UU Chemistry 216 chapter 4 Reactions in aqueous solutions

1. UU Chemistry 216 chapter 4 Reactions in aqueous solutions

2. Introduction: Reactions in aqueous solutions • A large amount of important chemistry takes place in water – aqueous solutions • Precipitation reaction- Soluble reactants yield an insoluble product: Pb(NO3)2(aq) + 2KI (aq) -> 2KNO3(aq) + PbI2(s) • Acid – Base neutralization reactions – acid reacts with base to yield water and a salt. HCl(aq) + NaOH(aq)  H2O (l) + NaCl(aq) • Oxidation – reduction reactions – electrons are transferred between reaction partners. Shown by change of charges / oxidation numbers Mg(s) +2HCl(aq)  MgCl2 (aq)+ H2(g)

3. Water solutions • Molecular substances (sugar) in water it contains neutral sucrose molecules surrounded by water. Nonelectrolyte. • Ionic substances (salt) when dissolves in water, the solutions contain separate Na+ and Cl- ions (dissociates) surrounded by water. Electrolyte. If dissociates a large extent (70 – 100%) then strong electrolyte - KCl. Small extent then weak electrolyte – acetic acid (CH3CO2H). Use a double arrow to indicate reaction goes both directions.  Goes until equilibrium is reached.  (molecules and ion formation)

4. Electrolyte classification • Strong electrolytes: HCl, HBr, HI, HClO4, HNO3, N2SO4, KBr, NaCl, NaOH, KOH • Weak: acetic acid, HF • Nonelectrolytes: H2O, methyl alcohol, ethyl alcohol, sucrose, most compound of carbon (organics) • Beaker with solution – wires / battery / bulb

5. Precipitation reactions • We’ve been writing molecular equations- substances written using their complete formulas as if they were molecules. For ionic’s it is more accurate to write precipitation reactions as ionic equations. Pb+2 + 2NO3- + 2K+ +2I- -> 2 K+ + 2 NO3- +PbI2 (s) Spectator ions (NO3- and K+) undergo no change, balance the charge, found on both sides of equation. Net ionic shows only the ions undergoing change

6. Calculations: [concentration] ions in solution • What is the total molar concentration of ions in a 0.350 M solutions of the strong electrolyte Na2SO4?

7. Properties of water • Only substance that exists in large quantities in all three states (solid, liquid, gas) • Less dense as a solid than a liquid • Polar molecule due to electronegativity of oxygen • Hydrogen bonding – Hydrogen of one compound bond with O, F, N of another • Surface tension – force needed to overcome intermolecular attraction, cohesion • Capillary action- adhesion • High specific heat – raise temp. of 1 g, 1*C

8. Calculation: Writing a net ionic equation • Aqueous hydrochloric acid reacts with zinc metal to yield hydrogen gas and aqueous zinc chloride. Write a net ionic equation for the process.

9. Solubility • To predict whether a precipitation reaction will occur you must know the probable solubility of each potential product. • 1. A compound is probably soluble if it contains one of the following cations: Group 1; ammonium • 2. A compound is probably soluble if it contains one of the following anions: Halides – Cl, Br, I except Ag, Hg2,Pb; Nitrate; perchlorate; acetate, sulfate – except Ba,Hg2, Pb, Ca • 3. If not contains one of above, then usually not soluble

10. Calculation: predict products • Will a precipitation reaction occur when aqueous solutions of CdCl2 and (NH4)2S are mixed? If so, write the net ionic equation.

11. Calculations: Using a precipitation reaction to prepare a substance • How might you use a precipitation reaction to prepare a sample of CuCO3? Write net ionic equation.

12. Calculation: Identify precipitation reactions • Predict if the compound is likely to be soluble. a. CdCO3 b. Na2S • Predict whether a precipitation reaction will occur. Write a net ionic equation for it. NiCl2 (aq)+ (NH4)2S(aq) 

13. Acid, base history • Antoine Lavoisier- acids contain a common element: oxygen • Sir Humphrey Davy – muriatic acid (HCl), not oxygen but Hydrogen • Svante Arrhenius – acids dissociate in water to produce hydrogen ions, bases hydroxide ions. • Bronsted – Lowery – acids donate a proton (Hydrogen ion – from Hydronium ion), base is proton acceptor

14. Acids – Bases neutralization: acid properties • Produce hydrogen ions in water, a sour taste, Corrode some metals, Turn blue litmus red, Are electrolytes, Neutralizes bases • Strong acids – dissociates completely into ions • Perchloric acid, sulfuric acid, Hydrobromic acid, Hydrochloric acid, Nitric acid • Weak acid – do not ionize very much • Phosphoric acid, Hydrofluoric acid, acetic acid

15. Acid – Base Neutralization:Base properties • Produce hydroxide ions in water, Taste bitter, Feel slippery, soapy, Turn red litmus blue, Are electrolytes, Neutralizes acids • Strong bases - separates completely into metal ions and hydroxide ions. • Sodium hydroxide, Potassium hydroxide, Barium hydroxide, Calcium hydroxide • Weak base – only a few hydroxide ions • Ammonia

16. Kw and the pH scale • Kw (ion product of water) is a constant formed from the multiplication of the hydronium and hydroxide concentration in pure water • Kw = [H3O+] [OH-] • (1 X 10 -14M) = (1 x 10-7M) (1 x 10-7M) • pH scale = - log [H3O+] : the log of a base 10 is the exponent. The negative turn it to a positive. • Strong acid 0 – 6 weak acid {7 neutral} weak base 8 – 14 strong base

17. Calculation pH

18. Acid – Base titration • Titration is a procedure for determining the concentration of a solution by allowing a carefully measured volume to react with a solution of another substance (standard solution) whose concentration is known. • Measure out a known volume of unknown concentration - HCl. Add indicator . Fill buret with known concentration of NaOH. Add until indicator (phenolphthalein) just begins to turn pink. Read volume of NaOH and calculate molarity of HCl.

19. Calculations: acid – base titrations • NaOH + HCl  NaCl + H2O • If we take 20.0 mL of a HCl solution and find that we have to add 48.6 mL of a 0.1 M NaOH from a buret to obtain complete reaction, what is the molarity of the HCl?

20. Acid – Base neutralization • When an acid and a base are mixed in the right stoichiometric proportions, both acid and basic properties disappear because of a neutralization reaction, producing water and a salt. The anion of salt (A-) comes from the acid, the cation of salt (M+) comes from the base. • HA (aq) + MOH (aq)  H2O (l) + MA (aq) Acid plus base yields water and salt

21. Calculations: writing ionic and net ionic equations • Write both an ionic equation and a net ionic equation for the neutralization reaction of aqueous HBr and aqueous Ba(OH)2.

22. Buffers • Resists a change in pH when small amounts of acid or base are added. • Must contain a weak acid and a salt of that acid OR weak base and a salt of that base. • Example: For blood (7.4 pH) – weak acid (carbonic acid) and salt of that acid (sodium bicarbonate). Add hydroxide (base) to blood and weak acid neutralizes it. Add hydronium ion (acid) to blood, it reacts with bicarbonate ion in blood to form carbonic acid.

23. Oxidation – Reduction (Redox):oxidation numbers • Oxidation is the loss of one or more electrons. • Reduction is the gain of one or more electrons • Oxidize  A-2 – A-1 – A – A+ – A+2  Reduce • The superscripts are now called oxidations number. It’s not always the same value as the ionic charge. • Rules for assigning oxidation numbers:

24. Oxidation rules • 1. An atom in its elemental state has an oxidation number of 0. (Na, H2, Br2, S, Ne) • 2. An atom in a monatomic ion has an oxidation number identical to its charge. (Na+1, Ca+2, Al+3, O -2, Cl-1) • 3. An atom in a polyatomic ion or in a molecular compound usually has the same oxidation number it would have if it were a monatomic ion. (H – O – H, [O-H]-1)

25. Oxidation rules con’t. • A. Hydrogen can be either +1 or -1 (when bonded to a metal like Calcium) • B. Oxygen usually has an oxidation number of -2 (peroxides [O2-2, O-O; oxygen is -1 ) • C. Halogens usually have an oxidation number of -1. (bonded to oxygen, +1 : Cl – O - Cl) • 4. The sum of the oxidation numbers is 0 for a neutral compound and is equal to the net charge for a polyatomic number.

26. Calculations: Assigning oxidation numbers • Assign oxidation numbers to each atom in the following substances. • A) CdS – • B) AlH3- • C) S2O3-2- • D) Na2Cr2O7-

27. Identify redox reactions • 4 Fe(s) + 3 O2 (g)  2 Fe2O3 (s) (g) • Assign oxidations numbers, and if a change (more positive / less negative – oxidation; less positive / more negative – reduction) Both always occur together. • Fe goes from 0 to +3 – loses 3 electron (x 4 by coefficient = 12 electrons total) • Oxygen from O to -2 (x2 for diatomic, x 3 by coefficient = 12 electrons total)

28. Oxidizing and reducing agents • Reducing agents – Cause reduction, loses one or more electrons, undergoes oxidation, oxidation number of atoms increases. In general metals. • Oxidizing agent – causes oxidation, gains one or more electrons, undergoes reduction, oxidation number of atom decreases. In general reactive nonmetals such as oxygen and the halogens.

29. Calculations: identify oxidizing and reducing agents • Assign oxidation numbers to all atoms, tell in each case which substance is undergoing oxidation and which reduction, and identify the oxidizing and reducing agents. • Ca(s) + 2 H+ (aq)  Ca+2 (aq) + H2 (g) • 2 Fe+2 (aq) + Cl2 (aq)  2 Fe+3 (aq) + 2 Cl- (aq)

30. Activity series • Whether or not a reaction occurs between a given ion and a given element depends on the relative ease with which they gain or lose electron. It’s possible to construct an activity series which rank the elements in order of their reducing ability in aqueous solution. Those on top give up electrons readily and are reducing agents, those at bottom give up electrons less. Any element higher in the activity series will reduce the ion of element below

31. A partial activity series of the elements • Strongly reducing: Li, K, Ba, Ca, Na, (these element react rapidly with acids or liquid water to release hydrogen gas) • Mg, Al, Mn, Zn, Cr, Fe, (these react with acids or steam to release hydrogen gas) • Co, Ni, Sn, (These react with acids to release hydrogen gas) • H2 • Rest does not react with acids Cu, Ag, Hg, Pt, Au : Weakly reducing

32. Calculations: predicting the product of redox reactions • Predict whether the following redox reactions will occur. • A) Hg+2 (aq) + Zn (s)  Hg (l) + Zn+2 (aq) • B) 2 H+ (aq) + 2 Ag (s)  H2 (g) + 2 Ag+ (aq)

33. Balancing Redox: oxidation – number method • 1. Write the unbalanced net ionic equation. • 2. Balance the equation for all atoms other than H and O. • 3. Assign oxidation numbers to all atoms. • 4. Decide which atoms have changed oxidation numbers and by how much • 5. Make the total increase in oxidation number for oxidizing atoms equal to the total decrease in oxidation number for reduced.

34. Balancing Redox: oxidation – number method con’t • 6. Balance the equation for O by adding water to the side with less O and then balance for H by adding H+ to the side with less H. • 7. (for bases) Add OH- (to balance H+) to each side of the product. This will “neutralize“ the H+, giving water on product and base in reactant side. • 8. (for bases) Cancel out extra waters that occurs on both sides. This is your final net ionic equation that is balanced for charges.

35. Calculation: Balancing a reaction in a acid and bases • Potassium permanganate (KMnO4) with sodium bromide in aqueous solution. • MnO4- (aq) + Br- (aq)  Mn+2 (aq) + Br2 (aq) • Potassium permanganate reacts with aqueous sodium sulfite (Na2SO3) in basic solution to yield the green manganate and sulfate ion. • MnO4- (aq) + SO3-2 (aq)  MnO4-2 (aq) + SO4-2 (aq)

36. Balancing Redox Reactions: the half- reaction method. • 1. Write the unbalanced net ionic equation • 2. Decide which atoms are oxidized and which are reduced, and write the two unbalanced half-reaction. • 3. Balance both half-reactions for all atoms except O and H. • 4. Balance each half-reactions for O by adding water to the side with less O, and balance for H by adding H+ to the side with less H.

37. Balancing Redox Reactions: the half-reaction method • 5. Balance each half-reaction for charge by adding electrons to the side with greater positive charge, and then multiply be suitable factors to make the electrons count the same in both half – reactions • 6. Add the two balanced half-reactions together, and cancel electrons and other species that appear on both sides of the equation. • If Base, add OH- to both side to neutralize H+

38. Calculation: writing half - reactions • A) Mn+2(aq) + ClO3-(aq)  MnO2(s) + ClO2(aq) • B) Cr2O7-2(q) + Fe+2(aq)  Cr+3(aq) + Fe+3(aq)

39. Calculation: Balancing an equation in an acid • Aqueous potassium dichromate (K2Cr2O7) with aqueous NaCl. • Cr2O7-2 (aq) + Cl- (aq)  Cr+3 (aq) + Cl2 (aq)

40. Calculations: Balancing a reaction in a base • Aqueous sodium hypochlorite (NaOCl) is a strong oxidizer that reacts with chromite ion [Cr(OH)4-] in a basic solution to yield chromate (CrO4-2 ) and chloride ion. • ClO- (aq) + Cr(OH)4- (aq)  CrO4-2 (aq) + Cl- (aq)

41. Redox titration • Very similar to acid / base titration. The substance whose concentration you want to determine undergo an oxidation or reduction reaction is 100% and there be some means (color change) to indicate the reaction is complete. Indicator maybe in the reaction itself or a due to added redox indicator.

42. Calculations: Using redox titration to determine a solution’s concentration • The concentration of an aqueous I3- solution can be determined by titration with aqueous sodium thiosulfate (Na2S2O3) in the presence of a starch indicator (turns from blue to colorless when all I3- has reacted). What is the molar concentration of I3- if 24.55mL of a 0.102M Na2S2O3 is needed for complete reaction with 10.0 mL of I3- solution. • 2S2O3-2 (aq) + I3- (aq)  S4O6-2 (aq) + 3I- (aq)

43. Redox applications • Combustion – burning of carbon hydrogen (fuel) with oxygen in air. • Bleaching – redox to decolorize / lighten materials. Hydrogen peroxide (H2O2) for hair, sodium hypochlorite (NaOCl) for clothes, elemental chlorine for wood pulp • Metallurgy – extracting and purifying metals form ores. Zinc is made from reduction of ZnO with coke (carbon). • ZnO(s) + C(s)  Zn(s) + CO(g)

44. More Redox application • Batteries – Zinc, ammonium chloride paste, MnO2 paste (graphite rod sticks) are connected by wires, sending electrons flowing through the wire toward the MnO2. • Zn(s) + 2 MnO2(s) + 2NH4Cl(s)  ZnCl2(aq) + Mn2O3(s) + 2NH3(aq) + H2O(l) • Corrosion – is the deterioration of a metal by oxidation (rust). ¼ Fe used to replace rusted. • 4 Fe(s) + 3O2(g) –(H2O) 2Fe2O3*H2O(s)