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Chapter 4: Types of Chemical Reactions and Solution Stoichiometry 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9 4.10 4.11 4.12 Water, the Common Solvent The Nature of Aqueous Solutions: Strong and Weak Electrolytes The Composition of Solutions (MOLARITY!) Types of Chemical Reactions Precipitation Reactions Describing Reactions in Solution Selective Precipitation Midterm Exam 1 Stoichiometry of Precipitation Reactions Acid-Base Reactions Oxidation-Reduction Reactions Balancing Oxidation-Reduction Equations Simple Oxidation-Reduction Titrations 1 Definitions – Solutes, Solvents and Solutions Solutes • The thing being dissolved, mixed, diluted! Solvents • The material doing the dissolving, mixing, dilution! Solution • The final combination of the dissolution, mixing, and dilution! 2 1 Definitions – Solutes, Solvents and Solutions Solutes • Compounds extracted from coffee grounds. Solvents • Water Solution • Morning coffee 3 WATER • water is an important solvent – dissolves many substances • water is a POLAR molecule • “hydration” breaks ionic compounds into anions and cations • water dissolves different ionic compounds to different degrees (more in Ch 8) • water also dissolves some nonionic substances if they are polar (ethanol-water) • nonpolar substances are not dissolved (fats, oils) 4 2 Electronegativity and Polarity 5 Figure 4.1: A space-filling model of the water molecule. More on molecular shapes in Ch 13. 6 3 7 Polar water molecules dissolve salts (ionic compounds) through hydration Anion Zumdahl Figure 4.2 Cation 8 4 Ethanol Molecules are Polar (contain a directional O-H bond) “LIKE DISSOLVING LIKE” Zumdahl Figure 4.3 9 The Role of Water as a Solvent: the Solubility of Ionic Compounds Electrical conductivity The flow of electrical current in a solution is an indicator of the presence of ions in solution and the solubility of ionic compounds. Electrolyte A substance that conducts a current when dissolved in water. Soluble, ionic compounds that dissociate completely conduct a large current and are called strong electrolytes. NaCl(s) H2O Na+(aq) + Cl -(aq) When sodium chloride dissolves in water the ions become solvated/hydrated, and are surrounded by water molecules. These ions are labeled “aqueous”, are free to move throughout the solution, and conduct electricity (they help help electrons move through out the 10 solution). 5 Which of the following solutions of strong electrolytes contains the greatest number of total ions? 1. 2. 3. 4. One mole of potassium chloride dissolved in 1.0 L of solution One mole of iron(II) nitrate dissolved in 1.0 L of solution One mole of potassium hydroxide dissolved in 1.0 L of solution One mole of sodium phosphate dissolved in 1.0 L of solution 11 Electrical Conductivity of Ionic Solutions 12 6 Strong Electrolytes • Produce ions in aqueous solution and conduct electricity well. • Strong electrolytes are soluble salts, strong acids, and strong bases. • Strong acids produce H+ ions when they dissolve in water. HCl, HNO3, and H2SO4 are strong acids: HNO3(aq) → H+(aq) + NO3-(aq) • Strong bases produce OH- ions when they dissolve in water: NaOH and KOH are strong bases: NaOH(s) → Na+(aq) + OH-(aq) All of the above species are ionized nearly 100% 13 Figure 4.5: HCl (aq) is completely ionized. Strong acids fully dissociate, forming the anion and a hydrated proton. 14 7 SIMPLIFIED NOTATION: + H 15 Acids - A group of covalent molecules which lose hydrogen ions to water molecules in solution When gaseous hydrogen iodide dissolves in water, the attraction between the oxygen atom of the water molecule and the hydrogen atom in HI is greater that the attraction of the of the iodide atom for the hydrogen atom, so… H+ is lost to the water molecule to form a hydronium ion and an iodide ion in solution. We can write the hydronium ion in solution as either H+(aq) or H3O+(aq) - they mean the same thing. The water (H2O) can be written as a reactant or above the arrow indicating that it is the solvent in which the HI was dissolved. HI(g) + H2O(l) HI(g) H3O+(aq) + I -(aq) H2O H+(aq) + I -(aq) 16 8 Figure 4.6: An aqueous solution of sodium hydroxide (NaOH). Strong bases fully dissociate, forming a cation and the hydroxide anion. 17 Weak Electrolytes • Produce relatively few ions in aqueous solutions • The most common weak electrolytes are weak acids and weak bases: Acetic acid is a typical weak acid: HC2H3O2(aq) ⇌ H+(aq) + C2H3O2-(aq) Ammonia is a common weak base: NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH-(aq) Both of these species are ionized only ~1% 18 9 Figure 4.7: Acetic acid (HC2H3O2) exists in water mostly as undissociated molecules. Weak acids partially dissociate, forming only a small number of anions and hydrated protons. 19 Figure 4.8: The reaction of NH3 in water. Weak bases partially dissociate (or react with water to a limited extent), forming only a small number of cations and hydroxide anions. 20 10 Nonelectrolytes • Dissolve in water but produce no ions in solution. • Nonelectrolytes do not conduct electricity because when a sample of the compound dissolves, the substance remains intact as whole molecules and no ions are produced. • Common nonelectrolytes include: ethanol (CH3CH2OH) table sugar (sucrose, C12H22O11) 21 The Solubility of Covalent Compounds in Water Polar covalent compounds are very soluble in water. They often have OH groups that can form “hydrogen bonds” with water. Examples are table sugar (C12H22O11), ethanol (C2H5OH), ethylene glycol (C2H6O2) in antifreeze, and methanol (CH3OH). These also are written with “(aq)” (i.e., aqueous) when dissolved in water. Example: C2H5OH(aq) Nonpolar covalent compounds can’t form “hydrogen bonds” and have little or no interactions with water molecules. Examples are the hydrocarbons in gasoline and oil, which don’t mix with water. octane = C8H18 and benzene = C6H6 22 11 Which of the following statements is true? 1. 2. 3. 4. If a substance is soluble, it must be an electrolyte. If a substance is an electrolyte, it must be soluble. Weak electrolytes must be less soluble than strong electrolytes. Nonelectrolytes are nonsoluble. 23 Determining Moles of Ions in Aqueous Solutions of Ionic Compounds - I Problem: How many moles of each ion are formed when the following compounds are dissolved in water: a) b) c) d) 4.0 moles of sodium carbonate 46.5 g of rubidium fluoride 9.32 x 1021 formula units of iron(III) chloride 7.8 moles of ammonium sulfate H2 O a) Na2CO3 (s) 2 Na+(aq) + CO32-(aq) moles of Na+ = 4.0 moles Na2CO3 2 mol Na+ 1 mol Na2CO3 = 8.0 moles Na+ (and 4.0 moles of CO32-) are present 24 12 Determining Moles of Ions in Aqueous Solutions of Ionic Compounds - II b) RbF(s) H2 O moles of RbF = 46.5 g RbF Rb+(aq) + F -(aq) = 0.445 moles RbF 1 mol RbF 104.47 g RbF thus, 0.445 mol Rb+ and 0.445 mol F - are present 25 How many moles of iron and chloride ions are formed by the dissolution of 9.32 x 1021 formula units of iron(III) chloride? 1. Fe+ = 0.0155 mol, Cl- = 0.0155 mol 2. Fe2+ = 0.0155 mol, Cl- = 0.0310 mol 3. Fe3+ = 0.0155 mol, Cl- = 0.0465 mol 4. Fe3+ = 0.155 mol, Cl- = 0.465 mol 26 13 Determining Moles of Ions in Aqueous Solutions of Ionic Compounds - III c) FeCl3 (s) H2 O Fe3+(aq) + 3 Cl -(aq) moles of FeCl3 = 9.32 x 1021 formula units = 0.0155 mol FeCl3 1 mol FeCl3 6.022 x 1023 formula units FeCl3 0.0155 mol Fe3+ are present moles of Cl - = 0.0155 mol FeCl3 3 mol Cl = 0.0465 mol Cl 1 mol FeCl3 27 Determining Moles of Ions in Aqueous Solutions of Ionic Compounds - IV d) 7.8 moles of ammonium sulfate dissolved in water (NH4)2SO4 (s) H2 O Moles of NH4+ = 7.8 moles (NH4)2SO4 2 NH4+(aq) + SO42-(aq) 2 mol NH4+ = 15.6 mol NH4+ 1 mol(NH4)2SO4 7.8 mol SO42- are also present 28 14 Concentration of Solutions: Molarity = M M = moles of solute liters of solution = moles L Units of concentration are mol/L or M solute = material dissolved in the solvent In sea water, WATER is the solvent and salts (magnesium chloride, sodium chloride, etc.) are the solutes When you add sugar to coffee, water is the solvent and the solutes are sugar and all the other compounds extracted from the roasted beans (acids, carbohydrates, proteins, lipids, caffeine, and other organic compounds). 29 Which solution do you expect to light the bulb the most brightly? 1. 2. 3. 4. 5. 6. 7. H2 O HCl (3M) HCl (1M) HC2H3O2 (acetic acid) (1M) C2H5OH (ethanol) NH4OH (1M) HCl (0.1M) 30 15 Determining Moles of Ions in Aqueous Solutions of Ionic Compounds Example problem: How many moles of each ion are in the following: 75.0 mL of 0.56 M scandium bromide in water (ScBr3) ScBr3(s) H2 O Sc3+(aq) + 3 Br -(aq) Converting from volume to moles: Moles of ScBr3 = 75.0 mL 1L 0.56 mol ScBr3 1000 mL L = 0.042 mol ScBr3 Moles of Br- = 0.042 mol ScBr3 x (3 mol Br -/mol ScBr3) = 0.13 mol Br Moles of Sc3+ = 0.042 mol ScBr3 x (1 mol Sc3+/mol ScBr3) = 0.042 mol Sc3+ 31 A sports drink is about 0.14 M NaCl. Calculate the volume (in mL) of the beverage that would contain 2.5 mg of NaCl? MM Na = 22.99 g/mol, MM Cl = 35.45 g/mol 1. 2. 3. 4. 5. 3.1x10-4 mL 0.31 mL 0.31 L 0.14 mL 0.14 mol 32 16 A sports drink is about 0.14 M NaCl. Calculate the volume (in mL) of the beverage that would contain 2.5 mg of NaCl? Convert 2.5 mg NaCl to moles: 2.5 mg NaCl 1 g NaCl 1 mol NaCl 1000 mg NaCl 58.45g NaCl = 4.28 x 10-5 mol NaCl What volume of 0.14 M NaCl would contain this amount of NaCl? V x M = moles Vx 0.14 mol NaCl = 4.28 x 10-5 mol NaCl L solution Solving for volume gives: V = 4.28 x 10-5 mol NaCl L solution = 3.1 x 10-4 L 0.14 mol NaCl or 0.31 mL of beverage 33 Preparing a Solution - I Example problem: A solution of sodium phosphate is prepared by dissolving 3.95 g of sodium phosphate in water and diluting it to 300.0 mL. What is the molarity of the salt and each of the ions? Strategy (1) Write the chemical equation showing the process of dissolution. (2) Calculate the moles of each species. (3) Divide # moles by # L solution to obtain molarity. 34 17 (1) Write the chemical equation showing the process of dissolution. Na3PO4 (s) H2O(solvent) → 3 Na+ (aq) + PO43- (aq) (2) Calculate the moles of each species. Molar mass of Na3PO4 = 163.94 g/mol mol Na3PO4 = 3.95 g / (163.94 g/mol) = 0.0241 mol mol Na+ = 3 x mol Na3PO4 = 0.0723 mol mol PO4-3 = 1 x mol Na3PO4 = 0.0241 mol (3) Divide # moles by # L solution to obtain molarity. M(Na3PO4) = 0.0241 mol Na3PO4 / 0.3000 L = 0.0803 M M(Na+) = 0.0723 mol Na+ / 0.3000 L = 0.241 M M(PO4-3) = 0.0241 mol PO43- / 0.3000 L = 0.0803 M 35 Figure 4.9: Steps involved in the preparation of a standard solution. 36 18 How many grams of ammonium carbonate are required to prepare 1.00 L of a 0.375 M solution? MM N = 14.01 g/mol, MM H = 1.008 g/mol, MM C = 12.01 g/mol, MM 16.00 g/mol 1. 2. 3. 4. 29.3 g 36.0 g 0.375 mol 0.375 g 37 Like Example 4.4 How would you prepare 1.00 L of a 0.375 M solution of ammonium carbonate? First, determine the moles of ammonium carbonate required: 1.00 L 0.375 mol (NH4)2CO3 = 0.375 mol (NH4)2CO3 L solution This amount can be converted to grams by using the molar mass: 0.375 mol (NH4)2CO3 96.09 g (NH4)2CO3 = 36.0 g (NH4)2CO3 mol (NH4)2CO3 To make 1.00 L of solution, 35.3 g of (NH4)2CO3 are weighed out and transferred to a 1.00 L volumetric flask. DI water is added to dissolve the solid and dilute the solution to the mark on the neck of the flask. 38 19 Dilution of Solutions If we dilute 25.00 mL of 0.0400 M KMnO4 to a final volume of 500. mL , what is the new molarity (M) of the diluted solution? A strategy for calculating final concentration: The number of moles of solute is the same before and after dilution! (The difference is the ratio of moles to volume.) M1 x V1 = moles solute = M2 x V2 M1V1 = M2 V2 0.0400 M* 25.00 mL = 2.00 x 10-3 M 500 mL 39 GRADUATED and VOLUMETRIC pipets. Zumdahl Figure 4.10 40 20 Figure 4.11: Dilution of Solutions KNOWN volume is PIPETTED into a larger container of KNOWN volume. M1 (a) A measuring pipette (b) Water is added to the flask. (c) The resulting molarity is M2 = M1 x V1 V2 V1 V2= M2 41 Make a Solution of Potassium Permanganate Potassium Permanganate (KMnO4) has a molar mass of 158.04 g/mol Problem: Prepare a solution by dissolving 1.58 grams of KMnO4 into sufficient water to make 250.00 ml of solution. Calculate the molarity of the ions in solution. 1.58 g KMnO4 1 mole KMnO4 158.04 g KMnO4 Molarity = = 0.0100 moles KMnO4 0.0100 moles KMnO4 = 0.250 liters 0.0400 M Molarity = [K+] ion = [MnO4-] ion = 0.0400 M 42 21 25.00 mL of 0.0400 M potassium permanganate solution is diluted to a final volume of 1.00 L. What is the new concentration? M1V1=M2V2 1. 2. 3. 4. 0.00100 M 0.001 M 10.0 M 10 M 43 Dilution of Solutions Take 25.00 mL of the 0.0400 M KMnO4 Dilute to 1.000 L What is the resulting molarity of the diluted solution? L x M = # moles 0.02500 L x 0.0400 M = 0.00100 moles 0.00100 mol / 1.000 L = 0.00100 M 44 22 MM (g/mol) 45 TYPES of CHEMICAL REACTIONS • Precipitation Reactions – SOLID is formed • Acid-Base (Neutralization) Reactions – Often WATER and a SALT are formed PROTONS are transferred!! • Reduction-Oxidation (REDOX) Reactions – ELECTRONS are transferred 46 23 PRECIPITATION REACTIONS • Double displacement reactions • Insoluble compound (precipitate) is formed from two or more solutes • Precipitate separates from solution • Small fraction of compound remains ionized • Compound formed has a net charge of zero • “Solubility product” is a constant (Ch 8) 47 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 solvent molecules (often water). 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, consider the solubility (in g/L) of the following compounds in water at 20oC : Solubility of NaCl Solubility of MgCl2 Solubility of AlCl3 Solubility of PbCl2 Solubility of AgCl Solubility of CuCl = 365 = 542.5 = 699 = 9.9 = 0.009 = 0.0062 48 24 Figure 4.12: When yellow aqueous potassium chromate is added to a colorless barium nitrate solution, yellow barium chromate precipitates. 49 Figure 4.13: Reactant solutions K2CrO4(aq) Ba(NO3)3(aq) 50 25 Fig. 4.14: Reaction of K2CrO4 (aq) and Ba(NO3)2 (aq) Clues: Precipitate has ZERO net charge. Chromate is yellow. Potassium forms soluble salts. BaCrO4 precipitates! 51 Precipitation Reactions: A solid product is formed Whenever two aqueous solutions are mixed, there is the possibility that an insoluble compound will form. We can predict the result of adding two different solutions together. Initially: Pb(NO3)2 (aq) + NaI(aq) Pb2+(aq) + 2 NO3-(aq) + Na+(aq) + I-(aq) The ions can combine in the way they came into the solution OR they can exchange partners. In this case, the product options are lead(II) nitrate and sodium iodide OR lead(II) iodide and sodium nitrate. To determine which will happen, we must look at the solubility rules. The rules indicate that the lead iodide will be insoluble, so a precipitate will form! Pb(NO3)2 (aq) + 2 NaI(aq) PbI2 (s) + 2 NaNO3 (aq) 52 26 Table 4.1: Simple Rules for Solubility of Salts in Water 1. Most nitrate (NO3-) salts are soluble. 2. Most salts of Na+, K+, and NH4+ are soluble. 3. Most chloride salts are soluble. Notable exceptions are AgCl, PbCl2, and Hg2Cl2. 4. Most sulfate salts are soluble. Notable exceptions are BaSO4, PbSO4, and CaSO4. 5. Most hydroxide salts are only slightly soluble. The important soluble hydroxides are NaOH, KOH, and Ca(OH)2 (marginally soluble). 6. Most sulfide (S2-), carbonate (CO32-), and phosphate (PO43-) salts are only slightly soluble. Know these!!! In general, 1-4 are “solubles” and 5-6 are “insolubles”. 53 Precipitation Reactions: Will a Precipitate form? Split parent species up into ions and look at all potential new combinations that could be formed. Fe(NO3)3(aq) + 3KOH(aq) → Fe3+(aq) + 3NO3- (aq) + 3K+(aq) + 3OH-(aq) RULE: If any of the possible new species formed by combining anions with cations is insoluble, then a precipitate will form. USE TABLE 4.1 In this case, Fe(OH)3 is insoluble and a precipitate forms. 54 27 Precipitation Reactions: Will a Precipitate form? Example: If a solution containing potassium chloride is added to a solution containing ammonium nitrate, will a precipitate form? KCl(aq) + NH4NO3(aq) → K+(aq) + Cl-(aq) + NH4+(aq) + NO3-(aq) Possible reaction products are KCl and NH4NO3, or NH4Cl and KNO3. All are soluble, so there is no precipitate. KCl(aq) + NH4NO3 (aq) = No Reaction! Example: If a solution containing sodium sulfate is added to a solution containing barium nitrate, will a precipitate form? Na2SO4 (aq) + Ba(NO3)2 (aq) → 2 Na+(aq) + SO42- (aq) + Ba2+(aq) + 2 NO3- (aq) Barium sulfate is insoluble; therefore a precipitate will form. Na2SO4 (aq) + Ba(NO3)2 (aq) BaSO4 (s) + 2 NaNO3 (aq) 55 Quantitative Precipitation Problem: Calculate the mass of solid sodium iodide that must be added to 2.50 L of a 0.125 M lead(II) nitrate solution to precipitate ALL of the lead as PbI2 (s). The chemical equation for the reaction is: Pb(NO3)2 (aq) + 2 NaI(aq) PbI2 (s) + 2 NaNO3 (aq) 56 28 What mass (in g) of solid sodium iodide must be added to 2.50 L of a 0.125 M lead(II) nitrate solution to precipitate ALL of the lead as PbI2 (s)? MM Na = 22.99 g/mol, MM I = 126.9 g/mol Pb(NO3)2 (aq) + 2 NaI(aq) 1. 2. 3. 4. PbI2 (s) + 2 NaNO3 (aq) 0.625 g 93.7 g 6.25x10-4 g 9.37 x10-2 g 57 Quantitative Precipitation Problem: Calculate the mass of solid sodium iodide that must be added to 2.50 L of a 0.125 M lead(II) nitrate solution to precipitate ALL of the lead as PbI2 (s). The chemical equation for the reaction is: Pb(NO3)2 (aq) + 2 NaI(aq) PbI2 (s) + 2 NaNO3 (aq) The moles of I- needed to precipitate PbI2 is 2 x moles of Pb2+. The number of moles of iodide needed: 2.50 L 0.125 mol Pb2+ 2 mol I1.00 L soln. 1 mol Pb2+ = 0.625 mol I- The mass of sodium iodide required is: 0.625 mol I- 1 mol NaI 149.9 g NaI = 93.7 g NaI 1 mol I- 1 mol NaI 58 29 Predicting Whether a Precipitation Reaction Occurs and Writing Descriptive Equations a) calcium nitrate and sodium sulfate solutions are added together Molecular Equation Ca(NO3)2 (aq) + Na2SO4 (aq) CaSO4 (s) +2 NaNO3 (aq) Complete Ionic Equation Ca2+(aq)+2 NO3-(aq) + 2 Na+(aq)+ SO4-2(aq) CaSO4 (s) + 2 Na+(aq)+ 2 NO3-(aq) Net Ionic Equation Ca2+(aq) + SO4-2(aq) CaSO4 (s) Spectator ions are Na+ and NO3b) ammonium sulfate and magnesium chloride are added together In exchanging ions, no precipitates will be formed, so no chemical 59 reaction occurs All ions are spectator ions! Limiting reagent problems in solution Only difference: using molarity and volume to convert to moles via M = moles L solution 60 30 Example problem: Lead has been used as a glaze for pottery for years and, if not fired properly, is leachable from the pottery. Vinegar is used in leaching tests, followed by the precipitation of lead as a lead sulfide. If 257.8 mL of a 0.0468 M solution of lead(II) acetate is added to 156.00 mL of a 0.0950 M solution of sodium sulfide, what mass of solid lead sulfide will be formed? This is a limiting-reactant problem because: the amounts of two reactants are given. Strategy: (1) (2) (3) (4) Write the balanced chemical reaction. Determine the limiting reactant. Calculate the moles of product. Convert moles of product to mass of the product using molar mass. 61 (1) Write the balanced equation Pb(C2H3O2)2 (aq) + Na2S (aq) → PbS (s) + 2 NaC2H3O2 (aq) (2) Determine the limiting reactant. mol Pb(C2H3O2)2 = V x M = 0.2578 L x (0.0468 mol/L) = 0.0121 mol mol Na2S = V x M = 0.15600 L x (0.0950 mol/L) = 0.0148 mol 0.0121 mol Pb(C2H3O2)2 1 mol Na2S 1 mol Pb(C2H3O2)2 = 0.0121 mol Na2S needed Therefore, lead(II) acetate is the limiting reactant. 62 31 (3) Calculate the moles of product. 0.0121 mol Pb(C2H3O2)2 x (1 mol PbS / mol Pb(C2H3 O2)2) = 0.0121 mol PbS (4) Calculate mass of product 0.0121 mol PbS x (239.3 g PbS/mol PbS) = 2.90 g PbS 63 Like Example 4.8 When aqueous solutions of silver nitrate and sodium chloride are mixed, silver chloride precipitates. What mass of silver chloride is formed by the addition of 75.00 mL of 3.17 M NaCl to 128 mL of 2.44 M silver nitrate? The stoichiometric relationship comes from the chemical equation: AgNO3 (aq) + NaCl(aq) AgCl(s) + NaNO3 (aq) There is a 1:1 relationship, therefore, the number of moles are the same, but which is in the lowest quantity? VAgNO3 x MAgNO3=0.128 L x 2.44 M =0.312 mol AgNO3 =0.312 mol Ag+ VNaCl x MNaCl = 0.07500 L x 3.17 M = 0.238 mol NaCl =0.238 mol ClSince the amount of chloride ion is smaller, it is limiting, and we use it to calculate the mass of AgCl produced. Since we can only obtain 0.238 mol of AgCl: 0.238 mol x 143.35 g AgCl/ mol = 34.1 g AgCl 64 32 Gravimetric Analysis for % Ca in Phosphate Rock 0.2920 g CaC2O4 .H2O 1 mol CaC2O4 .H2O = 1.998 x 10-3 mol 146.12 g CaC2O4 . H2O CaC2O4 . H2O 1.998 x 10-3 mol Ca2+ 40.08 g Ca2+ = 8.009 x 10-2 g Ca2+ 1 mol Ca2+ Mass % Ca is: 8.009 x 10-2 g Ca x 100% = 18.34% 0.4367 g sample 65 Figure 4.16: Selective precipitation of Ag/Ba/Fe Separating a mixture of silver, barium and iron ions. 66 33 TYPES of CHEMICAL REACTIONS • Precipitation Reactions – SOLID is formed • Acid-Base (Neutralization) Reactions – Often WATER and a SALT are formed PROTONS are transferred!! • Reduction-Oxidation (REDOX) Reactions – ELECTRONS are transferred 67 Acids = Covalent molecules that donate H+ cations (or protons) in solution When gaseous hydrogen iodide dissolves in water, the water molecules help the HI dissociate into H+ and I- ions. We can write the hydrogen ion in water as either H+(aq) or H3O+(aq). They mean the same thing. A molecule which has a H atom that is easily lost to the water solution as H+ (or proton) is called an “acid”, (PROTON DONOR) and the resulting solution is called an “acidic” solution. The acid is said to be “deprotonated” in solution: HI(g) H+(aq) + I -(aq) HI(g) + H2O(l) H3O+(aq) + I -(aq) The water (H2O) could also be written above the arrow indicating that the solvent was water in which the HI was dissolved: H2 O HI(g) H+(aq) + I -(aq) 68 34 H3 O+ is called the hydronium ion 69 What is the molarity of the nitrate and hydronium ions in a solution prepared by dissolving 155 g of concentrated nitric acid in sufficient water to produce 2.30 L of acid solution? Step 1: Write the balanced equation. 1. 2. 3. 4. 1.07 M 2.46 M 1.43 M 3.30 M 70 35 Strong Acids and the Molarity of H+ Ions in Aqueous Solutions of Acids Problem: In aqueous solutions, each molecule of nitric acid will lose one proton to yield one H+ ion and one nitrate ion. What is the molarity of the nitrate and hydronium ions in a solution prepared by dissolving 155 g of concentrated nitric acid in sufficient water to produce 2.30 L of acid solution? Plan: Determine the number of moles of nitric acid; divide the moles by the volume to get the molarity of the acid and each ion. Solution: One mole of H+ is released for every mole of acid: HNO3 (l) + H2O(l) H3O+(aq) + NO3- (aq) Moles HNO3 = 155 g HNO3 1 mol HNO3 = 2.46 moles HNO3 63.02 g HNO3 Molarity of NO3- = 2.46 mol NO3 = 1.07 M in NO32.30 L solution 71 Molarity of H3O+ = 1 x 1.07 M = 1.07 M in H3O+ (or H+) Acids and Bases An acid is a substance that produces H+ (H3O+) ions when dissolved in water, and is a proton donor. A base is a substance that produces OH - ions when dissolved in water: Example: NaOH(aq) → Na+(aq) + OH-(aq) The OH- ions react with the H+ ions (if an acid is present) to produce water, H2O, and are, therefore, proton acceptors. Acids and bases are electrolytes. Their strength is categorized in terms of their degree of dissociation in water to make hydronium or hydroxide ions. Strong acids and bases dissociate completely and are strong electrolytes. Weak acids and bases dissociate partially (some small % of the molecules dissociate) and are weak electrolytes. 72 36 Selected Acids and Bases Acids Bases Strong: H+(aq) + A-(aq) hydrochloric, HCl hydrobromic, HBr hydroiodoic, HI nitric acid, HNO3 sulfuric acid, H2SO4 perchloric acid, HClO4 Strong: M+(aq) + OH-(aq) lithium hydroxide, LiOH sodium hydroxide, NaOH potassium hydroxide, KOH calcium hydroxide, Ca(OH)2 strontium hydroxide, Sr(OH)2 barium hydroxide, Ba(OH)2 *(M is Group I or II metal) Weak hydrofluoric, HF phosphoric acid, H3PO4 acetic acid, CH3COOH (or HC2H3O2) Weak ammonia, NH3 accepts proton from water to make NH4+(aq) and OH-(aq) 73 Acid - Base Reactions: Neutralization Rxns. A generalized reaction between an acid and a base is: HA(aq) + Acid + MOH(aq) Base MA(aq) + → Salt + H2 O(l) Water The salt product can either be dissolved as ions or form a precipitate. Not ALL A/B reactions produce water. Look for the transfer of H+. 74 37 Writing Balanced Equations for Neutralization Reactions - I Problem: Write balanced descriptive equations (molecular, total ionic, and net ionic) for the following chemical reactions: a) calcium hydroxide(aq) and hydroiodic acid(aq) b) lithium hydroxide(aq) and nitric acid(aq) c) barium hydroxide(aq) and sulfuric acid(aq) Plan: These are all strong acids and bases, therefore, the products will be water and the corresponding salts. Solution: a) Ca(OH)2 (aq) + 2HI(aq) CaI2 (aq) + 2H2O(l) Ca2+(aq) + 2 OH -(aq) + 2 H+(aq) + 2 I -(aq) → Ca2+(aq) + 2 I -(aq) + 2 H2O(l) 2 OH -(aq) + 2 H+(aq) 2 H2O(l) 75 Writing Balanced Equations for Neutralization Reactions - II b) LiOH(aq) + HNO3 (aq) LiNO3 (aq) + H2O(l) Li+(aq) + OH -(aq) + H+(aq) + NO3-(aq) → Li+(aq) + NO3-(aq) + H2O(l) OH -(aq) + H+(aq) H2O(l) 76 38 What are the coefficients for the neutralization reaction between barium hydroxide(aq) and sulfuric acid(aq)? MOH(aq) + 1. 2. 3. 4. 5. 6. HA(aq) MA(aq) + H2 O(l) 1,1,1,1 1,1,1,2 1,2,1,2 2,1,1,2 2,2,1,1 2,2,2,4 77 Writing Balanced Equations for Neutralization Reactions - III c) Ba(OH)2 (aq) + H2SO4 (aq) BaSO4 (s) + 2 H2O(l) Ba2+(aq) + 2 OH -(aq) + 2 H+(aq) + SO42-(aq) → BaSO4 (s) + 2 H2O(l) 2 OH -(aq) + 2 H+(aq) 2 H2O(l) 78 39 Like Example 4.10 What volume of 0.125 M HCl is needed to neutralize 200.0 mL of a 0.00955 M Ca(OH)2 solution? Calculate the number of moles of base: Vbase x Mbase = 0.2000 L x 0.00955 M = 0.00191 mol Ca(OH)2 From the balanced equation find the moles of acid needed: Ca(OH)2(aq) + 2 HCl (aq) 2 H2O(l) + CaCl2 (aq) Since there are two hydroxide ions per molecule of base and one proton per molecule of acid, we need twice as much acid as we have base: 0.00382 mol HCl Volume of acid: moles acid 0.00382 mol Vacid = = = 0.0306 L HCl Macid 0.125 mol 79 L TITRATIONS • Volumetric analysis - technique by which one solution is used to analyze another solution • Titrant, also called a standard solution (M known exactly), is delivered from a buret. • The sample and an indicator are in an Erlenmeyer flask. • The indicator changes color at the equivalence point, the point at which titrant and sample are stoichiometrically equal. • The concentration of standard must be known exactly (in order to determine moles, millimoles, or equivalents added). • The concentration of an unknown sample is calculated from the volume and concentration of the standard used to reach the equivalence point, along with the volume of the sample. 80 40 81 Determining the Concentration of Acid by an Acid-base Titration with Known Base Volume (L) of base used to titrate (difference in buret readings) M (mol/L) of base Moles of base used to titrate molar ratio Moles of acid that were titrated volume (L) of acid M (mol/L) of original acid solution 82 41 Finding the Concentration of a Base from an Acid - Base Titration - I Problem: A titration is performed between sodium hydroxide and potassium hydrogen phthalate (KHP) to standardize the base solution. 50.00 mg of solid KHP are placed in a flask with deionized water and a few drops of an indicator. A buret is filled with the base and the initial buret reading is 0.55 mL. At the end of the titration the buret reading is 33.87 mL. What is the concentration of the base? Plan: 1) Use the molar mass of KHP (204.2 g/mol) to calculate the number of moles of the acid. 2) From the balanced chemical equation, the reaction is equal molar (1:1), so we know the moles of base needed. 3) From the difference in the buret readings, we can calculate the volume of base added and, therefore, the molarity of the base. KHC8H4O4 (aq) + OH -(aq) K+(aq) + C8H4O42-(aq) + H2O(l) 83 Potassium Hydrogen Phthalate KHC8H4O4 O O C O K+ C K+ O O C O H O C H+ O ONE molecule of KHP releases ONE proton 84 42 Finding the Concentration of a Base from an Acid - Base Titration - II Solution: moles KHP = 50.00 mg KHP 1.00 g 1000 mg 1 mol KHP 204.2 g KHP = 0.0002449 mol KHP Volume of base = Final buret reading - Initial buret reading = 33.87 mL - 0.55 mL = 33.32 mL of base delivered one mole of acid = one mole of base therefore 0.0002449 moles of acid will NEUTRALIZE 0.0002449 moles of base in a volume of 33.32 mL molarity of base = 0.0002449 moles 0.03332 L = 0.07349 M 85 Example: Aluminum hydroxide reacts with hydrochloric acid according to the balanced equation Al(OH)3 (s) + 3 HCl (aq) → 3 H2O(l) + AlCl3 (aq) What volume of 1.50 M HCl(aq) is required to neutralize 10.0 g Al(OH)3(s)? Strategy: (1) Calculate moles of Al(OH)3(s). (2) Calculate moles of HCl needed using the balanced equation. (3) Calculate volume HCl from moles of HCl and known molarity. 86 43 What volume of 1.50 M HCl(aq) is required to neutralize 10.0 g Al(OH)3(s)? Al(OH)3 (s) + 3 HCl (aq) → 3 H2O(l) + AlCl3 (aq) 1. 2. 3. 4. 0.128 L 0.385 L 0.256 L 0.0853 L 87 (1) Calculate moles of Al(OH)3(s). 10.0 g Al(OH)3 = 0.128 mol Al(OH)3 78.00 g/mol (2) Calculate moles of HCl needed using the balanced equation. Al(OH)3 (s) + 3 HCl (aq) 0.128 mol Al(OH)3 → 3 H O(l) + AlCl 3 mol HCl mol Al(OH)3 2 3 (aq) = 0.385 mol HCl (3) Calculate volume HCl from moles of HCl and known molarity. V= mol HCl molarity = 0.385 mol HCl 1.50 mol/L = 0.256 L 88 44 TYPES of CHEMICAL REACTIONS • Precipitation Reactions – SOLID is formed • Acid-Base (Neutralization) Reactions – Often WATER and a SALT are formed Protons are transferred • Reduction-Oxidation (REDOX) Reactions – ELECTRONS are transferred 89 Formation of NaCl is a REDOX Reaction • 2 Na(s) + Cl2(g) → 2 NaCl(s) • NaCl contains the ions Na+ and Cl• Electrons are transferred • Energy is released 90 45 Oxidation-Reduction Reactions Many important chemical reactions involve oxidation and reduction. In fact, most reactions used for energy production are redox reactions: In humans the oxidation of sugars, fats, and proteins provides the energy necessary for life. Glucose + O2 → Energy Combustion reactions, which provide most of the energy to power our civilization, also involve oxidation and reduction. CxHy + O2 → CO2 + H2O + Energy 91 Metallic Sodium Reacts with Chlorine Gas!!! Zumdahl Fig 4.19 92 46 Oxidation - A substance gives up electrons to another substance. 2 Mg + O2 → 2 Mg2+ + 2 O2- Mg is oxidized Reduction - A substance accepts electrons from another substance. 2 Mg + O2 → 2 Mg2+ + 2 O2- O is reduced In a chemical reaction, the total number of electrons and the number of charges are conserved. It is convenient to assign fictitious charges to the atoms in a molecule and call them “oxidation states” (O.S.) or “oxidation numbers” (O.N.). Oxidation numbers are chosen so that: (a) charges are conserved (b) in ionic compounds the sum of oxidation numbers on the atoms coincides with the charge on the ion 93 See Table 4.3 in Zumdahl 94 47 Period IA H 1 +1 -1 2 3 4 5 6 Possible OXIDATION NUMBERS IIA IIIA IVA Li Be B +1 +2 +3 Na Mg Al Si +1 +2 +3 +4,-4 K Ca Ga +1 +2 Rb Sr +1 +2 Cs +1 VA C N +4,+2 all from -1,-4 +5 -3 Ge +4,+2 +3, +2 -4 In +3,+2 +1 Sn +4,+2, -4 Ba Tl +2 +3,+1 Pb +4,+2 VIA VIIA O F -1,-2 -1 VIIIA He Ne P +5,+3 -3 S -1 Cl +6,+4 +7,+5 +2,-2 +3,+1 Ar As +5,+3 -3 Se -1 Br +6,+4 +7,+5 -2 +3,+1 Kr +2 Sb Te -1 I +5,+3 +6,+4 +7,+5 -3 -2 +3,+1 Bi +3 Po -1 At +6,+4 +7,+5 +2,-2 +3,+1 Xe +6,+4 +2 Rn +2 95 Transition Metals Possible Oxidation States IIIB Sc +3 Y +3 La +3 VIIIB IVB VB VIB VIIB Ti V Cr +2 Mn Fe Co +4,+3 +5,+4 +6,+3 +7,+6 +3,+2 +3,+2 +2 +3+2 +2 +4,+3 Zr Ni +2 IB Cu IIB Zn +2,+1 +2 Nb Mo Tc Ru Rh Pd Ag +5,+4 +6,+5 +7,+5 +8,+5 +4,+3 +4,+3 +4,+2 +1 +2 +4,+3 +4 +4,+3 Cd +2 Hf Ta W Re +2 Os Ir Pt Au Hg +5,+4 +6,+5 +7,+5 +8,+6 +4,+3 +4,+3 +4,+2 +3,+1 +2,+1 +3 +4 +4 +4,+3 +1 96 48 Determine the oxidation number (O.N.) of each element in the following compounds. Ex 1: iron(III) chloride Ex 2: nitrogen dioxide Ex 3: sulfuric acid Strategy: We apply the rules in Table 4.3, always making sure that the O.N. values add up to zero in a compound or, for a polyatomic ion, to the ion’s charge. Solutions: Ex 1: FeCl3 This compound is composed of monoatomic ions. The O.N. of Cl- is -1, for a total of -3. Therefore, the O.N. of Fe is +3. Ex 2: NO2 The O.N. of oxygen is -2 for a total of -4. Since the O.N. 97 in a compound must add up to zero, the O.N. of N is +4. What are the oxidation numbers for H, S, and O in H2SO4 (sulfuric acid)? 1. 2. 3. 4. +1, -2, -2 +1, -2, +2 +1, +6, -2 +1, -6, +2 98 49 Ex 3: H2SO4 The O.N. of H is +1, so the SO42- group must sum to -2. The O.N. of each O is -2 for a total of -8. Therefore S has the O.N. +6. Ex 4: BaI2 I = -1 Ba = +2 Ex 5: NH4NO3 H = +1 N = -3 O = -2 N = +5 99 Mnemonic devices “LEO says GER” “OIL RIG” 100 50 Recognizing Oxidizing and Reducing Agents - I Problem: Identify the oxidizing and reducing agent in each of the rxns a) Zn(s) + 2 HCl(aq) ZnCl2 (aq) + H2 (g) b) S8 (s) + 12 O2 (g) 8 SO3 (g) c) NiO(s) + CO(g) Ni(s) + CO2 (g) Plan: First we assign an oxidation number (O.N.) to each atom (or ion) based on the rules in Table 4.3. The reactant is the reducing agent if it contains an atom that is oxidized (O.N. increased in the reaction). The reactant is the oxidizing agent if it contains an atom that is reduced ( O.N. decreased). Solution: a) Assigning oxidation numbers: -1 +1 0 Zn(s) + 2 HCl(aq) -1 0 +2 ZnCl2 (aq) + H2 (g) HCl is the oxidizing agent, and Zn is the reducing agent! 101 Recognizing Oxidizing and Reducing Agents - II b) Assigning oxidation numbers: 0 +6 0 S8 (s) + 12 O2 (g) -2 8 SO3 (g) S [0] S[+6] S is Oxidized O[0] O[-2] O is Reduced S8 is the reducing agent and O2 is the oxidizing agent c) Assigning oxidation numbers: -2 -2 +2 +2 NiO(s) + CO(g) 0 +4 Ni[+2] Ni[0] Ni is Reduced C[+2] C[+4] C is Oxidized -2 Ni(s) + CO2 (g) CO is the reducing agent and NiO is the oxidizing agent 102 51 Demo: Activity Series Mg(s) + 2 HCl(aq) → Mg2+(aq) + 2 Cl– (aq) + H2(g). Mg is a very active metal (reduces H in both water and acid). Zn(s) + 2 HCl(aq) → Zn2+(aq) + 2 Cl–(aq) + H2(g). Zn is an active metal (reduces H in acid but not in water). Cu(s) + HCl(aq) → no reaction Cu is an inactive metal (does not reduce H in acid nor water). 103 Activity Series Lab 3: You use Mg instead of Zn 104 52 Balancing REDOX Equations: The oxidation states (number) method Step 1) Assign oxidation numbers to all elements in the equation. Step 2) From the changes in oxidation numbers, identify the oxidized and reduced species. Step 3) Compute the number of electrons lost in the oxidation and gained in the reduction from the oxidation number changes. Draw tie-lines between these atoms to show electron changes. Step 4) Choose coefficients for these species to make the electrons lost equal the electrons gained; in other words… total increase in O.N. = total decrease in O.N. Step 5) Complete the balancing by inspection. 105 REDOX Balancing using Ox. No. Method - I 0 +2 e- per O = 4 e- ___ 2 H2 (g) +___ O2 (g) 0 -2 ___ 2 H2O(g) - 1 e- per H +1 4 x (- 1 e- per H) = 4 eelectrons lost must = electrons gained Therefore, we need 4 H = 2 * H2. 106 53 Demo: Thermite Reaction Reaction Equation: __ Fe2O3(s) + __ Al(s) → __ Fe(s) + __ Al2O3(s) ΔHo = –849 kJ/mol ΔSo = –37.48 J/mol-K ΔGo = –838 kJ/mol Uses: welding, purifying an ore (U in the Manhattan Project) In this reaction, unfavorable entropy (ΔS) is offset by a very large negative ΔH (heat is released). The heat is sufficient to raise the temperature of the products past the melting point of Fe (1530 oC). Can go horribly wrong…the heat has to go somewhere and a sand pit is the best. “Mythbusters”: thermite and ice…ice chunks flew 150 feet, or 107 half a football field. What are the coefficients of the balanced reaction of thermite? __ Fe2O3(s) + __ Al(s) → __ Al2O3(s) + __ Fe(s) Step 1: Assign Oxidation States Step 2: Identify oxidized and reduced elements Step 3: Balance - total O.N. increase = total O.N. decrease 1. 2. 3. 4. 1,1,1,1 1,2,1,2 2,1,2,1 2,2,2,2 108 54 Balance the thermite reaction: __ Fe2O3(s) + __ Al(s) → __ Al2O3(s) + __ Fe(s) Assign O.N. to each element: Reactants O: -2 Fe:+3 Al:0 Products O: -2 Al:+3 Fe:0 Identify the elements that are oxidized and reduced: Fe3+ →Fe0 : iron is reduced (O.N. decreases by 3) Al0 →Al3+ : aluminum is oxidized (O.N. increases by 3) Balance atoms and ensure total O.N. increase = total O.N. decrease: Fe2O3 + 2Al → Al2O3 + 2Fe 109 Balancing the Thermite Reaction -2 Fe2O3(s) + 2 Al(s) → +3 -2 Al2O3(s) + 2 Fe(s) 0 → +3 2 x (- 3e- per Al) 0 2 x (+3e- per Fe) BALANCED! 110 55 REDOX Balancing Using Ox. No. Method - II -1e- per Fe +2 +3 +1 -2 Fe2+(aq) + MnO4-(aq) + H+(aq) +7 +1 -2 Fe3+(aq) + Mn2+(aq) + H2O(l) +5 e- per Mn +2 Balance the number of each redox element and then # electrons. Multiply Fe+2 & Fe+3 by five to balance the electrons gained by Mn: 5 Fe2+(aq) + MnO4-(aq) + H+(aq) 5 Fe3+(aq) + Mn2+(aq) + H2O(l) Balance O: Need 4 H2O on right to balance 4 O from the MnO4-. Balance H: Need 8 H+ on the left to balance 8 H in the 4 H2O. 5 Fe2+(aq) + MnO4-(aq) +8 H+(aq) 5 Fe3+(aq) + Mn2+(aq) +4 H2O(l) BALANCED! 111 Balancing REDOX Equations: The half-reaction method Step 1) Write the half-reactions for the chemical equation. Step 2) For each reaction, balance the atoms other than O and H. Step 3) Add H2O to balance O, then H+ to balance H. Step 4) Balance the charge by adding electrons. The net charge of the reactants should equal the net charge of the products. Step 5) Add the two half-reactions together, making sure e- lost equal e- gained, and canceling any species that appear on both sides of the reaction. The reaction is now balanced in an acidic solution. Step 6) If you need to balance in a basic solution, first balance in acidic solution (!), then add OH- to both sides to neutralize any H+ present. Cancel any species that appear on both sides of the reaction. 112 56 Balancing redox eqn using half-cell method in acidic solutions 113 Cu(s) + HNO3(aq) → Cu2+(aq) + NO(g) Identify half-reactions → one is ox., other is red. Cu(s) → Cu2+(aq) HNO3(aq) → NO(g) Balance all atoms that are neither H nor O OK as is Balance O by adding H2O to side deficient in O Cu(s) → Cu2+(aq) HNO3(aq) → NO(g) + 2H2O(l) 114 57 Balance H by adding H+ to side deficient in H Cu(s) → Cu2+(aq) 3H+ +HNO3(aq) → NO(g) + 2H2O(l) Balance charge by adding e- to side that has + charge Cu(s) → Cu2+(aq) + 2e+ 3e- + 3H + HNO3(aq) → NO(g) + 2H2O(l) Multiply each equation by factors so electrons cancel out 3[ Cu(s) → Cu2+(aq) + 2e- ] 2[ 3e- + 3H+ + HNO3(aq) → NO(g) + 2H2O(l) ] Add equations and cancel spectators (none here) 3Cu(s)+ 6H+ (aq) + 2HNO3(aq) → 3Cu2+(aq) + 2NO(g) + 4H2O(l) 115 Balancing redox equations in basic solutions: First balance in acidic solution 116 58 Balancing in Basic Solution 3Cu(s)+ 6H+(aq)+ 2HNO3(aq) → 3Cu2+(aq) + 2NO(g) + 4H2O +6 OH-(aq) +6 OH-(aq) 3Cu(s)+ 6H2O(l)+ 2HNO3(aq) → 3Cu2+(aq) + 2NO(g) + 4H2O +6 OH-(aq) 3Cu(s)+ 2H2O(l)+ 2HNO3(aq) → 3Cu2+(aq) + 2NO(g) + 4H2O +6 OH-(aq) 117 REDOX Balancing by Half-Reaction Method-I Fe2+(aq) + MnO4-(aq) Fe3+(aq) + Mn2+(aq) [acid solution] Identify Oxidation and Reduction Half Reactions Fe2+(aq) Fe3+(aq) + e- MnO4-(aq) Mn2+(aq) what shall we do with oxygen? iron is oxidized manganese is reduced Add H2O to the products to balance O, add H+ to balance the H added, add electrons to balance the charge. 5e- + MnO4-(aq) + 8H+(aq) Mn2+(aq) + 4H2O(l) Sum the two half-reactions { Fe2+(aq) MnO4-(aq) + 8H+(aq) +5eMnO4-(aq) + 8H+(aq) +5e- +5Fe2+(aq) Fe3+(aq) +e- } x5 Mn2+(aq) + 4H2O(l) 5Fe3+(aq) +5e- + Mn2+(aq)+ 4H2O(l) 118 59