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Chapter 4

Chapter 4. Types of Chemical Reactions and Solution Stoichiometry. Chapter 4: Types of Chemical Reactions and Solution Stoichiometry. 4.1 Water, the Common Solvent 4.2 The Nature of Aqueous Solutions: Strong and Weak Electrolytes 4.3 The Composition of Solutions

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Chapter 4

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  1. Chapter 4 Types of Chemical Reactions and Solution Stoichiometry

  2. Chapter 4: Types of Chemical Reactions and Solution Stoichiometry 4.1 Water, the Common Solvent 4.2 The Nature of Aqueous Solutions: Strong and Weak Electrolytes 4.3 The Composition of Solutions 4.4 Types of Chemical Reactions 4.5 Precipitation Reactions 4.6 Describing Reactions in Solution 4.7 Selective Precipitation 4.8 Stoichiometry of Precipitation Reactions 4.9 Acid-Base Reactions 4.10 Oxidation-Reduction Reactions 4.11 Balancing Oxidation-Reduction Equations 4.12 Simple Oxidation-Reduction Titrations

  3. Precipitation of silver chromate by adding potassium chromate to a solution of silver nitrate. K2CrO4 (aq) + 2 AgNO3 (aq) Ag2CrO4 (s) + 2 KNO3 (aq)

  4. Figure 4.1: A space-filling model of the water molecule.

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

  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.

  7. The Role of Water as a Solvent: The solubility of Ionic Compounds Electrical conductivity - The flow of electrical current in a solution is a measure of the solubility of ionic compounds or a measurement of the presence of ions in solution. Electrolyte - A substance that conducts a current when dissolved in water. Soluble ionic compound dissociate completely and may conduct a large current, and are called Strong Electrolytes. NaCl(s) + H2O(l) Na+(aq) + Cl -(aq) When Sodium Chloride dissolves into water the ions become solvated, and are surrounded by water molecules. These ions are called “aqueous” and are free to move through out the solution, and are conducting electricity, or helping electrons to move through out the solution

  8. Electrical Conductivity of Ionic Solutions

  9. Figure 4.4: Electrical Conductivity

  10. Figure 4.5: HCL (aq) is completely ionized.

  11. Figure 4.6: An aqueous solution of sodium hydroxide.

  12. Figure 4.7: Acetic acid (HC2H3O2) exists in water mostly as undissociated molecules.

  13. Figure 4.8: The reaction of NH3 in water.

  14. Carbohydrates Molecules that contain carbon and water! CxH2yOy H CH2OH O C H CH2OH C O H H HO C H OH C C C OH H C OH O CH2OH C C H H OH Sucrose C12H22O11 , C12(H2O)11 a disaccharide

  15. Molarity (Concentration of Solutions)= M Moles of Solute Moles Liters of Solution L M = = solute = material dissolved into the solvent In air , Nitrogen is the solvent and oxygen, carbon dioxide, etc. are the solutes. In sea water , Water is the solvent, and salt, magnesium chloride, etc. are the solutes. In brass , Copper is the solvent (90%), and Zinc is the solute(10%)

  16. LIKE EXAMPLE 4.1 (P 93) Calculate the Molarity of a solution prepared by bubbling 3.68g of Gaseous ammonia into 75.7 ml of solution. Solution: Calculate the number of moles of ammonia: 1 mol NH3 17.03g 3.68g NH3 X = 0.216 mol NH3 Change the volume of the solution into liters: 1 L 1000 mL 75.7 ml X = 0.0757 L Finally, we divide the number of moles of solute by the volume of the solution: 0.216 mol NH3 0.0757 L Molarity = = ____________ M NH3

  17. Preparing a Solution - I • Prepare a solution of Sodium Phosphate by dissolving 3.95g of Sodium Phosphate into water and diluting it to 300.0 ml or 0.300 l ! • What is the Molarity of the salt and each of the ions? • Na3PO4 (s) + H2O(solvent) = 3 Na+(aq) + PO4-3(aq)

  18. Preparing a Solution - II • Mol wt of Na3PO4 = 163.94 g / mol • 3.95 g / 163.94 g/mol = 0.0241 mol Na3PO4 • dissolve and dilute to 300.0 ml • M = 0.0241 mol Na3PO4 / 0.300 l = 0.0803 M Na3PO4 • for PO4-3 ions = ______________ M • for Na+ ions = 3 x 0.0803 M = ___________ M

  19. Like Example 4.3 (P 95) An isotonic solution, one with the same ionic content as blood is about 0.14 M NaCl. Calculate the volume of blood that would contain 2.5 mg Of NaCl? Find the moles in 1.0 mg NaCl: 1 g NaCl 1000 mg NaCl 1 mol NaCl 58.45g NaCl 2.5 mg NaCl x x = 4.28 x 10-5 mol NaCl What volume of 0.14 M NaCl that would contain the amount of NaCl (4.28 x 10-5 mol NaCl): 0.14 M NaCl L solution V x = 4.28 x 10-5 mol NaCl Solving for Volume gives: 4.28 x 10-5 mol NaCl 0.14 mol NaCl L solution V = = ______________________ L Or _________ ml of Blood!

  20. Figure 4.9: Steps involved in the preparation of a standard solution.

  21. Like Example 4.4 (P 97) A Chemist must prepare a 1.00 L of a 0.375 M solution of Ammonium Carbonate, what mass of (NH4)2CO3 must be weighed out to prepare this solution? First, determine the moles of Ammonium Carbonate required: 0.375 M (NH4)2CO3 L solution 1.00 L x = 0.375 M (NH4)2CO3 This amount can be converted to grams by using the molar mass: 94.07 g (NH4)2CO3 mol (NH4)2CO3 0.375 M (NH4)2CO3 x = 35.276 g (NH4)2CO3 Or, to make 1.00L of solution, one must weigh out 35.3 g of (NH4)2CO3, put this into a 1.00 L volumetric flask, and add water to the mark on the flask.

  22. Make a Solution of Potassium Permanganate Potassium Permanganate is KMnO4 and has a molecular mass of 158.04 g / mole Problem: Prepare a solution by dissolving 1.58 grams of KMnO4 into sufficient water to make 250.00 ml of solution. 1 mole KMnO4 158.04 g KMnO4 1.58 g KMnO4 x = 0.0100 moles KMnO4 0.0100 moles KMnO4 0.250 liters Molarity = = ______________ M Molarity of K+ ion = [K+] ion = [MnO4-] ion = _____________ M

  23. Figure 4.10: (a) A measuring pipette(b) A volumetric pipette.

  24. Figure 4.11: (a) A measuring pipette (b) Water is added to the flask. (c) The resulting solution is 1 M acetic acid.

  25. Dilution of Solutions • Take 25.00 ml of the 0.0400 M KMnO4 • Dilute the 25.00 ml to 1.000 l - What is the resulting Molarity of the diluted solution? • # moles = Vol x M • 0.0250 l x 0.0400 M = 0.00100 Moles • 0.00100 Mol / 1.00 l = _______________ M

  26. Figure 4.13:Reactant solutions: (a) Ba(NO3)3(aq)

  27. Figure 4.13:Reactant solutions: (b) K2CrO4(aq).

  28. Figure 4.12:When yellow aqueous potassium chromate is added to a colorless barium nitrate solution, yellow barium chromate precipitates.

  29. Figure 4.14: Reaction of K2CrO4 (aq) and Ba(NO3)2 (aq).

  30. Figure 4.15:Precipitation of silver chloride by mixing solutions of silver nitrate and potassium chloride.

  31. Figure 4.16:Photos and molecular-level representations illustrating the reaction of KCL(aq) with AgNO3(aq) to form AgCl(s).

  32. Table 4.1 (P102) Simple Rules for Solubility of Salts in Water • Most nitrate (NO3-) salts are soluble. • Most salts of Na+, K+, and NH4+ are soluble. • Most chloride salts are soluble. Notable exceptions are AgCl, • PbCl2, and Hg2Cl2. • Most sulfate salts are soluble. Notable exceptions are BaSO4, • PbSO4, and CaSO4. • Most hydroxide salts are only slightly soluble. The important • soluble hydroxides are NaOH, KOH, and Ca(OH)2 • (marginally soluble). • Most sulfide (S2-), carbonate (CO32-), and phosphate (PO43-) • salts are only slightly soluble.

  33. The Solubility of Ionic Compounds in Water The solubility of Ionic Compounds in water depends upon the relative strengths of the electrostatic forces between ions in the ionic compound and the attractive forces between the ions and water molecules in the solvent. There is a tremendous range in the solubility of ionic compounds in water! The solubility of so called “insoluble” compounds may be several orders of magnitude less than ones that are called “soluble” in water, for example: Solubility of NaCl in water at 20oC = 365 g/L Solubility of MgCl2 in water at 20oC = 542.5 g/L Solubility of AlCl3 in water at 20oC = 699 g/L Solubility of PbCl2 in water at 20oC = 9.9 g/L Solubility of AgCl in water at 20oC = 0.009 g/L Solubility of CuCl in water at 20oC = 0.0062 g/L

  34. The Solubility of Covalent Compounds in Water The covalent compounds that are very soluble in water are the ones with -OH group in them and are called “Polar” and can have strong polar (electrostatic)interactions with water. Examples are compound such as table sugar, sucrose (C12H22O11); beverage alcohol, ethanol (C2H5-OH); and ethylene glycol (C2H6O2) in antifreeze. H Methanol = Methyl Alcohol H C O H H Other covalent compounds that do not contain a polar center, or the -OH group are considered “Non-Polar” , and have little or no interactions with water molecules. Examples are the hydrocarbons in Gasoline and Oil. This leads to the obvious problems in Oil spills, where the oil will not mix with the water and forms a layer on the surface! Octane = C8H18 and / or Benzene = C6H6

  35. When a solution of Na2SO4 (aq) is added to a solution of Pb(NO3)2, the white solid PbSO4(s) forms.

  36. Determining Moles of Ions in Aqueous Solutions of Ionic Compounds - I Problem: How many moles of each ion are in each of the following: a) 4.0 moles of sodium carbonate dissolved in water b) 46.5 g of rubidium fluoride dissolved in water c) 5.14 x 1021 formula units of iron (III) chloride dissolved in water d) 75.0 ml of 0.56M scandium bromide dissolved in water e) 7.8 moles of ammonium sulfate dissolved in water a) Na2CO3 (s) 2 Na+(aq) + CO3-2(aq) moles of Na+ = 4.0 moles Na2CO3 x = 8.0 moles Na+ and 4.0 moles of CO3-2 are present H2O 2 mol Na+ 1 mol Na2CO3

  37. Determining Moles of Ions in Aqueous Solutions of Ionic Compounds - II H2O b) RbF(s) Rb+(aq) + F -(aq) 1 mol RbF 104.47 g RbF moles of RbF = 46.5 g RbF x = 0.445 moles RbF thus, 0.445 mol Rb+and 0.445 mol F -are present H2O c) FeCl3 (s) Fe+3(aq) + 3 Cl -(aq) moles of FeCl3 = 9.32 x 1021 formula units 1 mol FeCl3 6.022 x 1023 formula units FeCl3 x = 0.0155 mol FeCl3 3 mol Cl - 1 mol FeCl3 moles of Cl - = 0.0155 mol FeCl3 x = _________ mol Cl - and ____________ mol Fe+3 are also present.

  38. Determining Moles of Ions in Aqueous Solutions of Ionic Compounds - III H2O d) ScBr3 (s) Sc+3(aq) + 3 Br -(aq) Converting from volume to moles: 1 L 103 ml 0.56 mol ScBr3 1 L Moles of ScBr3 = 75.0 ml x x = 0.042 mol ScBr3 3 mol Br - 1 mol ScBr3 Moles of Br - = 0.042 mol ScBr3 x = 0.126 mol Br - 0.042 mol Sc+3 are also present H2O e) (NH4)2SO4 (s) 2 NH4+(aq) + SO4- 2(aq) 2 mol NH4+ 1 mol(NH4)2SO4 Moles of NH4+ = 7.8 moles (NH4)2SO4 x = ____ mol NH4+ and ______ mol SO4- 2are also present.

  39. Solid Fe(OH)3 forms when aqueous KOH and Fe(NO3)3 are mixed.

  40. Precipitation Reactions: Will a Precipitate form? If we add a solution containing Potassium Chloride to a solution containing Ammonium Nitrate, will we get a precipitate? KCl(aq) + NH4NO3 (aq) = K+(aq) + Cl-(aq) + NH4+(aq) + NO3-(aq) By exchanging cations and anions we see that we could have Potassium Chloride and Ammonium Nitrate, or Potassium Nitrate and Ammonium Chloride. In looking at the solubility table it shows all possible products as soluble, so there is no net reaction! KCl(aq) + NH4NO3 (aq) = No. Reaction! If we mix a solution of Sodium sulfate with a solution of Barium Nitrate, will we get a precipitate? From the solubility table it shows that Barium Sulfate is insoluble, therefore we will get a precipitate! Na2SO4 (aq) + Ba(NO3)2 (aq) BaSO4 (s) + 2 NaNO3 (aq)

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