Types of Chemical Reactions and Solution Stoichiometry 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 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

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

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

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

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.

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 compounds 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 throughout the solution, and are conducting electricity, or helping electrons to move throughout the solution.

Electrical Conductivity of Ionic Solutions

Figure 4.4: Electrical Conductivity

Figure 4.5: HCL (aq) is completely ionized.

Figure 4.6: An aqueous solution of sodium hydroxide.

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

Figure 4.8: The reaction of NH3 in water.

Carbohydrates Molecules that contain carbon and water! CxH2yOy H CH2OH Sucrose C12H22O11 , C12(H2O)11 a disaccharide

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. In brass , Copper is the solvent (90%), and Zinc is the solute(10%)

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

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)

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

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!

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

Like Example 4.4 (P 97) A Chemist must prepare 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.

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

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

Figure 4. 11: (a) A measuring pipette (b) Water is added to the flask Figure 4.11: (a) A measuring pipette (b) Water is added to the flask. (c) The resulting solution is 1 M acetic acid.

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

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

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

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

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

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

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

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.

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

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

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

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

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.

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- 2 are also present.

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

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)

Precipitation Reactions: A solid product is formed When ever two aqueous solutions are mixed, there is the possibility of forming an insoluble compound. Let us look at some examples to see how we can predict the result of adding two different solutions together. Pb(NO3)2 (aq) + NaI(aq) Pb+2(aq) + 2 NO3-(aq) + Na+(aq) + I-(aq) When we add These two solutions together, the ions can combine in the way they came into the solution, or they can exchange partners. In this case we could have Lead Nitrate and Sodium Iodide, or Lead Iodide and Sodium Nitrate formed, to determine which will happen we must look at the solubility table(P 141) to determine what could form. The table indicates that Lead Iodide will be insoluble, so a precipitate will form! Pb(NO3)2 (aq) + 2 NaI(aq) PbI2 (s) + 2 NaNO3 (aq)

Predicting Whether a Precipitation Reaction Occurs; Writing Equations: a) Calcium Nitrate and Sodium Sulfate solutions are added together. Molecular Equation Ca(NO3)2 (aq) + Na2SO4 (aq) CaSO4 (s) +2 NaNO3 (aq) Total 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 NO3- b) Ammonium Sulfate and Magnesium Chloride are added together. In exchanging ions, no precipitates will be formed, so there will be no Chemical reactions occurring! All ions are spectator ions!

Figure 4.17: Selective precipitation of Ag+, Ba2+, and Fe3+ ions.

Species present, Balanced net ionic equation.

Like Example 4.7 (P 108) Calculate the mass of solid sodium iodide that must be added to 2.50 L of a 0.125 M lead 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) Two times as much sodium iodide is needed to precipitate the Lead ions. The number of moles of sodium iodide needed is: 0.125 Mol Pb2+ 1.0 L soln. 2 mol I- 1 mol Pb2+ 2.50 L x x = 0.625 mol I- The mass of sodium iodide is: 149.9 g NaI 1 mol NaI 1 mol NaI 1 mol I- 0.625 mol I- x x = __________ g NaI

Like Example 4.8 (P 108) When aqueous solutions of silver nitrate and sodium chloride are mixed, silver chloride is precipitated. What mass of silver chloride would be formed by the addition of 75.00 ml to 3.17 M NaCl and 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 one to one 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 Ag+ VNaCl x MNaCl = 0.07500 L x 3.17 M = 0.238 mol Cl- Since the Chloride ion is smaller, it is limiting, and we use it to calculate the mass of AgCl, since we can only obtain 0.238 mol of AgCl: Mass AgCl = 0.238 mol x 143.35 g AgCl/ mol = _________ g

Like Example 4.9 (P 109) What mass of Pb2+ could by precipitated from a solution by the addition of 0.785 L of 0.0015 M Sodium Iodide solution? Find the stoichiometric relationship from the chemical equation: Pb2+(aq) + 2 I-(aq) PbI2 (s) It will take twice the iodide ion to precipitate the Lead ions: Moles I - = VNaI x MNaI = 0.785 L x 0.0015 Moles = 0.00118 mol I- Liter 0.00118 mol I- 2 mol I-/ mol Pb2+ Moles Lead ion = = 0.000590 mol Pb2+ Mass of Lead = 207.2g Pb x 0.000590 moles = ____________ g Pb mol Pb

Acids - A group of Covalent molecules which lose Hydrogen ions to water molecules in solution When gaseous hydrogen Iodide dissolves in water, the attraction of the oxygen atom of the water molecule for the hydrogen atom in HI is greater that the attraction of the of the Iodide ion for the hydrogen atom, and it is lost to the water molecule to form an Hydronium ion and an Iodide ion in solution. We can write the Hydrogen atom in solution as either H+(aq) or as H3O+(aq) they mean the same thing in solution. The presence of a Hydrogen atom that is easily lost in solution is an “Acid” and is called an “acidic” solution. The water (H2O) could also be written above the arrow indicating that the solvent was water in which the HI was dissolved. HI(g) + H2O(L) H+(aq) + I -(aq) HI(g) + H2O(L) H3O+(aq) + I -(aq) H2O HI(g) H+(aq) + I -(aq)

Strong Acids and the Molarity of H+ Ions in Aqueous Solutions of Acids Problem: In aqueous solutions, each molecule of sulfuric acid will lose two protons to yield two Hydronium ions, and one sulfate ion. What is the molarity of the sulfate and Hydronium ions in a solution prepared by dissolving 155g of concentrate sulfuric acid into sufficient water to produce 2.30 Liters of acid solution? Plan: Determine the number of moles of sulfuric acid, divide the moles by the volume to get the molarity of the acid and the sulfate ion. The hydronium ions concentration will be twice the acid molarity. Solution: Two moles of H+ are released for every mole of acid: H2SO4 (l) + 2 H2O(l) 2 H3O+(aq) + SO4- 2(aq) 1 mole H2SO4 98.09 g H2SO4 Moles H2SO4 = = 1.58 moles H2SO4 155 g H2SO4 x 1.58 mol SO4-2 2.30 l solution Molarity of SO4- 2 = = 0.687 Molar in SO4- 2 Molarity of H+ = 2 x 0.687 mol H+ = 1.37 Molar in H+

The gravimetric procedure.(P109-110) 1 mol CaC2O4 H2O 146.12g CaC2O4 H2O . 0.2920 g CaC2O4 H2O X = 1.998 x 10-3 mol CaC2O4 H2O 1.998 x 10-3 mol Ca2+ X = 8.009 x 10-2g Ca2+ Mass % Ca is: x 100% = ______________% . . 40.08 g Ca2+ 1 mol Ca2+ 8.009 x 10-2 0.4367 g

Species present, Balanced net ionic equation.

Acid - Base Reactions : Neutralization Rxns. 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. the OH- ions react with the H+ ions to produce water, H2O, and are therefore proton acceptors. Acids and Bases are electrolytes, and 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 weakly and are weak electrolytes. The generalized reaction between an Acid and a Base is: HX(aq) + MOH(aq) MX(aq) + H2O(L) Acid + Base = Salt + Water

Selected Acids and Bases Acids Bases Strong Strong Hydrochloric, HCl Sodium hydroxide, NaOH Hydrobromic, HBr Potassium hydroxide, KOH Hydroiodoic, HI Calcium hydroxide, Ca(OH)2 Nitric acid, HNO3 Strontium hydroxide, Sr(OH)2 Sulfuric acid, H2SO4 Barium hydroxide, Ba(OH)2 Perchloric acid, HClO4 Weak Weak Hydrofluoric, HF Ammonia, NH3 Phosphoric acid, H3PO4 Acetic acid, CH3COOH (or HC2H3O2)

Writing Balanced Equations for Neutralization Reactions - I Problem: Write balanced chemical reactions (molecular, total ionic, and net ionic) for the following Chemical reactions: a) Calcium Hydroxide(aq) and Hydroiodoic 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 they will make 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)

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) 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) Ba2+(aq) + 2 OH -(aq) + 2 H+(aq) + SO42-(aq) BaSO4 (s) + 2 H2O(l)

Figure 4.18: The titration of an acid with a base.

Like Example 4.10 (P 113) What volume of 0.468 M H2SO4 is needed to neutralize 215.00 ml of a 0.125 M LiOH solution? Calculate the number of moles of base: Vbase x Mbase = 0.21500 L x 0.125 M = 0.0269 mol LiOH From the balanced equation find the moles of acid needed: 2 LiOH(aq) + H2SO4 (aq) 2 H2O(l) + Li2SO4 (aq) Since there are two protons per molecule, we will need half as much sulfuric acid as we have lithium hydroxide: or 0.0134 mol H2SO4 Volume of acid: Moles acid Macid 0.0134 moles 0.468 Mol L Vacid = = = 0.0286 L H2SO4

A firefighter in a protective suit with an oxygen tank neutralizes an acid spill.

Finding the Concentration of Base from an Acid - Base Titration - I Problem: A titration is performed between Sodium Hydroxide and Potassium Hydrogenphthalate (KHP) to standardize the base solution, by placing 50.00 mg of solid Potassium Hydrogenphthalate in a flask with 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: Use the molar mass of KHP (204.2 g/mol) to calculate the number of moles of the acid, from the balanced chemical equation, the reaction is equal molar, so we know the moles of base, and from the difference in the buret readings, we can calculate the molarity of the base. Solution: HKC8H4O4 (aq) + OH -(aq) KC8H4O4-(aq) + H2O(aq)

Potassium Hydrogenphthalate KHC8H4O4

Finding the Concentration of Base from an Acid - Base Titration - II 50.00 mg KHP 204.2 g KHP 1 mol KHP 1.00 g 1000 mg moles KHP = x = 0.00024486 mol KHP Volume of base = Final buret reading - Initial buret reading = 33.87 ml - 0.55 ml = 33.32 ml of base one mole of acid = one mole of base; therefore 0.00024486 moles of acid will yield 0.00024486 moles of base in a volume of 33.32 ml. 0.00024486 moles 0.03332 L molarity of base = = __________ moles per liter molarity of base = ___________________ M

Like Example 4.12 (P114-115) A powered residue contains some ascorbic acid(Vitamin C, mol wt = 176g/mol) and the rest is a non acidic compound. If 10.0g of the powder is neutralized by 20.00 ml of 1.5 M sodium hydroxide, a strong base, and the remaining base titrated with hydrochloric acid using 10.50 ml of 1.80 M. What is the percentage of ascorbic acid? Mol acid = 0.01050 L x 1.80 Mol/L = 0.0189 mol HCl Mol base = 0.0200 L x 1.50 Mol/L = 0.0300 mol NaOH The difference between the base and acid will be the moles of ascorbic acid! Reacted base = 0.0300 – 0.0189 = 0.0111 mol Ascorbic acid Mass ascorbic acid = 0.0111 mol x 176g/mol = 1.95 g Ascorbic acid 1.95g 10.00g % ascorbic acid = x 100% = ____________%

Figure 4.19: Reaction of solid sodium and gaseous chlorine to form solid sodium chloride.

Oxidation of copper metal by nitric acid.

Highest and Lowest oxidation numbers of Chemically reactive main-group Elements 1A 2A 3A -4 4A 5A 6A 7A 1 +1 -1 H Period +4 +4 +5 +6 +7 +1 +2 +3 -3 -2 -1 2 Li Be B C N O F 3 Na Mg Al Si P S Cl 4 K Ca Ga Ge As Se Br non-metals 5 Rb Sr In Sn Sb Te I metalloids 6 Cs Ba Tl Pb Bi Po At metals 7 Fr Ra

Main Group Elements Period IA VIIIA H He 1 +1 -1 IIA IIIA IVA VA VIA +1 -1 IIA IIIA IVA VA VIA VIIA Li Be B C N O F Ne 2 +4,+2 -1,-4 all from +1 +2 +3 -1,-2 -1 +5 -3 Na Mg Al Si P S -1 Cl Ar 3 +5,+3 -3 +6,+4 +2,-2 +7,+5 +3,+1 +1 +2 +3 +4,-4 Kr K Ca Ga Ge As Se -1 Br 4 +4,+2 -4 +5,+3 -3 +6,+4 -2 +7,+5 +3,+1 +2 +1 +2 +3, +2 Xe Rb Sr In Sn Sb Te -1 I 5 +3,+2 +1 +4,+2, -4 +5,+3 -3 +6,+4 -2 +7,+5 +3,+1 +6,+4 +2 +1 +2 Rn Cs Ba Tl Pb Bi Po -1 At 6 +6,+4 +2,-2 +7,+5 +3,+1 +1 +4,+2 +2 +2 +3,+1 +3

Transition Metals Possible Oxidation States VIIIB IIIB IVB VB VIB VIIB Sc Ti V Cr +2 Mn Fe Co Ni Cu Zn +4,+3 +2 +5,+4 +3+2 +6,+3 +2 +7,+6 +4,+3 +3 +3,+2 +3,+2 +2 +2,+1 +2 Y Zr Nb Mo Tc Ru Rh Pd Ag Cd +5,+4 +2 +6,+5 +4,+3 +7,+5 +4 +8,+5 +4,+3 +3 +4,+3 +4,+3 +4,+2 +1 +2 La Hf Ta W Re +2 Os Ir Pt Au Hg +5,+4 +3 +6,+5 +4 +7,+5 +4 +8,+6 +4,+3 +4,+3 +1 +3 +4,+3 +4,+2 +3,+1 +2,+1

Determining the Oxidation Number of an Element in a Compound Problem: Determine the oxidation number (Ox. No.) of each element in the following compounds. a) Iron III Chloride b) Nitrogen Dioxide c) Sulfuric acid Plan: We apply the rules in Table 4.3, always making sure that the Ox. No. values in a compound add up to zero, and in a polyatomic ion, to the ion’s charge. Solution: a) FeCl3 This compound is composed of monoatomic ions. The Ox. No. of Cl- is -1, for a total of -3. Therefore the Fe is +3. b) NO2 The Ox. No. of oxygen is -2 for a total of -4. Since the Ox. No. in a compound must add up to zero, the Ox. No. of N is +4. c) H2SO4 The Ox. No. of H is +1, so the SO42- group must sum to -2. The Ox. No. of each O is -2 for a total of -8. So the Sulfur atom is +6.

Method to remove chlorinated organic molecules from Ground water Fe(s) + RCl(aq) + H+(aq) Fe2+(aq) + RH(aq) + Cl-(aq)

Figure 4.20: A summary of an oxidation-reduction process, in which M is oxidized and X is reduced.

Recognizing Oxidizing and Reducing Agents - I Problem: Identify the oxidizing and reducing agent in each of the Rx: 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 +1 +2 Zn(s) + 2 HCl(aq) ZnCl2 (aq) + H2 (g) HCl is the oxidizing agent, and Zn is the reducing agent!

Recognizing Oxidizing and Reducing Agents - II b) Assigning oxidation numbers: S [0] S[+6] -2 S is Oxidized +6 O[0] O[-2] S8 (s) + 12 O2 (g) 8 SO3 (g) O is Reduced S8 is the reducing agent and O2 is the oxidizing agent c) Assigning oxidation numbers: C[+2] C[+4] C is oxidized Ni[+2] Ni[0] Ni is Reduced -2 -2 -2 +2 +4 +2 NiO(s) + CO(g) Ni(s) + CO2 (g) CO is the reducing agent and NiO is the oxidizing agent

Activity Series of the Metals Strongly reducing Weakly Li Li+ + e- K K+ + e- These elements react rapidly with aqueous H+ ions Ba Ba2+ + 2 e- (acid) or with liquid H2O to release H2 gas. Ca Ca2+ + 2 e- Na Na+ + e- Mg Mg2+ + 2e- Al Al3+ + 3e- These elements react with aqueous H+ ions or with Mn Mn2+ + 2e- steam to release H2 gas. Zn Zn2+ + 2e- Cr Cr3+ + 3e- Fe Fe2+ + 2e- Co Co2+ + 2e- Ni Ni2+ + 2e- These elements react with aqueous H+ ions to Sn Sn2+ + 2e- release H2 gas. H2 2 H+ + 2e- Cu Cu2+ + 2e- Ag Ag+ + e- These elements do not react with aqueous H+ ions Hg Hg2+ + 2e- to release H2 gas. Pt Pt2+ + 2e- Au Au3+ + 3e-

Examples of Activity Series Problems Cu(s) + 2 Ag+(aq) Cu2+(aq) + 2 Ag(s) 2 Fe(s) + 3 Cu2+(aq) 2 Fe3+(aq) + 3 Cu(s) Mg(s) + Zn2+(aq) Mg2+(aq) + Zn(s) 2 K(s) + Sn2+(aq) 2 K+(aq) + Sn(s) Pt(s) + Ni2+(aq) N. R. 2 Al(s) + 6 H+(aq) 2 Al3+(aq) + 3 H2 (g) Au(s) + H+(aq) N. R.

Problem: Calculate the mass of metallic Iron that must be added to 500.0 liters of a solution containing 0.00040M of Pt2+(aq) ions in solution to reclaim all of the Platinum. Solution: V x M = # moles 500.0L x 0.00040 Mol/L = 0.20 Mol Pt2+ assume that the Iron goes to Fe3+ therefore we will need only 2 moles of Iron for every 3 moles of Platinum 0.20 mol Pt2+ x = 0.133 mol Fe 0.133 mol Fe x = ____________ g Fe 2 mol Fe 3 mol Pt2+ 55.85 g Fe mol Fe

Balancing REDOX Equations: The oxidation 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) Multiply one or both of these numbers by appropriate factors to make the electrons lost equal the electrons gained, and use the factors as balancing coefficients. Step 5) Complete the balancing by inspection, adding states of matter.

REDOX Balancing using Ox. No. Method - I -2 ___ H2 (g) +___ O2 (g) ___ H2O(g) 2 2 - 1 e- +1 electrons lost must = electrons gained therefore multiply Hydrogen reaction by 2! and we are balanced!

REDOX Balancing Using Ox. No. Method - II +2 -1e- +3 Fe+2(aq) + MnO4-(aq) + H3O+(aq) Fe+3(aq) + Mn+2(aq) + H2O(aq) +5 e- +2 +7 Multiply Fe+2 & Fe+3 by five to correct for the electrons gained by the Manganese. 5 Fe+2(aq) + MnO4-(aq) + H3O+(aq) 5 Fe+3(aq) + Mn+2(aq) + H2O(aq) Make four water molecules from protons from the acid, and the oxygen from the MnO4-, this will require 8 protons, or Hydronium ions. This will give a total of 12 water molecules formed. 5 Fe+2(aq) + MnO4-(aq) +8 H3O+(aq) 5 Fe+3(aq) + Mn+2(aq) +12 H2O(aq)

Balancing Oxidation-Reduction Equations Occurring in Acidic Solution by the Half-Reaction Method.

Balancing Oxidation-Reduction Equations Occurring in Basic Solution by the Half-Reaction Method.

Balancing Redox Equations in Aqueous Acid and Base Solutions : ACID : You may add either H+ ( H3O+ ), or water ( H2O ) to either side of the chemical equation. BASE : You may add either OH -, or water to either side of the chemical equation. H+ + OH - H2O H+ + OH - H2O H+ + H2O H3O+

REDOX Balancing by Half-Reaction Method-I Fe+2(aq) + MnO4-(aq) Fe+3(aq) + Mn+2(aq) [acid solution] Identify Oxidation and Reduction Half Reactions Fe+2(aq) Fe+3(aq) + e- [oxidation half-reaction] MnO4-(aq) Mn+2(aq) add H+ to the reactants and that will give water as a product! MnO4-(aq) + 8H3O+(aq) +5e- Mn+2(aq) + 12H2O(l) [reduction half-reaction] Sum the two half-reactions { Fe+2(aq) Fe+3(aq) +e- } x5 MnO4-(aq) + 8H3O+(aq) +5e- Mn+2(aq) + 12H2O(l) MnO4-(aq)+ 8H3O+(aq)+5e- +5Fe+2(aq) 5Fe+3(aq)+5e- + Mn+2(aq)+ 12H2O(l)

REDOX Balancing by Half-Reaction Method -II MnO4-(aq) + SO32-(aq) MnO2 (s) + SO42-(aq) [basic solution] Oxidation: SO32- SO42-(aq) + 2e - Add OH- to the reactant side, and water to the product side to get oxygen to balance since we have one more oxygen on sulfate than on sulfite. SO32-(aq) + 2 OH-(aq) SO42-(aq) + H2O(l) + 2e - Reduction: MnO4-(aq) + 3e - MnO2 (s) Add water to the reactant side and OH- to the product side to take up the oxygen lost when MnO4- goes to MnO2 and loses two oxygen atoms. MnO4-(aq) + 2 H2O(l)+ 3e - MnO2 (s) + 4 OH-(aq) Multiply the oxidation equation by 3 to make the electrons 6. Multiply the reduction equation by 2 to make the electrons 6, and add the two. 3 SO3-2(aq) + 2 MnO4-(aq) + H2O(l) 3 SO4-2(aq) + 2 MnO2 (s) + 2 OH-(aq)

REDOX Balancing by Half-Reaction Method-III MnO4-(aq) + SO32-(aq) MnO2 (s) + SO42-(aq) [acidic solution] Oxidation: SO32-(aq) SO42-(aq) + 2 e - Add water to the reactant side to supply an oxygen and add two protons to the product side that will remain plus the two electrons. SO32-(aq) + H2O(l) SO42-(aq) + 2 H+(aq) + 2 e - Reduction: MnO4-(aq) + 3 e- MnO2 (s) Add water to the product side to take up the extra oxygen from Mn cpds, and add Hydrogen to the reactant side . MnO4-(aq) + 3 e- + 4H+ MnO2 (s) + 2 H2O(l) Multiply the oxidation equation by 3, and the reduction equation by 2, and add them canceling out the electrons, protons and water molecules. 3SO32-(aq) + 2MnO4-(aq) + 2H+(aq) 3 SO42-(aq) + 2MnO2 (s) + H2O(l)

REDOX Balancing using Ox. No. Method - III +7 + 3 e - +4 ( Acidic Solution ) MnO4-(aq) + SO32-(aq) MnO2 (s) + SO42-(aq) +4 - 2 e - +6 To balance the electrons, we must multiply the sulfite by 3, and the permanganate by 2. We then have to account for the oxygen imbalance by adding acid to the reactant side, and water to the product side. 2 MnO4-(aq) + 3 SO32-(aq) + H3O+(aq) 2 MnO2 (s) + 3 SO42-(aq) + H2O(aq) For the final balance it is necessary to realize that protons needed to bind up the oxygen atoms must be balanced, and since we have called H+ion - hydronium ions,therefore water will be formed! 2 MnO4-(aq)+ 3 SO32-(aq)+2 H3O+(aq) 2 MnO2 (s) + 3 SO42-(aq) +3 H2O(aq)

REDOX Balancing by Half-Reaction Method-IV MnO4-(aq) +SO32-(aq) MnO2(s) + SO42-(aq) [basic solution] balance the equation as if it were in acid, and then convert it to base: 2MnO4-(aq) + 3SO32-(aq) + 2H+(aq) 2MnO2(s) + 3SO42-(aq) + H2O(l) 2MnO4-(aq) + 3SO32-(aq) + 2H+(aq) 2MnO2(s) + 3SO42-(aq) + H2O(l) To convert to base, add two OH- to each side of the equation: 2MnO4-(aq)+ 3SO32-(aq)+ 2H+(aq) + 2OH-(aq) 2MnO2(s)+ 3SO42-(aq)+ H2O(l)+2OH-(aq) On the reactant side, the H+ and the OH- cancel to give water. 2MnO4-(aq)+ 3SO32-(aq)+2H2O(l) 2MnO2(s)+ 3SO42-(aq)+ H2O(l)+2OH-(aq) Cancel out the water on each side of the equation, and you are done! 2MnO4-(aq) + 3SO32-(aq) + H2O(l) 2MnO2(s) + 3SO42-(aq) +2OH-(aq)

REDOX Balancing Using Ox. No. Method-IV Zinc metal is dissolved in Nitric Acid to give Zn2+ and the ammonium ion from the reduced Nitric acid, write the balanced chemical equation! Zn(s) + H+(aq) + NO3-(aq) Zn2+(aq) + NH4+(aq) Oxidation # method - 2 e- Zn(s) + H+(aq) + NO3-(aq) Zn2+(aq) + NH4+(aq) +5 +8 e- -3 Multiply Zinc and Zn2+ by 4, and ammonia by unity. Since we have no oxygen on the product side, add 3 water molecules to the product side, requiring 10 H+ on the reactant side. 4 Zn(s) +10 H+(aq) + NO3-(aq) 4 Zn2+(aq) + NH4+(aq) + 3 H2O(l)

REDOX Balancing by Half-Reaction Method-V Given: Zn(s) + H3O+(aq) + NO3-(aq) Zn2+(aq) + NH4+(aq) Oxidation: Zn(s) Zn2+ + 2 e- Reduction: H3O+(aq) + NO3-(aq) + 8 e - NH4+(aq) + H2O(l) We will need three waters to pick up the oxygen from the nitrate ion, and for the hydrogen, we will need to have 10 hydrogen ions. Because the Hydrogen ions came as hydronium ions, we will need 10 more water molecules. 10 H3O+(aq) + NO3-(aq) + 8 e - NH4+(aq) + 13 H2O(l) Finally, if we are to add the two equations, we must multiply the Ox. one by 4 to be able to cancel out the electrons, so the final balanced equation is: 10 H3O+(aq) + NO3-(aq) + 4 Zn(s) 4 Zn+2(aq) + NH4+(aq) + 13 H2O(l)

REDOX Balancing by Half-Reaction Method -VI - A In acid Potassium dichromate reacts with ethanol(C2H5OH) to yield the blue-green solution of Cr+3, the reaction used in “breathalyzers”. H3O+(aq) + Cr2O72-(aq) + C2H5OH(l) Cr3+(aq) + CO2 (g) + H2O(l) Oxidation: C2H5OH(l) CO2 (g) We need to balance oxygen by adding water to the reactant side, and balance Hydrogen by adding protons to the product side. C2H5OH(l) + 3 H2O(l) 2 CO2 (g) + 12 H+(aq) Since we wish to consider H+ as the Hydronium ion - H3O+ , we must add 12 water molecules to the reactant side, and make the H+ into H3O+. C2H5OH(l) + 15 H2O(l) 2 CO2 (g) + 12 H3O+(aq) + 12 e -

REDOX Balancing by Half-Reaction Method - VI - B Reduction: Cr2O72-(aq) Cr+3(aq) Dichromate has two chromium atoms, therefore the products need to have two Cr+3, and 3 electrons per atom. The oxygen atoms from the dichromate need to be taken up as water on the product side by adding protons to the reactant side. 14H+(aq) + Cr2O72-(aq) Cr+3(aq) + 7 H2O(l) Each Chromium atom changes oxidation from a +6 to a +3 thereby accepting 6 electrons, so we add 6 electrons to the reactant side. 6e - + 14 H3O+(aq) + Cr2O72-(aq) 2 Cr+3(aq) + 21 H2O(l) Adding the two equations will give the final equation: Ox: C2H5OH(l) + 15 H2O(l) 2 CO2 (g) + 12 H3O+(aq) + 12 e - Rd: [6e - + 14 H3O+(aq) + Cr2O72-(aq) 2 Cr+3(aq) + 21 H2O(l)] x 2 C2H5OH(l) + 16 H3O+(aq) + 2 Cr2O72-(aq) 2 CO2 (g) + 4 Cr+3(aq) + 27 H2O(l)

REDOX Balancing by Half-Reaction Method -VII - A Silver is reclaimed from ores by extraction using basic Cyanide ion. OH- Ag(s) + CN-(aq) + O2 (g) Ag(CN)2-(aq) Oxidation: CN-(aq) + Ag(s) Ag(CN)2-(aq) Since we need two cyanide ions to form the complex, add two to the reactant side of the equation. Silver is also oxidized, so it loses an electron, so we add one electron to the product side. 2 CN-(aq) + Ag(s) Ag(CN)2-(aq) + e - Reduction: O2 (g) + H2O(aq) OH-(l) Since oxygen is to form oxide ions, 4 electrons need to be added to the reactant side, and 2 water molecules are needed to supply the hydrogen to make hydroxide ions, yielding 4 OH- ions. 4 e - + O2 (g) + 2 H2O(aq) 4 OH-(l)

REDOX Balancing by Half-Reaction Method - VII - B Adding the Reduction equation to the Oxidation equation will require the Oxidation one to be multiplied by 4 to eliminate the electrons. Ox (x4) 8CN-(aq) + 4 Ag(s) 4 Ag(CN)2-(aq) + 4 e - Rd 4 e - + O2 (g) + 2 H2O(l) 4 OH -(aq) 8 CN -(aq) + 4 Ag(s) + O2 (g) + 2 H2O(l) 4 Ag(CN)2-(aq) + 4 OH -(aq)

REDOX Balancing Using Ox. No. Method -V -1 e - +1 Ag(s) + CN -(aq) + O2 (g) Ag(CN)2-(aq) + OH -(aq) + 2 e - - 2 To balance electrons we must put a 4 in front of the Ag, since each oxygen loses two electrons, and they come two at a time! That requires us to put a 4 in front of the silver complex, yielding 8 cyanide ions. 4 Ag(s) + 8 CN -(aq) + O2 (g) 4 Ag(CN)2-(aq) + OH -(aq) We have no hydrogen on the reactant side therefore we must add water as a reactant, and since we also add oxygen, we must add two water molecules, that well give us 4 hydroxide anions, giving us a balanced chemical equation. 4 Ag(s) + 8 CN -(aq) + O2 (g) + 2 H2O(l) 4 Ag(CN)2-(aq) + 4 OH -(aq)

Redox Titration- Calculation outline - I Problem: Calcium Oxalate was precipitated from blood by the addition of Sodium Oxalate so that calcium ion could be determined in the blood sample. The sulfuric acid solution that the precipitate was dissolved in required 2.05 ml of 4.88 x 10-4 M KMnO4 to reach the endpoint. a) calculate the amount (mol) of Ca+2. b) calculate the Ca+2 ion conc. Plan: a) Calculate the molarity of Ca+2 in the H2SO4 solution. b) Convert the Ca+2 concentration into units of mg Ca+2/ 100 ml blood. Volume (L) of KMnO4 Solution a) M (mol/L) Moles of KMnO4 b) Molar ratio Moles of CaC2O4 c) Chemical Formulas Moles of Ca+2

Figure 4.21: Permanganate being introduced into a flask of reducing agent.

Redox Titration - Calculation - I Equation: 2 KMnO4 (aq) + 5 CaC2O4 (aq) + 8 H2SO4 (aq) 2 MnSO4 (aq) + K2SO4 (aq) + 5 CaSO4 (aq) + 10 CO2 (g) + 8 H2O(L) a) Moles of KMnO4 Mol = Vol x Molarity Mol = 0.00205 L x 4.88 x 10- 4mol/L Mol = 1.00 x 10 - 6mol KMnO4 b) Moles of CaC2O4 5 mol CaC2O4 Mol CaC2O4 = 1.00 x 10-6 mol KMnO4 x = 2 mol KMnO4 Mol CaC2O4 = 2.50 x 10 -6 mol CaC2O4 c) Moles of Ca+2 1 mol Ca+2 Mol Ca+2 = 2.50 x 10 -6 mol CaC2O4 x = 1 mol CaC2O4 Mol Ca+2 = 2.50 x 10 -6 mol Ca+2

Redox Titration - Calculation Outline - II Moles of Ca2+/ 1 ml of blood multiply by 100 a) Calc of mol Ca+2 per 100 ml Moles of Ca2+/ 100 ml blood M (g/mol) b) Calc of mass of Ca+2 per 100 ml Mass (g) of Ca2+/ 100 ml blood 1g = 1000mg c) convert g to mg! Mass (mg) of Ca2+ / 100 ml blood

Redox Titration - Calculation - II a) Mol Ca+2 per 100 ml Blood Mol Ca+2 Mol Ca+2 = x 100 ml Blood = 100 ml Blood 1.00 ml Blood Mol Ca+2 2.50 x 10 -6 mol Ca+2 = x 100 ml Blood = 100 ml Blood 1.00 ml Blood Mol Ca+2 = 2.50 x 10 -4 mol Ca+2 100 ml Blood b) mass (g) of Ca+2 Mass Ca+2 = Mol Ca+2 x Mol Mass Ca/ mol = Mass Ca+2 = 2.50 x 10 -4mol Ca+2 x 40.08g Ca/mol = 0.0100 g Ca+2 c) mass (mg) of Ca+2 Mass Ca+2 = 0.0100g Ca+2 x 1000mg Ca+2/g Ca+2 = 10.0 mg Ca+2 100 ml Blood

Types of Chemical Reactions - I I) Combination Reactions that are Redox Reactions a) A Metal and a Non-Metal form an Ionic compound b) Two Non-Metals form a Covalent compound c) Combination of an Element and a Compound II) Combination Reactions that are not Redox Reactions a) A Metal oxide and a Non-Metal form an ionic compound with a polyatomic anion b) Metal Oxides and water form Bases c) Non-Metal Oxides and water form Acids III) Decomposition Reactions a) Thermal Decomposition i) Many ionic compounds with oxoanions form a metal oxide and a gaseous non-metal ii) Many Metal oxides, Chlorates, and Perchlorates release Oxygen b) Electrolytic Decomposition

Types of Chemical Reactions - II IV) Displacement Reactions a) Single -Displacement Reactions - Activity Series of the Metals i) A metal displaces hydrogen from water or an acid ii) A metal displaces another metal ion from solution iii) A halogen displaces a halide ion from solution b) Double -Displacement Reactions - i) In Precipitation Reactions- A precipitate forms ii) In acid-Base Reactions - Acid + Base form a salt & water V) Combustion Reactions - All are Redox Processes a) Combustion of an element with oxygen to form oxides b) Combustion of Hydrocarbons to yield Water & Carbon Dioxide Reactants Products