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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 (MOLARITY!) 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 Figure 4.1: A space-filling model of the water molecule. The expression ( aq ) is used to indicate a solvated anion or cation (or molecule) Figure 4.2: Polar water molecules interact with the positive and negative ions of a salt, assisting with the dissolving process. 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 Electrical Conductivity of Ionic Solutions Figure 4.5: HCL (aq) is completely ionized. Figure 4.6: An aqueous solution of sodium hydroxide. 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) • NaOH and KOH are strong bases: NaOH(s) → Na+(aq) + OH-(aq) All of the above species are ionized nearly 100% Weak bases like NH3 and weak acids like acetic acid don’t conduct electricity well Figure 4.7: Acetic acid (HC2H3O2) exists in water mostly as undissociated molecules. Figure 4.8: The reaction of NH3 in water. Weak Electrolytes • Produce relatively few ions in aqueous solution • 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% Because of hydrogen bond formation, water boils at a much higher temperature than CH4 (90 K), which has a comparable molecular mass. Nonelectrolytes Dissolve in water but produce no ions in solution. Nonelectrolytes do not conduct electricity because they dissolve as whole molecules, and produce no ions. Common nonelectrolytes include ethanol and table sugar (sucrose, C12H22O11) The Solubility of Covalent Compounds in Water Polar covalent compounds are very soluble in water. They have -OH groups that can form hydrogen bonds with water. Examples are compounds table sugar, sucrose (C12H22O11), ethanol (C2H5-OH), ethylene glycol (C2H6O2) in antifreeze, and methanol (CH3-OH). These also are written with “(aq)” when dissolved in water (i.e., aqueous). 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 / or Benzene = C6H6 Interaction of Water and Ethanol Carbohydrates Molecules that contain carbon and water! H CH2OH C HO H C OHH C C OH H Sucrose CxH2yOy O H C O CH2OH O H C C OH H CH2OH C C H OH C12H22O11 , C12(H2O)11 a disaccharide 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) b) c) d) 4.0 moles of sodium carbonate dissolved in water 46.5 g of rubidium fluoride dissolved in water 5.14 x 1021 formula units of iron (III) chloride dissolved in water 7.8 moles of ammonium sulfate dissolved in water H2O a) Na2CO3 (s) moles of Na+ = 4.0 moles Na2CO3 x 2 Na+(aq) + CO3-2(aq) 2 mol Na+ 1 mol Na2CO3 = 8.0 moles Na+ and 4.0 moles of CO3-2 are present Determining Moles of Ions in Aqueous Solutions of Ionic Compounds - II b) RbF(s) H2O Rb+(aq) + F -(aq) moles of RbF = 46.5 g RbF x 1 mol RbF 104.47 g RbF = 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 x = 0.0155 mol FeCl3 moles of Cl - 1 mol FeCl3 6.022 x 1023 formula units FeCl3 3 mol Cl = 0.0155 mol FeCl3 x = 0.0465 mol Cl 1 mol FeCl3 and 0.0155 mol Fe+3 are also present. Determining Moles of Ions in Aqueous Solutions of Ionic Compounds - III d) (NH4)2SO4 (s) H2O 2 NH4+(aq) + SO4- 2(aq) + 2 mol NH 4 = 15.6 mol NH + Moles of NH4 = 7.8 moles (NH4)2SO4 x 4 1 mol(NH4)2SO4 + and 7.8 mol SO4- 2 are also present. Molarity (Concentration of Solutions) = M M= Moles of Solute = Liters of Solution mol L solute = material dissolved into the solvent In sea water, water is the solvent, and NaCl, MgCl2, etc are the solutes. In brass, copper is the solvent (90%), and zinc is the solute (10%). Determining Moles of Ions in Aqueous Solutions of Ionic Compounds – I+ Example problem: a) b) c) d) e) How many moles of each ion are in each of the following: 4.0 moles of sodium carbonate dissolved in water (Na: IA) 46.5 g of rubidium fluoride dissolved in water (Rb: IA) 9.32 x 1021 formula units of iron (III) chloride dissolved in water 7.8 moles of ammonium sulfate dissolved in water 75.0 mL of 0.56 M scandium bromide dissolved in water (Sc:IIIB) We already did a) through d) above. Let’s do e): Determining Moles of Ions in Aqueous Solutions of Ionic Compounds - IV e) ScBr3(s) H2O Sc3+(aq) + 3 Br -(aq) Converting from volume to moles: Moles of ScBr3 = 75.0 mL x (0.56mol ScBr3 / L) x (1 L / 1000 mL) = 0.042 mol ScBr3 Moles of Br- = 0.042 mol ScBr3 x (3 mol Br-/ mol ScBr3) = 0.13 mol BrMoles of Sc3+ = 0.042 mol ScBr3 x (1 mol Sc3+ / mol ScBr3) = 0.042 mol Sc3+ Molarity (Concentration of Solutions) = M M= Moles of Solute = Liters of Solution mol L solute = material dissolved into the solvent For electrolytes, we describe the molarity of both the compound itself and of each of the ions it produces: e.g., Na2CO3 (s) → 2 Na+(aq) + CO3-2(aq) = “Na2CO3 (aq)” The Na+ molarity or concentration is 2x that of the other 2 species here. 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, M, of the salt and each of the ions? Strategy (1) Write chemical equation showing process. (2) Calculate moles of each species. (3) Divide # moles by # L water to obtain molarity. (1) Na3PO4 (s) H2O(solvent) → 3 Na+ (aq) + PO4-3 (aq) (2) Calculate moles of each species. Mol wt 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 = mol Na3PO4 = 0.0241 mol (3) Divide # moles by # L water to obtain molarity. M(Na3PO4) = 0.0241 mol Na3PO4 / 0.3000 L = 0.0803 M M(PO4-3) = M(Na3PO4) = 0.0803 M M(Na+ ) = 3 x M(Na3PO4) = 0.241 M Figure 4.9: Steps involved in the preparation of a standard solution. 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: 1.00 L x 0.375 mol (NH4)2CO3 = 0.375 mol (NH4)2CO3 L solution This amount can be converted to grams by using the molar mass: 94.07 g (NH4)2CO3 0.375 mol (NH4)2CO3 x = 35.276 g (NH4)2CO3 mol (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. 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 2.5 mg NaCl: 1 g NaCl x 2.5 mg NaCl x 1000 mg NaCl 1 mol NaCl = 4.28 x 10-5 mol 58.45g NaCl NaCl What volume of 0.14 M NaCl would contain this amount of NaCl (4.28 x 10-5 mol NaCl): V x M = moles 0.14 M NaCl Vx = 4.28 x 10-5 mol NaCl L solution Solving for Volume gives: -5 V = 4.28 x 10 mol NaCl = 3.06 x 10-4 L 0.14 mol NaCl or 0.306 mL of Blood! L solution Dilution of Solutions Dilute 25.00 mL of 0.0400 M KMnO4 to a final volume of 500. mL. What is the resulting 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 V1 x M1 = moles solute = V2 x M2 V1M1 = V2 M 2 V1M1 M2 = V2 25.00mL × 0.0400M = 0.00200 M M2 = 500.mL Make a Standard Solution of 0.00100 M Potassium Permanganate Potassium Permanganate is KMnO4 and has a molecular mass of 158.04 g / mole Step 1: Prepare a 0.0400 M solution. For example, dissolve 1.00x10-2 moles of KMnO4 into water to a final volume of ¼ L (250.00 mL) of solution. 158.04 g KMnO4 = 1.58 g KMnO 0.0100 mol KMnO4 x 4 mole KMnO4 Molarity = 0.0100 moles KMnO4 = 0.0400 mol/L = 0.0400 M 0.250 L Molarity of K+ ion = [K+] ion = [MnO4-] ion = 0.0400 M Step 2: Dilute some of it to 0.00100 M. 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 = 0.00100 M Figure 4.10: (a) A measuring pipette or graduated pipette (b) A volumetric pipette: delivers one specific volume only. Figure 4.11: (a) A measuring pipette (b) Water is added to the flask. (c) The resulting molarity is M2 = M1 x V1/V2. V1 V2 = TYPES of CHEMICAL REACTIONS in LIQUID SOLUTIONS: • Precipitation reactions • Acid – Base reactions • Reduction – Oxidation (REDOX) reactions PRECIPITATION REACTIONS: a solid compound or “precipitate” forms from two (or more) solutes DEMO: Pb(NO3)2(aq) + 2 KI(aq) → PbI2(s) + 2 KNO3(aq) Lead iodide is a gold precipitate Show all ions explicitly: Pb2+(aq) + 2 NO3-(aq) + 2 K+(aq) + 2 I-(aq)→ PbI2(s) + 2 K+(aq) + 2 NO3-(aq) Remove spectator ions: Pb2+(aq) + 2 I-(aq) → PbI2(s) ( same as with KI ) DEMO: Purple precipitate CoCl2(aq)+ Na2CO3(aq) → CoCO3(s) + 2 NaCl(aq) A note on this one: You want an excess of carbonate in the cylinder first, then pour the cobalt in. Otherwise, the background is pink! Go HUSKIES!! Figure 4.12: When yellow aqueous potassium chromate is added to a colorless barium nitrate solution, yellow barium chromate precipitates. Figure 4.13: Reactant solutions: (a) Ba(NO3)3(aq) Figure 4.13: Reactant solutions: (b) K2CrO4(aq). Figure 4.14: Reaction of K2CrO4 (aq) and Ba(NO3)2 (aq). Table 4.1 Simple Rules for Solubility of Salts in Water (I will give you this table on tests.) 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. Predicting if a precipitate forms, and which? Pb(NO3)2(aq) + NaCl(aq) → Pb+2(aq) + 2 NO3- (aq) + Na+ (aq) + Cl-(aq) RULE: If any of the possible new species formed by combining anions with cations is insoluble, then that precipitate will form. USE TABLE 4.1 (I’ll give you this Table for the tests and quiz.) In this case, PbCl2 is insoluble, and a precipitate forms. Note: Like the demo with PbI2(s) above. 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, 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) → Ba2+(aq) + SO42- (aq) + 2 Na+(aq) + 2NO3- (aq) Barium sulfate is insoluble; therefore its precipitate will form. Aqueous KOH and Fe(NO3)3 are mixed. Solid Fe(OH)3 forms. Figure 4.16: Photos and molecular-level representations illustrating the reaction of KCl(aq) with AgNO3(aq) to form AgCl(s). Quantitative Precipitation Problems: Calculate the mass of solid sodium iodide that must be added to 2.50 L of a 0.125 M lead nitrate (Pb(NO3)2) 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 sodium iodide needed to precipitate PbI2 is twice the lead ions. The number of moles of sodium iodide needed is: 2.50 L 2+ 0.125 Mol Pb(NO ) 1 Mol Pb 3 2 x x 1.0 L soln. 1 Mol Pb(NO3)2 The mass of sodium iodide is: 0.625 mol I- x 1 mol NaI x 1 mol I- 2 mol I x = 0.625 mol I1 mol Pb2+ 149.9 g NaI = 93.68 g NaI 1 mol NaI Quantitative Precipitation Problems: Calculate the mass of solid sodium iodide that must be added to 2.50 L of a 0.125 M lead nitrate (Pb(NO3)2) 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 sodium iodide needed to precipitate PbI2 is twice the lead ions. The number of moles of sodium iodide needed is: 2.50 L 2+ 0.125 mol Pb(NO ) 1 mol Pb 3 2 x x 1.0 L soln. 1 mol Pb(NO3)2 1 mol NaI 2 mol I x x 2+ 1 mol I1 mol Pb An alternate method: 2.50 L x 0.125 mol Pb(NO3)2 x 2 mol NaI 1.0 L soln. 1 mol Pb(NO3)2 = Like Example 4.9 (P 112) 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 mol I- = 0.00118 mol IL 2+ 1 mol Pb 2+ Moles Lead ion = 0.00118 mol I- x = 0.000590 mol Pb 2 mol IMass of Lead = 207.2 g Pb x 0.000590 moles = 0.122 g Pb mol Pb 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) “Spectator Ions” here are Na+ and NO3- CaSO4 (s) DEMO: 0.1 M Potassium ferrocyanide = 4 K+(aq) + Fe(CN)6-4(aq) 0.1 M Ferric chloride = Fe3+(aq) + 3 Cl-(aq) 4 K+(aq) + Fe(CN)6-4(aq) + Fe+3 (aq) + 3 Cl-(aq) → K[Fe(CN)6](s) + 3 K+(aq) + 3 Cl-(aq) Prussian Blue (very dark purplish blue) Remove spectator ions: K+(aq) + Fe(CN)6-4(aq) + Fe+3 (aq) → K[Fe(CN)6](s) Limiting reagent problems in precipitation reactions. Solve as usual, using full equation or only net ionic equation. ÷ stoich. coeff. to get equivalents. Amounts: Just use molarities and volumes to get moles (when moles or grams are not given). 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 lead precipitation as a sulfide. If 257.8 ml of a 0.0468 M solution of lead(II) acetate (Pb(C2H3O2)2) is added to 156.00 mL of a 0.095 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 equation. Determine the limiting reactant. Calculate the moles of product. Convert moles of product to mass of the product using molar mass. (1) Write the balanced equation Pb(C2H3O2)2 (aq) + Na2S (aq) → PbS (s) + 2 Na(C2H3O2) (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 (0.095 mol/L) = 0.0148 mol Mol Pb(C2H3O2)2 / 1 mol = 0.0121 eq. Mol Mol Na2S / 1 mol = 0.0148 eq. Therefore lead(II) acetate is the limiting reactant. (3) Calculate the moles of product. 0.00121 mol Pb(C2H3O2)2 x (1 mol PbS / mol Pb(C2H3O2)2) = 0.0121 mol PbS (4) Calculate mass of product 0.0121 mol PbS x 239.3g PbS/mol PbS = 2.90 g PbS Precipitation problem: When aqueous silver nitrate and sodium chromate solutions are mixed, solid silver chromate forms in a solution of sodium nitrate. If 257.8 mL of a 0.0468 M solution of silver nitrate is added to 156.00 mL of a 0.0950 M solution of sodium chromate, what mass of silver chromate (M = 331.8 g/mol) will be formed? This is a limiting-reactant problem because the amounts of two reactants are given. Strategy: (1) Write the balanced equation. (2) Calculate the number of moles of each reactant. (3) Determine the limiting reactant. (4) Calculate the moles of product. (5) Convert moles of product to mass of the product using molar mass. Molecular equation, Total ionic equation, and net ionic equation Precipitation problem: When aqueous silver nitrate and sodium chromate solutions are mixed, solid silver chromate forms in a solution of sodium nitrate. If 257.8 mL of a 0.0468 M solution of silver nitrate is added to 156.00 mL of a 0.0950 M solution of sodium chromate, what mass of silver chromate (M = 331.8 g/mol) will be formed? • Write the balanced equation 2AgNO3(aq) + Na2CrO4(aq) → Ag2CrO4(s) + 2 NaNO3(aq) (2) Determine the number of moles of each reactant Mol AgNO3 = V x M = 0.2578 L x (0.0468 mol/L) = 0.0121 mol Mol Na2CrO4 = V x M = 0.15600 L x (0.095 mol/L) = 0.0148 mol 2) Determine the limiting reactant. 0.0121 mol AgNO3 / (2 mol/eq) = 0.00605 eq. AgNO3 0.0148 mol Na2CrO4 / (1 mol/eq) = 0.0148 eq. Na2CrO4 Therefore silver nitrate is the limiting reactant. 2AgNO3(aq) + Na2CrO4(aq) → Ag2CrO4(s) + 2 NaNO3(aq) (3) Calculate the moles of product. Moles Ag2CrO4(s) = 0.0121 moles AgNO3 x (1 mol Ag2CrO4 / 2 mol AgNO3 ) = 0.0121 mol/2 = 0.00605 mol Ag2CrO4 (4) Calculate mass of product 0.00605 mol Ag2CrO4 x (331.8 g Ag2CrO4 /mol Ag2CrO4) = 2.01 g Ag2CrO4(s) Like Example 4.8 (P 111) 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) + _________ mol Ag+ = mol Cl- = __________ is limiting, and we use it to calculate the mass of AgCl: Mass AgCl = Like Example 4.8 (P 111) 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 ClSince 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 = 34.1 g Limiting reagent problems in solution: Only difference is: using molarity and / or volume to connect to moles via Mx = (moles of x) / V. The gravimetric procedure.(P111-112) Determining an unknown amount by precipitation: Finding the Ca concentration in rock! Full balanced equation not needed. All you need to know is that ALL the Ca ends up as CaC2O4 . H2O (s). .H O 1 mol CaC O 2 4 2 -3 mol 0.2920 g CaC2O4 . H2O X = 1.998 x 10 . 146.12g CaC2O4 H2O of CaC2O4 .H2O 2+ 40.08 g Ca -3 2+ -2g Ca2+ = 8.009 x 10 1.998 x 10 mol Ca X 2+ 1 mol Ca Mass % Ca is: 8.009 x 10-2 g Ca2+ 0.4367 g rock x 100% = 18.34% Ca by mass Acid-Base Reactions Acids = Covalent molecules which lose H+ cations (= protons) to water molecules in solution When gaseous hydrogen iodide dissolves in water, the water molecules help the HI dissociate into H+ and I- ions, both solvated by water. 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”, 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: HI(g) H2O H+(aq) + I -(aq) H3O+ is called the hydronium ion Strong Acids and the Molarity of H+ Ions in Aqueous Solutions of Acids Problem: In aqueous solutions, each molecule of sulfuric acid will loose two protons to yield two H+ ions, and one sulfate ion. NOT TRUE What is the molarity of the sulfate and Hydronium ions in a solution prepared by dissolving 155 g of concentrated 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) Moles H2SO4 = Molarity of SO4- 2 = Molarity of H+ = Strong Acids and the Molarity of H+ Ions in Aqueous Solutions of Acids Problem: In aqueous solutions, each molecule of sulfuric acid will loose one proton to yield one H+ ion and one HSO4- ion. (Approx.) What is the molarity of the HSO4- and Hydronium ions in a solution prepared by dissolving 155 g of concentrated 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 HSO4-1 ion. The hydronium ion’s concentration will be the same molarity. Solution: One mole of H+ is released for every mole of acid: H2SO4 (l) + H2O(l) H3O+(aq) + HSO4-(aq) Moles H2SO4 = 155 g H2SO4 x 1 mole H2SO4 = 1.58 moles H2SO4 98.09 g H2SO4 -2 1.58 mol SO 4 Molarity of HSO4- = = 0.687 Molar in SO4- 2 2.30 L solution Molarity of H+ = 0.687 Molar in H+ (or H3O+) Approximately true only. 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- ion react with the H+ ions (if an acid is present) to produce water, H2O, and is therefore a proton acceptor. 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 dissociate partially (some small % of the molecules dissociate) and are weak electrolytes. Selected Acids and Bases Acids Bases Strong: H+(aq) + A-(aq) Strong: M+(aq) + OH-(aq) (100% dissociates in solution) 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 (small % dissociates, most intact) Hydrofluoric, HF Ammonia, NH3 Phosphoric acid, H3PO4 accepts proton from water to make Acetic acid, CH3COOH NH4+(aq) and OH-(aq) (or HC2H3O2) Acid - Base Reactions : Neutralization Rxns. The generalized neutralization reaction between an Acid and a Base is: HA(aq) + MOH(aq) MA(aq) + H2O(l) (Here “M” stands for anything.) Acid + Base → Salt + Water The salt product can either be dissolved as ions or form a precipitate. DEMO Carbonic acid (aq) + NaOH (aq) → Na bicarbonate + water (makes CO2(gas)) 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 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 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 LiNO3 (aq) + H2O(l) (aq) Li+(aq) + OH -(aq) + H+(aq) + NO3-(aq) Li+(aq) + NO3-(aq) + H2O(l) OH -(aq) + H+(aq) c) Ba(OH)2 (aq) + H2SO4 (aq) H2O(l) 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) Burette Like Example 4.10 (P 115) 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 = From the balance equation find the moles of acid needed: __ LiOH(aq) + __ H2SO4 (aq) __ H2O(l) + __ Li2SO4 (aq) Use their ratio of stoichiometric coefficients to convert mol LiOH to mol H2SO4. Then use it molarity to get volume. Volume of acid: Moles acid = Vacid = Macid Like Example 4.10 (P 115) 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 = 0.0268 mol LiOH From the balance equation find the moles of acid needed: 2 LiOH(aq) + H2SO4 (aq) 2 H2O(l) + Li2SO4 (aq) 0.0268 mol LiOH x (1 mol H2SO4 / 2 mol LiOH) = 0.0134 mol H2SO4 Volume of acid: Moles acid 0.0134 mol = 0.468 mol Vacid = Macid L = 0.0286 L H2SO4 On Exam Day: • Seat according to seating chart- check at home the day before the exam!! • If you arrive late on exam day- see TA nearest the door. • Bring a photo ID to the exam. Finding some unknown Concentration of Acid by an Acid - Base Titration with known base Volume (L) of base (difference in burette readings) needed to titrate M (mol/L) of base Moles of base needed to titrate molar ratio Moles of acid which were titrated volume (L) of acid M (mol/L) of original acid soln. Potassium Hydrogenphthalate KHC8H4O4 (called “KHP” for short) O O C C C O K+ O K+ O O O H C O An easy-to-weigh acid: Convenient for making a standard acid concentration for doing titrations H+ Example: A titration is performed between sodium hydroxide and KHP(204.2 g/mol) to “standardize” a base solution (i.e., to determine its exact concentration for use in later titrations). You place 50.00 mg of solid KHP in a flask with a little water and a few drops of an indicator. At the beginning of the titration, the burette reading of the base solution is 43.87 mL and its final burette reading is 10.55 mL . The reaction is: HKC8H4O4(aq) +Na+(aq) +OH-(aq) → KC8H4O4-(aq) +Na+(aq) +H2O(l) H+(aq) + OH-(aq) → H2O(l) What is the concentration of the base (i.e., of the NaOH soln.)? Strategy: (1)Use the molar mass of KHP to calculate the number of moles of the acid. (2) From the balanced chemical equation, calculate moles of base (3) From the difference in the burette readings, and the definition of molarity, calculate the molarity of the base. (1)Use the molar mass of KHP to calculate the number of moles of the acid. 50.00 mg 1g moles KHP = × = 0.0002449 mol -1 204.2 g mol 1000 mg (2) From the balanced chemical equation, calculate moles of base → KC8H4O4-(aq) )+Na+(aq)+H2O(l) HKC8H4O4(aq)+Na+(aq)+OH-(aq) 0.0002449 mol KHP x (1 mol OH- / 1 mol KHP) = 0.0002449 mol OH(3) From the difference in the burette readings (V of NaOH), and the definition of molarity, calculate the molarity of the base. moles of NaOH 0.0002449 mol molarity of NaOH = = = 0.007349 M V NaOH 0.3332 L 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 balanced equation (3) Calculate volume HCl from (2) and known molarity (1) Calculate moles of Al(OH)3(s). 10.0 g Al ( OH )3 = 0.128 mol Al ( OH )3 -1 78.00 g mol (2) Calculate moles of HCl needed using balanced equation Al(OH)3 (s) + 3 HCl (aq) → 3 H2O(l) + AlCl3 (aq) 3 mol HCl 0.128 mol Al ( OH )3 × = 0.385 mol HCl mol Al ( OH )3 (3) Calculate volume HCl from (2) and known molarity mol HCl 0.385 mol HCl V= = = 0.256 L -1 molarity 1.50 mol L 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 NOW: 4.10 Oxidation-Reduction Reactions (Redox rxns.) 4.11 Balancing Oxidation-Reduction Equations 4.12 Simple Oxidation-Reduction Titrations Figure 4.19: Reaction of solid sodium and gaseous chlorine to form solid sodium chloride: an oxidation – reduction reaction. Oxidation - A substance gives up electrons to another substance. Mg + ½ O2 → Mg2+ + O2- Mg is oxidized Reduction - A substance accept electrons from another substance. Mg + ½ O2 → Mg2+ + 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” or “oxidation numbers” (ON). Oxidation numbers are chosen so that (a) charge are conserved, and (b) in ionic compounds the sum of oxidation numbers on the atoms coincides with the charge on the ion. (also called Oxidation Numbers) (i.e., except w/ O and higher halogens) The Periodic Table of the Elements Most Probable Oxidation State 0 +1 +3 +_4 - 3 H +2 Li Be B C N +1 + 2 Al Si P Na Mg +3 +4 +5 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Fr Ra Ac Rf Db Sg Bh Hs Mt Ds +3 +3 -2 -1 He O F Ne S Cl Ar Se Br Kr Te I Xe Po At Rn Ce Pr Nd PmSm Eu Gd Tb Dy Ho Er TmYb Lu Th Pa U Np Pu AmCmBk Cf Es FmMd No Lr Determine the oxidation number (ON) 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 ON values in a compound add up to zero, and in a polyatomic ion, to the ion’s charge. Solution: Ex 1: FeCl3 This compound is composed of monoatomic ions. The ON of Cl- is -1, for a total of -3. Therefore the ON of Fe is +3. Ex 2: NO2 The ON of oxygen is -2 for a total of -4. Since the ON in a compound must add up to zero, the ON of N is +4. Ex 3: H2SO4 The ON of H is +1, so the SO42- group must sum to -2. The ON of each O is -2 for a total of -8. Therefore S has the ON +6. An aside: What is the ON for S in the SO32- ion? Examples (cont.): Ex 4: BaI2 I = -1 Ba = +2 Ex 5: NH4NO3 H = +1 O = -2 N = -3 N = +5 (i.e., except w/ O and higher halogens) Recognizing Oxidizing and Reducing Agents - I Problem: Identify the oxidizing and reducing agent in each Rxn: 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. Then apply RULE: RULE: 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! Recognizing Oxidizing and Reducing Agents - II b) Assigning oxidation numbers: 0 0 S8 (s) + 12 O2 (g) +6 S [0] -2 8 SO3 (g) 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: Ni[+2] Ni[0] Ni is Reduced C[+2] -2 +2 NiO(s) -2 +2 + CO(g) 0 Ni(s) +4 -2 + CO2 (g) C[+4] C is oxidized CO is the reducing agent and NiO is the oxidizing agent Activity Series of the Metals Strongly reducing Weakly reducing Li K Ba Ca Na Li+ + eK+ + eBa2+ + 2 eCa2+ + 2 eNa+ + e- Mg Al Mn Zn Cr Fe Mg2+ + 2eAl3+ + 3eMn2+ + 2eZn2+ + 2eCr3+ + 3eFe2+ + 2e- Co Ni Sn Co2+ + 2eNi2+ + 2eSn2+ + 2e- H2 2 H+ + 2e- Cu Ag Hg Pt Au Cu2+ + 2eAg+ + eHg2+ + 2ePt2+ + 2eAu3+ + 3e- These elements react rapidly with aqueous H+ ions (acid) or with liquid H2O to release H2 gas. These elements react with aqueous H+ ions or with steam to release H2 gas. These elements react with aqueous H+ ions to release H2 gas. These elements do not react with aqueous H+ ions to release H2 gas. Examples of Activity Series Predictions 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. 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 (or total increase in ON = total decrease in ON), remembering to first balance the numbers of atoms of those elements which change ON (red = my addition). Step 5) Complete the balancing by inspection. REDOX Balancing using Ox. No. Method - I +2 e- per O 0 ___ H2 (g) +___ O2 (g) 0 - 1 e- per H -2 ___ H2O(g) +1 electrons lost must = electrons gained Two O atoms each gain 2 eTherefore need 4 H atoms give these 4 e-s! REDOX Balancing using Ox. No. Method - I 2x +2 e- 0 ___ 2 H2 (g) +_1_ O2 (g) 0 4x -1 e- -2 ___ 2 H2O(g) +1 electrons lost must = electrons gained Two O atoms each gain 2 eTherefore need 4 H atoms give these 4 e-s! And we are balanced! (not necessarily emotionally) DEMO Balance the thermite reaction: __ Fe2O3(s) + __ Al(s) → __ Al2O3(s) + __ Fe(l) Assign ON to each element Reactants O: -2 Al:0 Fe:+3 Products O:-2 Al:+3 Fe:0 Identify the elements that are oxidized and reduced Fe3+ →Fe0 : Iron is reduced (ON decreases by 3) Al0 →Al3+ : Aluminum is oxidized (ON increases by 3) Make ON total increase = ON total loss Fe2O3 + 2Al → Al2O3 + 2Fe Thermite reaction _ Fe2O3(s) + __Al(s) → _ Al2O3(s) + _ Fe(l) +3 0 → +3 - 3e- per Al +3e- per Fe 0 Thermite reaction Fe2O3(s) + 2 Al(s) → Al2O3(s) + 2 Fe(l) +3 0 → +3 2x(- 3e- per Al) 2x(+3e- per Fe) BALANCED! 0 Problem: Calculate the mass of metallic Iron that must be added to 500.0 liters of a solution containing 0.00040 M Pt2+(aq) ions in solution to reclaim all Pt via: __ Fe(s) + __ Pt2+(aq) → __ Fe3+(aq) + __ Pt(s) Solution: V x M = # moles 500.0 L x 0.00040 mol Pt2+/ L = 0.20 mol Pt2+ Balance the equation to detn. how many moles of Fe needed for every mole of Pt. 0.20 mol Pt2+ x = 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 Pt via: 2 Fe(s) + 3 Pt2+(aq) → 2 Fe3+(aq) + 3 Pt(s) Solution: V x M = # moles 500.0L x 0.00040 mol Pt2+/ L = 0.20 mol Pt2+ Fe(s) → Fe3+ + 3 ePt2+ + 2 e- → Pt(s) Need 2 moles of Iron for every 3 moles of Platinum 0.20 mol Pt2+ 2 mol Fe x = 0.133 mol Fe 2+ 3 mol Pt 0.133 mol Fe x 55.85 g Fe = 7.4 g Fe mol Fe REDOX Balancing Using Ox. No. Method - II +2 -1e- per Fe +3 Fe+2(aq) + MnO4-(aq) + H+(aq) Fe+3(aq) + Mn+2(aq) + H2O(aq) +5 e- per Mn +2 +7 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 Fe+2(aq) + MnO4-(aq) + H+(aq) 5 Fe+3(aq) + Mn+2(aq) + H2O(aq) Balance O: Need 4 on H2O on right to balance 4 O from the MnO4-. Balance H: Need 8 on H+ on the left to balance 8 H in these 4 H2O s. 5 Fe+2(aq) + MnO4-(aq) +8 H+(aq) 5 Fe+3(aq) + Mn+2(aq) +4 H2O(aq) BALANCED! Note: Add 8 water molecules to both sides if you prefer to express the 8 H+ ions instead as H3O+ ions. REDOX Balancing Using Ox. No. Method - II +2 -1e- per Fe +3 Fe+2(aq) + MnO4-(aq) + H3O+(aq) +7 Fe+3(aq) + Mn+2(aq) + H2O(aq) +5 e- per Mn +2 5 Fe+2(aq) + MnO4-(aq) +8 H+(aq) 5 Fe+3(aq) + Mn+2(aq) +4 H2O(aq) 5 Fe+2(aq) + MnO4-(aq) +8 H3O+(aq) 5 Fe+3(aq) + Mn+2(aq) +12 H2O(aq) Balancing redox eqn using half-cell method in acidic solutions Cu(s) +HNO3(aq) → Cu2+(aq) + NO(g) Identify half-reactions, one is ox, other is red Cu(s) → Cu2+(aq) 0 → +2 HNO3(aq) → NO(g) +5 → +2 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 Balance H by adding H+ to side deficient in H Cu(s) → Cu2+(aq) 3H+ +HNO3(aq) → NO(g) + 2H2O Balance charge by adding e- to side that has + charge Cu(s) → Cu2+(aq) + 2e3H+ + HNO3(aq) + 3e- → NO(g) + 2H2O Multiply each eq by factors so electrons cancel out 3x(Cu(s) → Cu2+(aq) + 2e-) 2x(3H+ + HNO3(aq) + 3e- → NO(g) + 2H2O) Add equations and cancel spectators (none here) 3Cu(s)+ 6H+ + 2HNO3(aq) → 3Cu2+(aq) + 2NO(g) + 4H2O Balancing redox equations in basic solutions: First balance in acidic 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) The following redox equation is balanced in acidic solution. Balance it for a basic solution. Cr2O72-(aq)+2NO(g)+6H+(aq)→2Cr3+(aq)+2NO3- (aq) +3H2O(l) Add OH- equal to number of H+ on both sides 6OH-(aq) +Cr2O72-(aq)+2NO(g)+6H+(aq)→2Cr3+(aq)+2NO3-+3H2O(l) +6OH-(aq) Combine OH- and H+ to form water to max extent possible 6H2O(l)+Cr2O72-(aq)+2NO(g)→2Cr3+(aq)+2NO3-+3H2O(l) +6OH- (aq) Cancel H2O on both sides to max extent possible 3H2O(l)+Cr2O72-(aq)+2NO(g)→2Cr3+(aq)+2NO3- +6OH- (aq) 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. H+ + OH - H2O BASE : You may add either OH -, or water to either side of the chemical equation. H+ + OH H+ + H2O 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) MnO4-(aq) Fe+3(aq) + e- [oxidation half-reaction] Mn+2(aq) Add water as needed to balance O, then add H+ to balance H, then e-s. MnO4-(aq) + 8 H+(aq) +5e- Mn+2(aq) + 4 H2O(l) [reduction half-reaction] Sum the two half-reactions (multiplied to balance electrons): { Fe+2(aq) Fe+3(aq) +e- } x5 MnO4-(aq) + 8 H+(aq) +5eMn+2(aq) + 4H2O(l) MnO4-(aq)+ 8 H+(aq)+5e- +5Fe+2(aq) 5Fe+3(aq)+5e- + Mn+2(aq)+ 4 H2O(l) REDOX Balancing by Half-Reaction Method-III MnO4-(aq) + SO32-(aq) MnO2 (s) + SO42-(aq) [Acidic solution] Oxidation: SO32-(aq) SO42-(aq) Add water to balance O, then add protons to balance H, and finally add electrons to balance charge. SO32-(aq) + H2O(l) SO42-(aq) + 2 H+(aq) + 2 e Reduction: MnO4-(aq) MnO2 (s) Add water to balance O, then add protons to balance H, and finally add electrons to balance charge. MnO4-(aq) + 3 e- + 4H+ MnO2 (s) + 2 H2O(l) Multiply the oxidation equation by 3, and the reduction equation by 2 to balance electrons. Cancel 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- MnO4-(aq) + SO32-(aq) +4 ( Acidic Solution ) MnO2 (s) + SO42-(aq) +4 - 2 e+6 To balance the electrons, we must multiply the sulfite by 3, and the permanganate by 2. Then balance oxygen by adding water. Then add H+ as needed. 2 MnO4-(aq) + 3 SO32-(aq) + 2H+(aq) 2 MnO2 (s) + 3 SO42-(aq) + H2O(aq) If desired, you may write the H+ ions as hydronium ions instead: 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) To convert to base, add enough OH- to each side to cancel any H+: 2MnO4-(aq)+ 3SO32-(aq)+ 2 H+(aq) + 2 OH(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, so add 3 water molecules to the products. It requires 10 H+ on the reactant side to balance H. 4 Zn(s) +10 H+(aq) + NO3-(aq) 4 Zn2+(aq) + NH4+(aq) + 3 H2O(l) REDOX Balancing by Half-Reaction Method-V Given: Oxidation: Zn(s) + H3O+(aq) + NO3-(aq) Zn(s) Zn2+(aq) + NH4+(aq) Zn2+ + 2 e- Reduction: NO3-(aq) NH4+(aq) Other atoms (not O, H) are already balanced (N). We need 3 waters to balance the O from the nitrate ion. For H balance, we need 10 hydrogen ions. Charge balance then requires 8 e-. NH4+(aq) + 3 H2O(l) 10 H+(aq) + NO3-(aq) + 8 e - 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 H+(aq) + NO3-(aq) + 4 Zn(s) 4 Zn+2(aq) + NH4+(aq) + 3 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) Oxidation: Cr3+(aq) + CO2 (g) + H2O(l) C2H5OH(l) CO2 (g) Balance C first. Then balance O by adding water to the reactant side. Balance H by adding protons to the product side: C2H5OH(l) + 3 H2O(l) Add the electrons: C2H5OH(l) + 3 H2O(l) 2 CO2 (g) + 12 H+(aq) 2 CO2 (g) + 12 H+(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. The 7 oxygen atoms from the dichromate need to be balanced with water on the product side. Then add protons to the reactant side to balance H. 14H+(aq) + Cr2O72-(aq) 2 Cr+3(aq) + 7 H2O(l) Add 6 electrons to the reactant side to balance charge: 6e - + 14 H+(aq) + Cr2O72-(aq) 2 Cr+3(aq) + 7 H2O(l) Adding the two equations to cancel electrons: Ox: C2H5OH(l) + 3 H2O(l) 2 CO2 (g) + 12 H+(aq) + 12 e Rd: [6e - + 14 H+(aq) + Cr2O72-(aq) 2 Cr+3(aq) + 7 H2O(l)] x 2 C2H5OH(l) + 16 H+(aq) + 2 Cr2O72-(aq) 2 CO2 (g) + 4 Cr+3(aq) + 11 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 looses an electron, so we add one electron to the product side. 2 CN-(aq) + Ag(s) Ag(CN)2-(aq) + e Reduction: O2 (g) 2 H2O(l) Add 2 water molecules to bal. O. Add 4 H+ to bal. H. Needs 4e- . 4 e - + O2 (g) + 4 H+(aq) 2 H2O(aq) Basic: Add 4 OH- ions on each side to neutralize H+. 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 0 -1 e - Ag(s) + CN -(aq) + O2 (g) +1 Ag(CN)2-(aq) + OH -(aq) 0 + 2 e-2 To balance electrons we must put a 4 in front of the Ag, since each oxygen looses 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(CN)2-(aq) + OH -(aq) 4 Ag(s) + 8 CN -(aq) + O2 (g) We have no hydrogen's 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) A Redox Titration Example Problem: Ca in Blood In order to measure the Ca2+ conc. in blood, calcium oxalate was precipitated from a 1.00 mL sample of blood. (“oxalate” = C2O42-.) This precipitate was dissolved in a sulfuric acid solution, which then required 2.05 mL of 4.88 x 10-4 M KMnO4 to reach the endpoint via the rxn.: 2MnO4-(aq) + 5C2O42-(aq) + 16H+ → 2Mn 2+ (aq) + 10CO2(g) + 8H2O(l) a) Calculate the moles of C2O42- which = moles of Ca2+ in the blood sample. b) Calculate the Ca2+ conc. in blood, in M then in mg / 100 mL. Redox Titration- Calculation outline - I Volume (L) of KMnO4 Solution a) M (mol/L) Moles of KMnO4 b) Molar ratio Moles of CaC2O4 c) Chemical Formula Moles of Ca+2 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. Redox Titration - Calculation – cont. Solution: a) Calculate the number of moles of MnO4- used in titration. moles MnO-4 = V×M = 2.05×10-3 L×4.88×10-4 moles L-1 =1.00 ×10-6 moles b) Calculate the number of moles of calcium oxalate using balanced chemical equation 25 moles C O -6 22 4 1.00 ×10-6moles MnO-4 × = 2.50 × 10 moles C O 2 4 2 moles MnO4 moles Ca2+ = moles C2O42c) Calculate the concentration of Ca2+ in the blood sample. 2+ −6 moles Ca 2.50 × 10 moles 2+ molarity Ca = = volume blood 1.00 ×10−3 L = 2.50 ×10−3 M Redox Titration - Calculation – cont. Change this M concentration into the correct units: a) Mol Ca+2 per 100. mL (or 0.100 L) of Blood 2.50x10-3 mol Ca+2 x 0.100 L Blood = 2.50 x 10-4 mol Ca+2 L Blood 100. mL Blood 100. mL Blood b) mass (g) of Ca+2 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