation states of all elements in these species: NH 3 N 2 H 4 NH 2 OH N 2 N 2 O NO NO 2 – NO 2 NO 3 – Cl – Cl 2 ClO – ClO 2 – ClO 2 ClO 2 – ClO 3 – ClO 4 – CO H 2 C 2 O 4 C 2 O 4 2 – C 2 H 6 CH 4 CO 2 CO 3 2 – PH 3 P 4 H 3 PO 4 PO 4 3– H 2 SHS – S 2 – S 6 SO 2 SO 3 – SO 3 S 2 O 3 2– SO 4 2– H 2 O 2 MnO 2 KMnO 4 MnO 4 – K 2 CrO 4 CrO 4 – K 2 Cr 2 O 7 Cr 2 O 7 – Review stoichiometry

12. aqueous solution reactions12 Half Reactions These reactions explained during the lecture: Zn = Zn 2+ + 2 e – Cu 2+ + 2 e – = Cu net (electron transfer) Zn (s) + Cu 2+ (aq) = Zn 2+ (aq) + Cu (s) For these half reactions, Zn = Zn 2+ + 2 e – 2 H + + 2 e – = H 2 get and explain the net reaction yourself Explain how reactions proceed

13. aqueous solution reactions13 Balance Half Equations 1.Identify the key element that undergoes an oxidation state change. 2.Balance the number of atoms of the key element on both sides. 3.Add the appropriate number of electrons to compensate for the change of oxidation state. 4.Add H + (in acid medium), or OH - (in basic medium), to balance the charge on both sides of the half-reactions; and H 2 O, if necessary, to balance the equations. Balance MnO 4 – + ____ + _____  Mn 2+ + __ H 2 O Page 9 of handout

14. aqueous solution reactions14 Balance these half equations Zn (s)  Zn 2+ (aq) Cu 2+ (aq)  Cu (s) H 2 (g)  H + (aq) I – (aq)  I 2 Fe 2+ (aq)  Fe 3+ (aq) SO 3 2– (aq)  SO 4 2– (aq) 2 S 2 O 3 2– (aq)  S 4 O 6 2– (aq) + 2e –1 Cr 2 O 7 2– (aq)  Cr 3+ (aq) I will illustrate the balance of these half reaction equations and those equations in other slides during the lecture. If you are not at my lecture, you should practice their balance to acquire your skills. The textbook gives a slight different method to balance redox equations and please find out the difference. Both ways wok. You may use either method. Work on these from time to time

15. aqueous solution reactions15 Balance Redox Reaction equations Add two half reaction equations so that you can cancel all electrons to obtain a balanced redox reaction equation H 2 O 2 + I –  I 2 + H 2 O MnO 4 – + H 2 O 2  Mn 2+ + O 2 MnO 4 – + SO 3 2–  Mn 2+ + SO 4 2– MnO 4 – + Fe 2+  Mn 2+ + Fe 3+ Cr 2 O 7 2– + UO 2+  Cr 3+ + UO 2 2+ Work on these from time to time to refresh your skills.

16. aqueous solution reactions16 Disproportionation Reactions Balance disproportionation reaction (the same substance is both oxidized and reduced) H 2 O 2  2 H 2 O + O 2 S 2 O 3 2–  SO 4 2– + S (s) S 2 O 3 2–  SO 2 + S (s) Work on these from time to time

17. aqueous solution reactions17 Analyze and Learn the Skills Analyze this example and learn the skills to help you overcome any difficulty. Task: Balance the equation: S 2 O 3 2–  SO 4 2– + S (s) Identify the element oxidized and reduced: oxidation states of S in S 2 O 3 2– SO 4 2– and S are +4, +6, and 0 respectively. S is both oxidized and reduced. The oxidation half reaction: S 2 O 3 2–  SO 4 2– + 8 e – (2 S from +2  +6, (2×4=8)) The reduction half reaction: S 2 O 3 2– + 4 e – (2S from +2  0, (2×2=4))  S Balance the charge with H + S 2 O 3 2–  SO 4 2– + 8 e – + 10 H + (both sides have 2-) S 2 O 3 2– + 4 e – + 6 H +  S (both sides have 0) Add water to balance: S 2 O 3 2– + 5 H 2 O  SO 4 2– + 8 e – + 10 H + S 2 O 3 2– + 4 e – + 6 H +  S + 3 H 2 O Make # of e the same: S 2 O 3 2– + 5 H 2 O  SO 4 2– + 8 e – + 10 H + 2 S 2 O 3 2– + 8 e – + 12 H +  S + 6 H 2 O The balanced equation: 3 S 2 O 3 2– + 2 H +  SO 4 2– + 4 S + 10 H + Make sure you fill in the details too much to be included here!!!

18. aqueous solution reactions18 Balance Redox Reaction in Basic Solutions Redox reactions may have different products depends on the acidity (pH) of the solutions. In basic solutions, there are more OH – ions than H + ions. Thus, it is sensible to have OH – appearing in the equations than to have H + ions. Balance these reactions in a basic solution: MnO 4 – + CN –  MnO 2 + OCN – Any one of several ways to assign oxidation states for CN works. MnO 4 – + SO 3 2–  MnO 2 + SO 4 2– Practice balance these from time to time to polish you skills!

19. aqueous solution reactions19 Oxidizing and Reducing Agents Oxidizing agent or oxidant such as O 2 or F 2 is a substance that is able to oxidize other substances. It is reduced in the process (gains electron or decreases oxidation state). Describe reducing agent or reductant NH 3 N 2 H 4 NH 2 OH N 2 N 2 O NO NO 2 – NO 2 NO 3 – Species Cannot be reduced further Species Cannot be oxidized further O x I d a n t R e d u c t a n t

20. aqueous solution reactions20 Titrations Titration is a method used in volumetric analysis. The addition of a solution is carefully controlled so that stoichiometric amounts can be read from a burette. Titration can be carried out for instantaneous or rapid reactions such as neutralization and oxidation reactions. Explain these terms: neutralization, titration, quivalence point, half equivalence point, indicators

21. aqueous solution reactions21 Conductance in Titration conductance V of NaOH added Cl – Na + H+H+ OH – Conductance measurement of a HCl solution titrated by NaOH is shown: measured (total) conductance area due to the ions labeled Do all ions have the same conductance? Why or why not? Why does the total conductance vary? Is conductance of an ion depends on the concentration? Explain physical properties of chemicals

22. aqueous solution reactions22 Stoichiometry in Solution Chemistry For titration calculations, the amount of reactant m is evaluated from the concentration C and volume V by m = C * V For example, when m 1 amount of acid is neutralized by m 2 amount of base, m 1 = m 1. For redox reactions, similar relationship can also be used, but the stoichiometric relationship should be kept in mind. Amount in mmol = C in M * V in mL

23. aqueous solution reactions23 Volumetric Analysis A 5.00-mL sample of vinegar requires 38.08 mL of 0.1000 M NaOH solution to reach the equivalence point. What is the concentration of acetic acid in the vinegar by weight percent if its density is 1.01 g / mL? Solution : Net reaction: OH – + HC 2 H 3 O 2  H 2 O + C 2 H 3 O 2 – (aq) 38.08 mL OH – 5 mL vinegar 0.1000 mol 1000 mL 1 mol HC 2 H 3 O 2 1 mol OH – 60.05 g HC 2 H 3 O 2 1 mol HC 2 H 3 O 2 1 mL vinegar 1.01 g vinegar = 0.0453 HC 2 H 3 O 2 in vinegar = 4.53 % HC 2 H 3 O 2 in vinegar The vinegar has 4.53 % of acetic acid by mass. What’s the concentration of a 7% vinegar? If you find this solution difficult to understand, use your own method to solve it

24. aqueous solution reactions24 38.08 mL OH – 5 mL vinegar 0.1000 mol 1000 mL 1 mol HC 2 H 3 O 2 1 mol OH – 60.05 g HC 2 H 3 O 2 1 mol HC 2 H 3 O 2 1.01 g vinegar 1 mL vinegar = 0.0453 HC 2 H 3 O 2 in vinegar = 4.53 % HC 2 H 3 O 2 in vinegar What the concentration of a 7% vinegar? A 5.00-mL sample of vinegar requires 38.08 mL of 0.1000 M NaOH solution to reach the equivalence point. What is the concentration of acetic acid in the vinegar by weight percent if its density is 1.01 g / mL? Solution : a slight variation and hope you find it easier to follow Net reaction: OH – + HC 2 H 3 O 2  H 2 O + C 2 H 3 O 2 – (aq) Mass of acetic acid Mass of sample 4.53 % ofHC 2 H 3 O 2 in vinegar = Another way of thinking

25. aqueous solution reactions25 Chemical Analysis Application How much 0.1000 M KMnO 4 solution is required to reach the equivalence point in a titration of 1.00 g oxidize oxalic acid (oa = H 2 C 2 O 4.  2H 2 O)? Solution : Redox reaction: 2 MnO 4 – + 5 H 2 C 2 O 4 + 6 H + = 2 Mn 2+ + 8 H 2 O + 10 CO 2 (balanced? Chk pls) 1.00 g oa 1 mol oa 126 g oa 2 mol MnO 4 5 mol oa 1000 mL 0.1000 mol = 31.75 mL MnO 4 – Do problem 5.97 Molar mass of H 2 C 2 O 4.  2H 2 O = 126 g mol -1 Note mole ratio in the balanced equation.

26. aqueous solution reactions26 Back Titration A 1.00-g sample containing MnO 2 dissolved in solution is treated with 2.00 g of oxalic acid (oa = H 2 C 2 O 4.  2H 2 O). Then 50.00 mL 0.1000 M KMnO 4 is required for the titration of the excess oa. What is the % MnO 2 by mass? Solution : Redox reaction: 2 MnO 4 – + 5 H 2 C 2 O 4 + 6 H + = 2 Mn 2+ + 8 H 2 O + 10 CO 2 H 2 C 2 O 4 + MnO 2 + 2 H + = Mn 2+ + 2 H 2 O + 2 CO 2 (balanced? Chk pls) 50.00 mL MnO 4 0.1000 mol MnO 4 1000 mL 5 mol oa 2 mol MnO 4 86.9 g MnO 2 126 g oa 100 % 1.0 g Sample 126 g oa 1 mol oa = 1.575 g oa (2.000 - 1.575) g oa = 29.0 % MnO 2 by mass Cool head 4 complicated problem, eh!

27. aqueous solution reactions27 Review 1 Sufficient amount of AgNO 3 is placed in a 10.00 mL of tab water, and the AgCl solid is filtered and dried. The solid weighs 0.123 g. What is the concentration of chloride ion? Solution: access and limiting reagent, and ppt 1 mol AgCl 143.4 g AgCl 1 mol Cl - 1 mol AgCl 0.123 g AgCl 10.00 mL = 0.0858 mol / L of Cl – 1000 mL 1 L Express [Cl-] in mol and mass %.

28. aqueous solution reactions28 Review 2 A 0.1234-g sample (S) NaCl and sugar mixture contains 40% NaCl. When dissolved in water, the solution is treated with AgNO 3. How much dry AgCl is the theoretical yield? Solution: percentage analysis 40 g NaCl 100 g S 1 mol AgCl 58.5 g NaCl 0.1234 g S = 0.1323 g AgCl  theoretical yield 143.4 g AgCl 1 mol AgCl Use the factors to help you think!

29. aqueous solution reactions29 Review 3 A 0.2345-g sample (S) NaCl and CaCl 2 mixture contains 40% NaCl. When dissolved in water, the solution is treated with AgNO 3. How much dry AgCl is the theoretical yield? Solution: 0.2345*0.40 g NaCl = 0.2299 g AgCl 143.4 g AgCl 58.5 g NaCl Solve the same for a mixture of NaCl and KCl. Find the percentage of NaCl if 0.5000 g AgCl is collected. 0.2345*0.60 g CaCl 2 2*143.4 g AgCl 110 g CaCl 2 = 0.3668 g AgCl (0.2299 + 0.3668) g AgCl = 0.5967 g AgCl

30. aqueous solution reactions30 Review 4 When 0.2345-g NaCl and CaCl 2 mixture is dissolved in water, the solution is treated with AgNO 3. The mass of dry AgCl collected is 0.5967 g. What is the percentage of NaCl in the mixture? Solution : Assume the sample contains x fraction of NaCl 0.2345 ( x ) g NaCl 143.4 g AgCl 58.5 g NaCl Solve the same problem for 0.6000 instead of 0.5967 g AgCl. In this problem, what are the min. and max. masses of AgCl? (0.5748 – 0.6114 g) + 0.2345 (1 – x ) g CaCl 2 2*143.4 g AgCl 110 g CaCl 2 = 0.5967 g AgCl Solve for x in the equation: x = 0.40 = 40%

31. aqueous solution reactions31 Review 5 - skills obtain empirical formula from composition and combustion experiment find molecular formula from empirical formula and molar mass evaluate theoretical yield in a reaction, identify the limiting reagent Identify limiting reagent and evaluation percent yield in a reaction predict amounts of products or reagents required in reactions find percent yield of a reaction but pay attention to limiting reagent give concentration of common ions in a solution of several salts calculate masses of reagents or products (AgCl) involving common ions find percentage of a compound in a mixture by titration or precipitation

32. aqueous solution reactions32 Review 6 – skill 2 Identify oxidation states and recognize reagent oxidized balance redox reaction equations evaluate quantities of reagents or products involving redox reactions stoichiometry calculation in titration experiments paying attention to molar relationship such as H 3 PO 4, Ca(OH) 2, as well as redox reactions evaluate quantities involving isotope composition of elements calculate one of density, mass and volume in solid, liquid, and gas name inorganic compounds write proper net ionic equation of reaction (know your spectator ions)

33. aqueous solution reactions33 TA Hours & Rooms – fall 03 The Tutors handle the tutorial periods for CHEM 120/121 but they can also provide assistance to individuals or small groups on a drop-in basis according to the following schedule. (i.e. One of the Tutors will be in each room for the specified period. You can drop in at any time during that period to get some help.) DayTimeRoom* Monday 12:30-1:20PHY 313 Monday1:30-2:20MC 4062 Monday2:30-3:20 PHY 150 Wednesday12:30-3:20 PHY 313 Wednesday1:30-2:20 MC 4060 * These rooms are for one-on-one tutorial, not tutors office.