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Chapter 4 Type of Chemical Reactions and Solution Stoichiometric

Chapter 4 Type of Chemical Reactions and Solution Stoichiometric. Water, Nature of aqueous solutions, types of electrolytes, dilution. Types of chemical reactions: precipitation, acid-base and oxidation reactions. Stoichiometry of reactions and balancing the chemical equations.

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Chapter 4 Type of Chemical Reactions and Solution Stoichiometric

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  1. Chapter 4Type of Chemical Reactions and Solution Stoichiometric • Water, Nature of aqueous solutions, types of electrolytes, dilution. • Types of chemical reactions: precipitation, acid-base and oxidation reactions. • Stoichiometry of reactions and balancing the chemical equations.

  2. Aqueous Solutions Water is the dissolving medium, or solvent.

  3. Figure 4.1: (Left) The water molecule is polar. (Right) A space-filling model of the water molecule.

  4. Figure 4.2: Polar water molecules interact with the positive and negative ions of a salt assisting in the dissolving process.

  5. Some Properties of Water • Water is “bent” or V-shaped. • The O-H bonds are covalent. • Water is a polar molecule. • Hydration occurs when salts dissolve in water.

  6. Figure 4.3: (a) The ethanol molecule contains a polar O—H bond similar to those in the water molecule. (b) The polar water molecule interacts strongly with the polar O—H bond in ethanol. This is a case of "like dissolving like."

  7. A Solute • dissolves in water (or other “solvent”) • changes phase(if different from the solvent) • is present in lesser amount (if the same phase as the solvent)

  8. A Solvent • retains its phase(if different from the solute) • is present in greater amount (if the same phase as the solute)

  9. General Rule for dissolution • Like dissolve like • Polar dissolve polar (water dissolve ethanol) • Non-polar dissolve nonpolar (benzene dissolve fat)

  10. Figure 4.5: When solid NaCl dissolves, the Na+ and Cl- ions are randomly dispersed in the water.

  11. Electrolytes Strong - conduct current efficiently NaCl, HNO3 Weak- conduct only a small current vinegar, tap water Non - no current flows pure water, sugar solution

  12. Figure 4.4: Electrical conductivity of aqueous solutions.

  13. Acids Strong acids - dissociate completely to produce H+ in solution hydrochloric and sulfuric acid HCl , H2SO4 Weak acids - dissociate to a slight extent to give H+ in solution acetic and formic acid CH3COOH, CH2O

  14. Bases Strong bases- react completely with water to give OH ions. sodium hydroxide Weak bases- react only slightly with water to give OH ions. ammonia

  15. Figure 4.6: HCl(aq) is completely ionized.

  16. Figure 4.7: An aqueous solution of sodium hydroxide.

  17. Figure 4.8: Acetic acid (HC2H3O2) exists in water mostly as undissociated molecules. Only a small percentage of the molecules are ionized.

  18. Molarity Molarity (M) = moles of solute per volume of solution in liters:

  19. Common Terms of Solution Concentration Stock - routinely used solutions prepared in concentrated form. Concentrated - relatively large ratio of solute to solvent. (5.0 M NaCl) Dilute - relatively small ratio of solute to solvent. (0.01 M NaCl): (MV)initial=(MV)Final

  20. Figure 4.10: Steps involved in the preparation of a standard aqueous solution.

  21. Figure 4.12: Dilution Procedure (a) A measuring pipet is used to transfer 28.7mL of 17.4 M acetic acid solution to a volumetric flask. (b) Water is added to the flask to the calibration mark. (c) The resulting solution is 1.00 M acetic acid.

  22. Practice Example How many moles are in 18.2 g of CO2?

  23. Practice Example Consider the reaction N2 + 3H2 = 2NH3 How many moles of H2 are needed to completely react 56 g of N2?

  24. Practice Example How many grams are in 0.0150 mole of caffeine C8H10N4O2

  25. Practice Example A solution containing Ni2+ is prepared by dissolving 1.485 g of pure nickel in nitric acid and diluting to 1.00 L. A 10.00 mL aliquot is then diluted to 500.0 mL. What is the molarity of the final solution?(Atomic weight: Ni = 58.70).

  26. Practice Example Calculate the number of molecules of vitamin A, C20H30O in 1.5 mg of this compound.

  27. Practice Example What is the mass percent of hydrogen in acetic acid HC2H3O2

  28. Types of Solution Reactions • Precipitation reactions AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq) • Acid-base reactions NaOH(aq) + HCl(aq)  NaCl(aq) + H2O(l) • Oxidation-reduction reactions Fe2O3(s) + Al(s)  Fe(l) + Al2O3(s)

  29. Simple Rules for Solubility 1. Most nitrate (NO3) salts are soluble. 2. Most alkali (group 1A) salts and NH4+are soluble. 3. Most Cl, Br, and I salts are soluble(NOTAg+, Pb2+, Hg22+) 4. Most sulfate salts are soluble(NOTBaSO4, PbSO4, HgSO4, CaSO4) 5. Most OH salts are only slightly soluble(NaOH, KOH are soluble, Ba(OH)2, Ca(OH)2 are marginally soluble) 6. Most S2, CO32, CrO42, PO43 salts are only slightly soluble.

  30. Figure 4.13: When yellow aqueous potassium chromate is added to a colorless barium nitrate solution, yellow barium chromate precipitates.

  31. Describing Reactions in SolutionPrecipitation 1. Molecular equation(reactants and products as compounds) AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq) 2.Complete ionic equation(all strong electrolytes shown as ions) Ag+(aq) + NO3- (aq) + Na+ (aq) + Cl(aq) AgCl(s) + Na+ (aq) + NO3- (aq)

  32. Describing Reactions in Solution (continued) 3. Net ionic equation(show only components that actually react) Ag+(aq) + Cl(aq)  AgCl(s) Na+ and NO3 are spectator ions.

  33. Performing Calculations for Acid-Base Reactions 1. List initial species and predict reaction. 2. Write balanced net ionic reaction. 3. Calculate moles of reactants. 4. Determine limiting reactant. 5. Calculate moles of required reactant/product. 6. Convert to grams or volume, as required. Remember: n H+ = n OH- (MV) H+ = (MV) OH-

  34. Neutralization Reaction acid + base salt + water HCl (aq) + NaOH (aq) NaCl (aq) + H2O H+ + Cl- + Na+ + OH-- Na+ + Cl- + H2O H+ + OH- H2O 4.3

  35. Key Titration Terms Titrant - solution of known concentration used in titration Analyte - substance being analyzed Equivalence point - enough titrant added to react exactly with the analyte Endpoint - the indicator changes color so you can tell the equivalence point has been reached. movie

  36. Oxidation-Reduction Reactions 2Mg (s) + O2 (g) 2MgO (s) 2Mg 2Mg2+ + 4e- O2 + 4e- 2O2- 2Mg + O2 + 4e- 2Mg2+ + 2O2- + 4e- 2Mg + O2 2MgO (electron transfer reactions) Oxidation half-reaction (lose e-) Reduction half-reaction (gain e-)

  37. Redox Reactions • Many practical or everyday examples of redox reactions: • Corrosion of iron (rust formation) • Forest fire • Charcoal grill • Natural gas burning • Batteries • Production of Al metal from Al2O3 (alumina) • Metabolic processes combustion

  38. Rules for Assigning Oxidation States 1. Oxidation state of an atom in an element = 0 2. Oxidation state of monatomic element = charge 3. Oxygen = 2 in covalent compounds (except in peroxides where it = 1) 4. H = +1 in covalent compounds 5. Fluorine = 1 in compounds 6. Sum of oxidation states = 0 in compounds Sum of oxidation states = charge of the ion

  39. Cu (s) + 2AgNO3 (aq) Cu(NO3)2 (aq) + 2Ag (s) Zn (s) + CuSO4 (aq) ZnSO4 (aq) + Cu (s) Cu2+ + 2e- Cu Copper wire reacts with silver nitrate to form silver metal. What is the oxidizing agent in the reaction? Cu Cu2+ + 2e- Zn Zn2+ + 2e- Ag+ + 1e- Ag Zn is the reducing agent Zn is oxidized Cu2+ is reduced Cu2+ is the oxidizing agent Ag+is reduced Ag+ is the oxidizing agent

  40. Oxidation numbers of all the elements in the following ? IF7 F = -1 7x(-1) + ? = 0 I = +7 K2Cr2O7 NaIO3 O = -2 K = +1 O = -2 Na = +1 3x(-2) + 1 + ? = 0 7x(-2) + 2x(+1) + 2x(?) = 0 I = +5 Cr = +6

  41. Balancing by Half-Reaction Method 1. Write separate reduction, oxidation reactions. 2. For each half-reaction:  Balance elements (except H, O) Balance O using H2O  Balance H using H+  Balance charge using electrons

  42. Balancing by Half-Reaction Method (continued) 3. If necessary, multiply by integer to equalize electron count. 4. Add half-reactions. 5. Check that elements and charges are balanced.

  43. Half-Reaction Method - Balancing in Base 1. Balance as in acid. 2. Add OH that equals H+ ions (both sides!) 3. Form water by combining H+, OH. 4. Check elements and charges for balance.

  44. Balancing Redox Equations Example: Balance the following redox reaction: Cr2O72- + Fe2+ Cr3+ + Fe3+ (acidic soln) 1) Break into half reactions: Cr2O72- Cr3+ Fe2+ Fe3+

  45. 2) Balance each half reaction: Cr2O72- Cr3+ Cr2O72- 2Cr3+ Cr2O72- 2 Cr3+ + 7 H2O Cr2O72- + 14 H+ 2 Cr3+ + 7 H2O 6 e- + Cr2O72- + 14 H+ 2 Cr3+ + 7 H2O Balancing Redox Equations

  46. Fe2+ Fe3+ Fe2+ Fe3+ + 1 e- Balancing Redox Equations 2) Balance each half reaction (cont)

  47. 6 e- + Cr2O72- + 14 H+ 2 Cr3+ + 7 H2O Fe2+ Fe3+ + 1 e- x 6 Balancing Redox Reactions 3) Multiply by integer so e- lost = e- gained

  48. Balancing Redox Reactions 3) Multiply by integer so e- lost = e- gained 6 e- + Cr2O72- + 14 H+ 2 Cr3+ + 7 H2O 6 Fe2+ 6 Fe3+ + 6 e- 4) Add both half reactions Cr2O72- + 6 Fe2+ + 14 H+ 2 Cr3+ + 6 Fe3+ + 7 H2O

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