Ch 4 Major classes of chemical reactions PDF

Ch. 4 Three Major Classes of Chemical Reactions: Precipitation, Acid-Base Neutralization, and Redox Reactions Chemistry The Molecular Nature of Matter and Change Ninth Edition Martin S. Silberberg and Patricia G. Amateis ©McGraw-Hill Education. All rights reserved. Authorized only for instructor use...

Ch. 4 Three Major Classes of Chemical Reactions: Precipitation, Acid-Base Neutralization, and Redox Reactions Chemistry The Molecular Nature of Matter and Change Ninth Edition Martin S. Silberberg and Patricia G. Amateis ©McGraw-Hill Education. All rights reserved. Authorized only for instructor use in the classroom. No reproduction or further distribution permitted without the prior written consent of McGraw-Hill Education. Three Major Classes of Chemical Reactions 4.1 Solution Concentration and the Role of Water as a Solvent 4.2 Precipitation Reactions 4.3 Acid-Base Reactions 4.4 Oxidation-Reduction (Redox) Reactions 4.5 Elements in Redox Reactions ©McGraw-Hill Education. 4.1.Solution. Water as a Solvent. Aqueous Solutions A solution is a homogeneous mixture: uniform composition at molecular scale in any and all directions. a solute is one of the substances present in quantity smaller than 50%. the solvent is the substance present in the largest quantity. an aqueous solution is a homogeneous mixture where water serves as the solvent. Some solvents disperse the substances into individual molecules by weak forces and interactions. (More on this in Ch. 12) Water is much more reactive, interacting by stronger forces with the substances and even reacting with them in some cases. (More on this in Ch. 9 – 12). ©McGraw-Hill Education. Electron Distribution in Molecules of H2 and H2O Water is a polar molecule, meaning that it has an uneven distribution of electron density around it. The density is higher around O and diminished around H. At submolecular level, water has a negative pole at O and a positive pole half-way between the 2 H atoms. Because of this electric inhomogeneity, water is able to attract both positive ions and negative ions (see next slide). ©McGraw-Hill Education. An Ionic Compound Dissolving in Water: from solid ionic crystal to hydrated individual ions Figure 4.2 NaCl(s) Solvation: surrounding with solvent molecules ©McGraw-Hill Education. aq = aqua Na+(aq) + hydrated Na Cl (aq) hydrated Cl- The Electrical Conductivity of Ionic Solutions Figure 4.3 Electrolyte = a substance that conducts electric current when dissolved in a solvent ©McGraw-Hill Education. © McGraw-Hill Education/Stephen Frisch, photographer Covalent Compounds in Water: Nonelectrolytes Electrolyte = a substance that conducts electric current when dissolved in a solvent. – Why does it conduct electric current? Because it dissociates into solvated ions which possess electric charges. A soluble covalent compound dissolved in water does not typically conduct electricity. It is a nonelectrolyte. – Covalent compounds do not dissociate into solvated ions. Covalent compounds simply dissolve away by interacting and being surrounded by solvent molecules. Examples: alcohol in water, sucrose (sugar) in water. – In cases where the covalent compound generates a conductive solution, either the solute or the solvent molecule dissociates into ions in the presence of the other. Common covalent compounds behaving this way is the acid-base class: 𝐻𝐶𝑙 𝑔 + 𝐻2 𝑂(𝑙) 𝐻𝐶𝑙 𝑎𝑞 ©McGraw-Hill Education. 𝑑𝑖𝑠𝑠𝑜𝑙𝑣𝑖𝑛𝑔 100% 𝑑𝑖𝑠𝑠𝑜𝑐𝑖𝑎𝑡𝑖𝑜𝑛 𝐻𝐶𝑙(𝑎𝑞) 𝐻 + 𝑎𝑞 + 𝐶𝑙 − (𝑎𝑞) © McGraw-Hill Education/Stephen Frisch, photographer Calculating the Number of Moles of Ions in Solution The subscripts of the ions in the chemical formula of an ionic compound indicate the number of moles of each ion produced when the compound dissolves and dissociates. Examples: 1) 1 mol of KBr dissociates into 2 mol of ions: 1 mol of K+ and 1 mol of Br – 𝐻2 𝑂 𝑙 𝐾𝐵𝑟 𝑠 𝐾 + 𝑎𝑞 + 𝐵𝑟 − (𝑎𝑞) 2) 1 mol of CaBr2 dissociates into 3 mol of ions: 1 mol of Ca2+ and 2 mol of Br – 𝐻2 𝑂 𝑙 𝐶𝑎𝐵𝑟2 (𝑠) 𝐶𝑎2+ 𝑎𝑞 + 2𝐵𝑟 − (𝑎𝑞) 3) 1 mol of Al2(SO4)3 dissociates into 5 mol of ions: 2 mol of Al3+ and 3 mol of 𝑆𝑂42− 𝐴𝑙2 𝑆𝑂4 3 (𝑠) ©McGraw-Hill Education. 𝐻2 𝑂 𝑙 2𝐴𝑙3+ 𝑎𝑞 + 3𝑆𝑂42− (𝑎𝑞) © McGraw-Hill Education/Stephen Frisch, photographer Sample Problem 4.1 - Problem Using Molecular Scenes to Depict an Ionic Compound in Aqueous Solution PROBLEM: The beakers shown below contain aqueous solutions of the strong electrolyte potassium sulfate. – (a) Which beaker best represents the compound in solution? (H2O molecules are not shown). – (b) If each particle represents 0.10 mol, what is the total number of particles in solution? ©McGraw-Hill Education. Sample Problem 4.1 – Plan and Solution (a) PLAN: (a) Determine the formula and write an equation for the dissociation of 1 mol of the compound. Potassium sulfate is a strong electrolyte; it therefore dissociates completely in solution. Remember that polyatomic ions remain intact in solution. (b) Count the number of separate particles in the relevant beaker, then multiply by 0.1 mol and by Avogadro’s number. ©McGraw-Hill Education. Sample Problem 4.1 – Solution (b) SOLUTION: (a) The formula is K2SO4, so the equation for dissociation is: K2SO4 (s) 𝑯𝟐 𝑶(𝒍) 2K+ (aq) + SO42- (aq) Ions should be dissolved (separated from the initial crystal) and dissociated. Beakers A and B incorrectly show associated ions. Beaker C represents this correctly. There should be 2 cations for every 1 anion. Beaker D violates this molar ratio, yielding an electric charge imbalance and a charged solution. A solution of ions is electrically neutral, just like the original solid ionic compound (the solute) and the solvent. (b) Beaker C contains 9 particles, 6 K+ ions and 3 SO42- ions. 6.022x1023 particles 9 × 0.1 mol × = 𝟓. 𝟒𝟐𝟎 × 𝟏𝟎𝟐𝟑 𝐩𝐚𝐫𝐭𝐢𝐜𝐥𝐞𝐬 1 mol ©McGraw-Hill Education. Sample Problem 4.2 – Problem and Plan Determining Amount (mol) of Ions in Solution PROBLEM: What amount (mol) of each ion is in each solution? – (a) 5.0 mol of ammonium sulfate dissolved in water – (b) 78.5 g of cesium bromide dissolved in water (cesium is in group 1A, bromine is in group 7A, M(cesium bromide)=212.8 g/mol) – (c) 7.42x1022 formula units of copper(II) nitrate dissolved in water PLAN: Write an equation for the dissociation of 1 mol of each compound. Use this information to calculate the actual number of moles represented by the given quantity of substance in each case. ©McGraw-Hill Education. Sample Problem 4.2 – Solution (a) SOLUTION: (a) The formula is (NH4)2SO4 so the equation for dissociation is: (NH4)2SO4 (s) → 2NH4+ (aq) + SO42- (aq) 2 mol NH4+ 5.0 mol (NH4)2SO4 x = 10. mol NH4+ 1 mol (NH4)2SO4 1 mol SO42− 5.0 mol (NH4)2SO4 x = 5.0 mol SO42− 1 mol (NH4)2SO4 ©McGraw-Hill Education. Sample Problem 4.2 – Solution (b) SOLUTION: (b) The formula is CsBr so the equation for dissociation is: CsBr (s) → Cs+ (aq) + Br- (aq) 1 mol CsBr 1 mol Cs+ 78.5 g CsBr x × = 0.369 mol Cs+ 212.8 g CsBr 1 mol CsBr There is one Cs+ ion for every Br- ion, so the number of moles of Br- is also equal to 0.369 mol. ©McGraw-Hill Education. Sample Problem 4.2 – Solution (c) SOLUTION: (c) The formula is Cu(NO3)2 so the formula for dissociation is: Cu(NO3)2 (s) → Cu2+ (aq) + 2NO3- (aq) 7.42x1022 formula units Cu(NO3)2 1 mol x = 0.123 mol Cu(NO3)2 6.022x1023 formula units 1 mol Cu2+ 0.123 mol Cu(NO3)2 x = 0.123 mol Cu2+ ions 1 mol Cu(NO3)2 There are 2 NO3- ions for every 1 Cu2+ ion, so there are 0.246 mol NO3- ions. ©McGraw-Hill Education. Concentration of Solutions in Molarity Many reactions are carried out in solution because the reactions achieve a higher rate. A solution consists of one or more solutes dissolved in a solvent. The concentration of a solution is given by the quantity of solute present in a given quantity of solution. Molarity (M) is often used to express concentration moles solute Molarity = liters of solution mol Unit: M = L ©McGraw-Hill Education. Sample Problem 4.3 – Problem and Plan Calculating the Molarity of a Solution PROBLEM: What is the molarity of an aqueous solution that contains 53.7g of glycine (H2NCH2COOH) in 495 mL solution? PLAN: Molarity is the number of moles of solute per liter of solution. ©McGraw-Hill Education. Sample Problem 4.3 – Solution SOLUTION: 1 mol glycine n(glycine, mol)=53.7 g glycine× =0.715 mol 75.07 g glycine glycine 𝑚𝑜𝑙 0.715 mol glycine 1000 mL 𝑀 𝑔𝑙𝑦𝑐𝑖𝑛𝑒, = × 𝐿 495 mL soln 1L = 𝟏. 𝟒𝟒 𝐌 𝐠𝐥𝐲𝐜𝐢𝐧𝐞 ©McGraw-Hill Education. Mass-mole-number-volume Relationships in Solution Figure 4.4 Molarity can be used as a conversion factor, either as mol/L or inverse molarity in L/mol. ©McGraw-Hill Education. Sample Problem 4.4 – Problem and Plan Calculating Mass of Solute in a Given Volume of Solution PROBLEM: How many grams of solute are in 1.75 L of 0.460 M sodium hydrogen phosphate solution? PLAN: Calculate the moles of solute using the given molarity and volume. Convert moles to mass using the molar mass of the solute. ©McGraw-Hill Education. Sample Problem 4.4 – Solution SOLUTION: 0.460 mol 1.75 L x = 0.805 mol Na2HPO4 1L 141.96 g Na2HPO4 0.805 mol Na2HPO4 x = 114 g Na2HPO4 1 mol Na2HPO4 (3 sig. figs) ©McGraw-Hill Education. Sample Problem 4.5 – Problem and Plan Determining Amount (mol) of Ions in a Solution PROBLEM: What amount (mol) of each ion is in 35 mL of 0.84 M zinc chloride? PLAN: We write an equation that shows 1 mol of compound dissociating into ions. We convert molarity and volume to moles of zinc chloride and use the dissociation equation to convert moles of compound to moles of ions. ©McGraw-Hill Education. Sample Problem 4.5 – Solution SOLUTION: ZnCl2 (aq) → Zn2+ (aq) + 2Cl– (aq) Converting from volume (mL) and molarity (mol/L) to amount (mol) of compound: 1L 0.84 mol ZnCl2 Amount (mol) of ZnCl2 = 35 mL x × = 1000 mL 1L =2.9 x 10–2 mol ZnCl2 – 2 mol Cl Amount (mol) of Cl– = 2.9 x 10–2 mol ZnCl2 x = 1 mol ZnCl2 = 5.8 x 10–2 mol Cl– 2.9 x 10–2 mol Zn2+ is also present CHECK: The relative numbers of moles of ions are consistent with the formula for ZnCl2: 0.029 mol Zn2+/0.058 mol Cl– = 1 Zn2+ / 2 Cl–, or ZnCl2. ©McGraw-Hill Education. Laboratory Preparation of Molar Solutions Step 1 (not shown). Weigh the solid to the amount desired on a weighing boat. Step 2. Transfer the solid quantitatively to a volumetric flask. Step 3. Add deionized (DI) water below the calibration mark and shake to completely dissolve the solute. Step 4. Add DI water to the final volume. Stopper the flask and shake by flipping repeatedly. Figure 4.5 ©McGraw-Hill Education. © McGraw-Hill Education/Stephen Frisch, photographer Diluting a Solution Diluting = the act of adding solvent to a solution of higher molarity, to decrease the concentration to lower molarity Figure 4.6 ©McGraw-Hill Education. © McGraw-Hill Education/Stephen Frisch, photographer Solving Dilution Problems Diluting = the act of adding solvent to a solution of higher molarity, to decrease the concentration to lower molarity: – The volume of solution increases – The amount (mol) of solute remains the same – Diluted solution has fewer solute particles per unit volume. Solving dilution problems: 𝑛(𝑠𝑜𝑙𝑢𝑡𝑒)𝑖𝑛𝑖𝑡𝑖𝑎𝑙(𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑒𝑑) = 𝑛(𝑠𝑜𝑙𝑢𝑡𝑒)𝑓𝑖𝑛𝑎𝑙(𝑑𝑖𝑙𝑢𝑡𝑒𝑑) = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 Since 𝑛 𝑠𝑜𝑙𝑢𝑡𝑒 = 𝑀 × 𝑉 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 → 𝑀𝑑𝑖𝑙 × 𝑉𝑑𝑖𝑙 = 𝑀𝑐𝑜𝑛𝑐 × 𝑉𝑐𝑜𝑛𝑐 = 𝑛 In general, volumes are not additive because the interaction between particles changes when going from a concentrated to a diluted solution. The volume of solvent needed is NOT Vdil – Vconc. ©McGraw-Hill Education. © McGraw-Hill Education/Stephen Frisch, photographer Sample Problem 4.6 – Problem and Plan Preparing a Dilute Solution from a Concentrated Solution PROBLEM: “Isotonic saline” is a 0.15 M aqueous solution of NaCl. How would you prepare 0.80 L of isotonic saline from a 6.0 M stock solution? PLAN: To dilute a concentrated solution, we add only solvent, so the moles of solute are the same in both solutions. We use the volume and molarity of the dilute solution to calculate the moles of solute. Then we calculate the volume of concentrated solution that contains the same number of moles. ©McGraw-Hill Education. Sample Problem 4.6 - Solution SOLUTION: Mdil x Vdil = # mol solute = Mconc x Vconc Using the volume and molarity for the dilute solution: 0.15 mol NaCl 0.80 L soln x = 0.12 mol NaCl 1 L soln Using the moles of solute and molarity for the concentrated solution: 1 L soln 0.12 mol NaCl x = 0.020 L soln 6.0 mol NaCl Measure out 0.020 L portion of the 6.0 M stock solution with a pipet and dilute it to a final volume of 0.80 L by adding deionized water in a graduated cylinder. ©McGraw-Hill Education. Sample Problem 4.6 - Solution SOLUTION: Mdil x Vdil = # mol solute = Mconc x Vconc ALTERNATIVE APPROACH, using the dilution formula as a linear equation and proportion: 𝑚𝑜𝑙 𝑀𝑑𝑖𝑙 𝐿 𝑉𝑐𝑜𝑛𝑐 𝐿 = 𝑉𝑑𝑖𝑙 𝐿 × 𝑚𝑜𝑙 𝑀𝑐𝑜𝑛𝑐 𝐿 𝑚𝑜𝑙 0.15 1 000 𝑚𝐿 𝐿 𝑉𝑐𝑜𝑛𝑐 𝐿 = 0.80 𝐿 × = 0.020 𝐿 × = 𝟐𝟎. 𝒎𝑳 𝑚𝑜𝑙 1𝐿 6.0 𝐿 ©McGraw-Hill Education. Sample Problem 4.7 – Problem and Plan Visualizing Changes in Concentration PROBLEM: The beaker and circle represents a unit volume of solution. Draw the solution after each of these changes: – (a) For every 1 mL of solution, 1 mL of solvent is added. – (b) One third of the volume of the solution is boiled off. PLAN: Only the volume of the solution changes; the total number of moles of solute remains the same. Find the new volume and calculate the number of moles of solute per unit volume. ©McGraw-Hill Education. Sample Problem 4.7 - Solution SOLUTION: Ndil x Vdil = Nconc x Vconc where N is the number of particles. (a) Ndil = Nconc × (b) Nconc = Ndil × ©McGraw-Hill Education. Vconc Vdil Vdil Vconc = 8 particles × 1 mL 2 mL = 𝟒 𝐩𝐚𝐫𝐭𝐢𝐜𝐥𝐞𝐬 1 mL 3mL = 𝟏𝟐 𝐩𝐚𝐫𝐭𝐢𝐜𝐥𝐞𝐬 = 8 particles × 2Τ Writing Equations for Aqueous Ionic Reactions Molecular equation = a standard, balanced chemical equation made up of electrically neutral formulas. It assumes no reaction participant is dissociated. Example: 2𝐴𝑔𝑁𝑂3 𝑎𝑞 + 𝑁𝑎2 𝐶𝑟𝑂4 𝑎𝑞 → 𝐴𝑔2 𝐶𝑟𝑂4 𝑠 + 2𝑁𝑎𝑁𝑂3 (𝑎𝑞) ©McGraw-Hill Education. © McGraw-Hill Education/Stephen Frisch, photographer Total Ionic Equation The total ionic equation shows all soluble ionic substances dissociated into ions. This gives the most accurate information about all species in solution: 2Ag+ (aq) + 2NO3– (aq) + 2Na+ (aq) + CrO42– (aq) → Ag2CrO4 (s)+ 2Na+ (aq) + 2NO3– (aq) Spectator ions are ions that are not involved in the actual chemical change. Spectator ions appear unchanged on both sides of the total ionic equation: 2Ag+ (aq) +2NO3−(aq)+ 2Na+ (aq) + CrO42– (aq) → Ag2CrO4 (s)+ 2Na+ (aq) +2NO3−(aq) ©McGraw-Hill Education. Net Ionic Equation The net ionic equation eliminates the spectator ions and shows only the actual chemical change: 2Ag+ (aq) + CrO42– (aq) → Ag2CrO4 (s) In a net ionic equation, the electric charges on both sides of the equation must balance, too, in addition to the chemical species. ©McGraw-Hill Education. © McGraw-Hill Education/Stephen Frisch, photographer 4.2 Precipitation Reactions In a precipitation reaction two soluble ionic compounds react to form an insoluble product, called a precipitate. – English meaning of “precipitate”: https://www.thefreedictionary.com/precipitate – To precipitate = to cause a solid substance to be suddenly separated from a solution A precipitate is an insoluble compound, either ionic or covalent. Not all ionic compounds are soluble in water. Because it is a rushed separation of solid from the solution, the resulting solid crystals are very small (microcrystals). In contrast, a slow separation of a solid from a solution yields larger crystals, a process called crystallization. ©McGraw-Hill Education. Precipitation Reactions The precipitate forms through the net removal of ions from solution: 2Ag+ (aq) + CrO42– (aq) → Ag2CrO4 (s) – Mixing a soluble Ag+(aq) solution with a soluble CrO42-(aq) solution yields insoluble Ag2CrO4(s), thus removing Ag+(aq) and CrO42-(aq) from the solution. It is possible for more than one precipitate to form in such a reaction. ©McGraw-Hill Education. Precipitation of PbI2, a Metathesis Reaction Molecular equation: 2NaI ( aq ) + Pb ( NO3 )2 ( aq ) → PbI 2 ( s ) + 2NaNO3 ( aq ) Total ionic equation: 2Na + ( aq ) + 2I − ( aq ) + Pb 2+ ( aq ) + 2NO3− ( aq ) → PbI 2 ( s ) + 2Na + ( aq ) + 2NO3− ( aq ) Net ionic equation: Pb 2+ ( aq ) + 2I − ( aq ) → PbI 2 ( s ) Precipitation reactions are also called double displacement reactions or metathesis reactions. Ions exchange partners and a precipitate forms. Figure 4.8 © McGraw Hill English pronunciation and meaning of metathesis: https://www.thefreedictionary.com/metathesis © McGraw-Hill Education/Stephen Frisch, photographer Precipitation of Calcium Fluoride ©McGraw-Hill Education. © McGraw-Hill Education/Richard Megna, photographer Solubility Rules for Ionic Compounds in Water Soluble Ionic Compounds Insoluble Exceptions All common compounds of Group lA(1) ions (Li+, Na+, None K+, etc.) All common compounds of ammonium ion (NH4+) None All common nitrates (NO3−), acetates (CH3COO− or C2H3O2−), and perchlorates (ClO4−) None All common chlorides (Cl−), bromides (Br−), and iodides (I−) Chlorides, bromides, and iodides of Ag+, Pb2+, Cu+, and Hg22+ : AgX(s), PbX2(s), CuX(s), Hg2X2(s) All common fluorides (F−) PbF2 and fluorides of Group 2A(2): MgF2, CaF2, etc. All common sulfates (SO42−) CaSO4, SrSO4, BaSO4, Ag2SO4, PbSO4 Insoluble Ionic Compounds Soluble Exceptions All common metal hydroxides Group 1A(1) hydroxides: NaOH, KOH, etc. Ca(OH)2, Sr(OH)2, and Ba(OH)2: limited solubility All common carbonates (CO32-) and phosphates (PO43–) Carbonates and phosphates of Group 1A(1) and NH4+ All common sulfides Sulfides of Group 1A(1), Group 2A(2), and NH4+ ©McGraw-Hill Education. Predicting Whether a Precipitate Will Form Steps: 1) Identify the ions present in the reactants. 2) Consider all possible cation-anion combinations. 3) Use the solubility rules to decide whether any of the ion combinations is insoluble. – Any insoluble combination identifies a precipitate that will form. ©McGraw-Hill Education. Sample Problem 4.8 – Problem and Plan Predicting Whether a Precipitation Reaction Occurs; Writing Ionic Equations PROBLEM: Predict whether a reaction occurs when each of the following pairs of solutions are mixed. If a reaction does occur, write balanced molecular, total ionic, and net ionic equations, and identify the spectator ions. – (a) potassium fluoride (aq) + strontium nitrate (aq) → – (b) ammonium perchlorate (aq) + sodium bromide (aq) → PLAN: Note reactant ions, write the possible cation-anion combinations, and use Table 4.1 to decide if the combinations are insoluble. Write the appropriate equations for the process. ©McGraw-Hill Education. Sample Problem 4.8 – Solution (a) SOLUTION: (a) The reactants are KF and Sr(NO3)2. The possible products are KNO3 and SrF2. KNO3 is soluble, but SrF2 is an insoluble combination. Molecular equation: 2KF (aq) + Sr(NO3)2 (aq) → 2KNO3 (aq) + SrF2 (s) Total ionic equation: 2K+ (aq) + 2F− (aq) + Sr2+ (aq) + 2NO3− (aq) → 2K+ (aq) + 2NO3− (aq) + SrF2 (s) K+ and NO3- are spectator ions Net ionic equation: Sr2+ (aq) + 2F− (aq) → SrF2 (s) ©McGraw-Hill Education. Sample Problem 4.8 – Solution (b) SOLUTION: (b) The reactants are NH4ClO4 and NaBr. The possible products are NH4Br and NaClO4. Both are soluble, so no precipitate forms. Molecular equation: NH4ClO4 (aq) + NaBr (aq) → NH4Br (aq) + NaClO4 (aq) Total ionic equation: NH4+ (aq) + ClO4− (aq) + Na+ (aq) + Br− (aq) → NH4+ (aq) + Br− (aq) + Na+ (aq) + ClO4− (aq) All ions are spectator ions and there is no net ionic equation. The compounds remain in solution as solvated ions. ©McGraw-Hill Education. Sample Problem 4.9 - Problem Using Molecular Depictions in Precipitation Reactions PROBLEM: The following molecular views show reactant solutions for a precipitation reaction (with H2O molecules omitted for clarity). – (a) Which compound is dissolved in beaker A: KCl, Na2SO4, MgBr2, or Ag2SO4? – (b) Which compound is dissolved in beaker B: NH4NO3, MgSO4, Ba(NO3)2, or CaF2? – (c) Name the precipitate and the spectator ions when solutions A and B are mixed, and write balanced molecular, total ionic, and net ionic equations for any reaction. – (d) If each particle represents 0.010 mol of ions, what is the maximum mass (g) of precipitate that can form (assuming complete reaction)? ©McGraw-Hill Education. Sample Problem 4.9 – Plan and Solution (a) and (b) PLAN: (a) and (b) Note the number and charge of each kind of ion and use Table 4.1 to determine the ion combinations that are soluble. SOLUTION: (a) Beaker A contains two 1+ ions for each 2- ion. Of the choices given, only Na2SO4 and Ag2SO4 are possible. Na2SO4 is soluble while Ag2SO4 is not. Beaker A therefore contains Na2SO4. (b) Beaker B contains two 1- ions for each 2+ ion. Of the choices given, only CaF2 and Ba(NO3)2 match this description. CaF2 is not soluble while Ba(NO3)2 is soluble. Beaker B therefore contains Ba(NO3)2. ©McGraw-Hill Education. Sample Problem 4.9 – Plan and Solution (c) PLAN: (c) Consider the cation-anion combinations from the two solutions and use Table 4.1 to decide if either of these is insoluble. SOLUTION: (c) The reactants are Ba(NO3)2 and Na2SO4. The possible products are BaSO4 and NaNO3. BaSO4 is insoluble while NaNO3 is soluble. Molecular equation: Ba(NO3)2 (aq) + Na2SO4 (aq) → 2NaNO3 (aq) + BaSO4 (s) Total ionic equation: Ba2+ (aq) + 2NO3− (aq) + 2Na+ (aq) + SO42− (aq) → 2Na+ (aq) + 2NO3− (aq) + BaSO4 (s) Na+ and NO3- are spectator ions Net ionic equation: Ba2+ (aq) + SO42− (aq) → BaSO4 (s) ©McGraw-Hill Education. Sample Problem 4.9 – Plan and Solution (d) PLAN: (d) Count the number of each kind of ion that combines to form the solid. Multiply the number of each reactant ion by 0.010 mol and calculate the mol of product formed from each. Decide which ion is the limiting reactant and use this information to calculate the mass of product formed. SOLUTION: (d) There are 4 Ba2+ particles and 5 SO42- particles depicted. 0.010 mol Ba2+ 1 mol BaSO4 4 particles x × = 0.040 mol BaSO4 1 particle 1 mol Ba2+ 2− 0.010 mol SO 1 mol BaSO4 4 2− 5 SO4 particles x × = 0.050 mol BaSO4 1 particle 1 mol SO42− Ba2+ ©McGraw-Hill Education. Sample Problem 4.9 – Solution (d), Cont’d Ba2+ ion is the limiting reactant, since the given amount yields less BaSO4. 233.4 g BaSO4 0.040 mol BaSO4 x = 9.3 g BaSO4 1 mol BaSO4 ©McGraw-Hill Education. Sample Problem 4.10 – Problem and Plan Calculating Amounts of Reactants and Products in a Precipitation Reaction PROBLEM: Magnesium is the second most abundant metal in sea water after sodium. The first step in its industrial extraction involves the reaction of Mg2+ with Ca(OH)2 to precipitate Mg(OH)2. What mass of Mg(OH)2 is formed when 0.180 L of 0.0155 M MgCl2 reacts with excess Ca(OH)2? PLAN: We are given the molarity (0.0155 M) and the volume (0.180 L) of MgCl2 solution that reacts with excess Ca(OH)2, and we must calculate the mass of the precipitate, Mg(OH)2. (i) Write the balanced equation. (ii) Find the amount (mol) of MgCl2 from its molarity and volume. (iii) Use molar (stoichiometric) ratios to calculate the amount (mol) of Mg(OH)2 precipitated. (iv) Use molar mass of Mg(OH)2 to convert amount (mol) to mass (g). ©McGraw-Hill Education. Sample Problem 4.10 – Plan ©McGraw-Hill Education. Sample Problem 4.10 – Solution SOLUTION: (i) Write the balanced equation. MgCl2(aq) + Ca(OH)2(aq) → Mg(OH)2(s) + CaCl2(aq) (ii) Find the amount (mol) of MgCl2 from its molarity and volume. 0.0155 mol MgCl2 0.180 L MgCl2 x = 0.00279 mol MgCl2 1 L MgCl2 (iii) Use molar (stoichiometric) ratios to calculate the amount (mol) of Mg(OH)2 precipitated. 1 mol Mg(OH)2 0.00279 mol MgCl2 x = 0.00279 mol Mg(OH)2 1 mol MgCl2 ©McGraw-Hill Education. Sample Problem 4.10 – Solution Cont’d (iv) Use molar mass of Mg(OH)2 to convert amount (mol) to mass (g). 58.33 g Mg(OH)2 0.00279 mol Mg(OH)2 x = 0.163 g Mg(OH)2 1 mol Mg(OH)2 CHECK: The answer seems reasonable; rounding to check the math shows that we have (0.2 L) x (0.02 M) = 0.004 mol MgCl2; due to the 1:1 molar ratio, 0.004 mol Mg(OH)2 is produced. ©McGraw-Hill Education. Amount-mass-number Relationships Figure 4.10 ©McGraw-Hill Education. Sample Problem 4.11 – Problem Solving Limiting-Reactant Problems for Precipitation Reactions PROBLEM: Iron(III) hydroxide, used to adsorb arsenic and heavy metals from contaminated soil and water, is produced by reacting aqueous solutions of iron(III) chloride and sodium hydroxide. (a) What mass of iron(III) hydroxide is formed when 0.155 L of 0.250 M iron(III) chloride reacts with 0.215 L of 0.300 M sodium hydroxide? (b) Write a reaction table for this process. ©McGraw-Hill Education. Sample Problem 4.11 –Plan PLAN: From the molarity and volume of each solution, we calculate the amount (mol) of each reactant. Then, we use the molar ratio to find the amount of product [Fe(OH)3] that each reactant forms. The limiting reactant forms fewer moles of Fe(OH)3, which we convert to mass (g) of Fe(OH)3 using its molar mass (see the road map). We use the amount of Fe(OH)3 formed from the limiting reactant in a reaction table. ©McGraw-Hill Education. Sample Problem 4.11 – Solution (a) SOLUTION: (a) FeCl3 (aq) + 3NaOH (aq) → Fe(OH)3 (s) + 3NaCl (aq) 0.250 mol FeCl3 1 mol Fe(OH)3 0.155 L soln x × = 0.0388 mol Fe(OH)3 1 L soln 1 mol FeCl3 0.300 mol NaOH 1 mol Fe(OH)3 0.215 L soln x × = 0.0215 mol Fe(OH)3 1 L soln 3 mol NaOH NaOH is the limiting reactant because the given amount yields less Fe(OH)3. 106.87 g Fe(OH)3 0.0215 mol Fe(OH)3 x = 2.3 g Fe(OH)3 1 mol Fe(OH)3 ©McGraw-Hill Education. Sample Problem 4.11 – Solution (b) (b) The reaction table is constructed using the amount of NaOH to determine the changes, since it is the limiting reactant. Amount (mol) FeCl3(aq) + 3NaOH(aq) → Fe(OH)3(s) + 3NaCl(aq) Initial 0.0388 0.0645 0 0 Change −0.0215 −0.0645 +0.0215 +0.0645 Final 0.0173 0 0.0215 +0.0645 ©McGraw-Hill Education. 4.3 Acids An acid is a substance that produces H+ ions when dissolved in H2O. HX → H+ (aq) + X− (aq) – An acid must have at least a H atom in the formula. – Not all H atoms of an acid may be transferable /acidic. Not all formulas with one or more H atoms are acids. – Acids are covalent compounds, not ionic. Dissociation of dissolved covalent compounds is rare. It happens primarily because of the properties of the water molecule (discussed later in the course). Some acids are strong: they dissociate completely into ions in aqueous solution. Examples: HCl, HNO3, H2SO4 Most acids are weak: they dissociate less than 10%, most of them under 1%. Examples: HF, HNO2, H2SO3, H3PO4 ©McGraw-Hill Education. H+ Ion as a Solvated Hydronium Ion H+ interacts strongly with H2O, forming H3O+ in aqueous solution. We will re-write the chemical equation of dissociation as follows: 𝐻𝐵𝑟 𝑔 + 𝐻2 𝑂 𝑙 → 𝐻3 𝑂+ 𝑎𝑞 + 𝐵𝑟 − (𝑎𝑞) Figure 4.10. The H3O+ ion further surrounds itself with H2O molecules in an interaction known as Hydrogen Bonding (Ch. 12). ©McGraw-Hill Education. Strong and Weak Acids Acids Strong: only 6 in this course Hydrochloric acid, HCl Hydrobromic acid, HBr Hydriodic acid, HI Nitric acid, HNO3 Sulfuric acid, H2SO4 Perchloric acid, HClO4 Weak (a few of many examples): everything else Hydrofluoric acid, HF Phosphoric acid, H3PO4 Acetic acid, CH3COOH (or HC2H3O2) + all organic acids ©McGraw-Hill Education. Bases A base is a substance that produces OH- ions when dissolved in H2O. Production of the OH-(aq) group may occur under different scenarios: 1) A pure compound already contains the OH- group. It must be ionic, and the cation is a metal. Example: NaOH MOH(s) 𝐻2 𝑂(𝑙) M+ (aq) + OH− (aq) – Not all compounds that contain the OH group in the formula are bases. In general, if M is a metal the compound is a base, but there are exceptions. 2) An ionic oxide of group 1(A) or group 2(A). These oxides react with water to form metal hydroxides: 𝐾2 𝑂 𝑠 + 𝐻2 𝑂 𝑙 → 2𝐾 + 𝑎𝑞 + 2𝑂𝐻− (𝑎𝑞) 3) Aqueous solutions of some covalent compounds, such as NH3, are bases. These do not have the OH group but are able to split the water molecule: 𝑁𝐻3 𝑎𝑞 + 𝐻2 𝑂 𝑙 → 𝑁𝐻4+ 𝑎𝑞 + 𝑂𝐻 − 𝑎𝑞 ©McGraw-Hill Education. 𝑦𝑖𝑒𝑙𝑑 𝑖𝑠 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑛 1% Strong and Weak Bases Strong Bases Group 1A(1) hydroxides: Lithium hydroxide, LiOH Sodium hydroxide, NaOH Potassium hydroxide, KOH Rubidium hydroxide, RbOH Cesium hydroxide, CsOH Heavy Group 2A(2) hydroxides: only 3 of them Calcium hydroxide, Ca(OH)2 Strontium hydroxide, Sr(OH)2 Barium hydroxide, Ba(OH)2 Weak Bases: everything else Ammonia, NH3; Magnesium hydroxide, Mg(OH)2 ©McGraw-Hill Education. Acids and Bases as Electrolytes Strong acids and strong bases dissociate completely into ions in aqueous solution. They are strong electrolytes and conduct well in solution. Figure 4.11A ©McGraw-Hill Education. © McGraw-Hill Education/Stephen Frisch, photographer Acids and Bases as Electrolytes, Cont’d Weak acids and weak bases dissociate very little into ions in aqueous solution. They are weak electrolytes and conduct poorly in solution. Figure 4.11B ©McGraw-Hill Education. © McGraw-Hill Education/Stephen Frisch, photographer Sample Problem 4.12 – Problem and Plan Determining the Number of H+ (or OH-) Ions in Solution PROBLEM: How many H+(aq) ions are in 25.3 mL of 1.4 M nitric acid? PLAN: Use the volume and molarity to determine the mol of acid present. Since HNO3 is a strong acid, moles acid = moles H+. ©McGraw-Hill Education. Sample Problem 4.12 - Solution SOLUTION: 1L 1.4 mol HNO3 25.3 mL soln x 3 × = 0.035 mol HNO3 10 mL 1 L soln One mole of H+(aq) is released per mole of nitric acid (HNO3). HNO3 (aq) → H+ (aq) + NO3− (aq) 1 mol H+ 6.022x1023 ions 0.035 mol HNO3 × × = 2.1 x 1022 H+ ions 1 mol HNO3 1 mol ©McGraw-Hill Education. Acid-Base Reaction: Neutralization Acids and bases react with each other easily. The reaction wipes both the “sour” taste of an acid and the “bitter” taste of a base. The reaction also terminates both the corrosive behavior of the acid and the caustic and slippery touch (sensation on the skin) of the base. For these reasons, an acid-base reaction was also called a neutralization reaction centuries ago. An acid-base reaction always produces water. Example: - Molecular equation: 𝐻𝐶𝑙 𝑎𝑞 + 𝑁𝑎𝑂𝐻 𝑎𝑞 → 𝑁𝑎𝐶𝑙 𝑎𝑞 + 𝐻2 𝑂(𝑙) - Total ionic equation: 𝐻 + 𝑎𝑞 + 𝐶𝑙 − 𝑎𝑞 + 𝑁𝑎+ 𝑎𝑞 + 𝑂𝐻− 𝑎𝑞 → 𝑁𝑎+ 𝑎𝑞 + 𝐶𝑙 − 𝑎𝑞 + 𝐻2 𝑂(𝑙) - Net ionic equation: 𝐻 + 𝑎𝑞 + 𝑂𝐻− 𝑎𝑞 → 𝐻2 𝑂(𝑙) Evaporating the water after neutralization crystallizes the salt. ©McGraw-Hill Education. Proton-transfer Process Figure 4.12. Acid-base reactions are double-displacement (metathesis) reactions. ©McGraw-Hill Education. Proton-transfer Process Figure 4.12 ©McGraw-Hill Education. A Gas-forming Reaction with a Weak Acid Carbonic Acid is unstable. Any reaction that produces carbonic acid will instead release CO2(g), visible as fizzing: 𝐻2 𝐶𝑂3 → 𝐶𝑂2 𝑔 + 𝐻2 𝑂(𝑙) Similarly, sulfurous acid decomposes to SO2(g) and H2O(l): [𝐻2 𝑆𝑂3 𝑎𝑞 ] → 𝑆𝑂2 𝑔 + 𝐻2 𝑂(𝑙) Figure 4.13 ©McGraw-Hill Education. © McGraw-Hill Education/Charles Winters/Timeframe Photography, Inc. Sample Problem 4.13 – Problem and Plan Writing Ionic Equations for Acid-Base Reactions PROBLEM: Write balanced molecular, total ionic, and net ionic equations for the following acid-base reactions and identify the spectator ions. (a) hydrochloric acid (aq) + potassium hydroxide (aq) → (b) strontium hydroxide (aq) + perchloric acid (aq) → (c) barium hydroxide (aq) + sulfuric acid (aq) → PLAN: All reactants are strong acids and bases (see Table 4.2). The product in each case is H2O and an ionic salt. Write the molecular reaction in each case and use the solubility rules to determine if the product is soluble or not. ©McGraw-Hill Education. Sample Problem 4.13 – Solution (a) SOLUTION: (a) Molecular equation: HCl (aq) + KOH (aq) → KCl (aq) + H2O (l) Total ionic equation: + − + − H (aq) + Cl (aq) + K (aq) + OH (aq) + − → K (aq) + Cl (aq) + H2O (l) Net ionic equation: + − H (aq) + OH (aq) → H2O (l) Spectator ions are K+ and Cl- ©McGraw-Hill Education. Sample Problem 4.13 – Solution (b) SOLUTION: (b) Molecular equation: Sr(OH)2 (aq) + 2HClO4 (aq) → Sr(ClO4)2 (aq) + 2H2O (l) Total ionic equation: + 2 Sr − + − (aq) + 2OH (aq) + 2H (aq) + 2ClO4 (aq) + − 2 → Sr (aq) + 2ClO4 (aq) + 2H2O (l) Net ionic equation: + − 2H aq + 2OH aq → 2H2O l + − or H (aq) + OH (aq) → H2O(l) Spectator ions are Sr2+ and ClO4- ©McGraw-Hill Education. Sample Problem 4.13 – Solution (c) SOLUTION: (c) Molecular equation: Ba(OH)2 (aq) + H2SO4 (aq) → BaSO4 (s) + 2H2O (l) Total ionic equation: + 2 Ba − + (aq) + 2OH (aq) + 2H (aq) + SO4 → BaSO4(s) + H2O (l) 2− (aq) The net ionic equation is the same as the total ionic equation since there are no spectator ions. This reaction is both a neutralization reaction and a precipitation reaction. ©McGraw-Hill Education. Sample Problem 4.13A - Problem Writing Proton-Transfer Equations for Acid-Base Reactions PROBLEM: Write balanced total and net ionic equations for the following reactions and use curved arrows to show how the proton transfer occurs. – (a) hydriodic acid (aq) + calcium hydroxide (aq) → Give the name and formula of the salt present when the water evaporates. – (b) potassium hydroxide (aq) + propionic acid (aq) → Note that propionic acid is a weak acid. Be sure to identify the spectator ions in this reaction. ©McGraw-Hill Education. Sample Problem 4.13A – Plan and Solution (a) PLAN: In (a) the reactants are a strong acid and a strong base. The acidic species is therefore H3O+, which transfers a proton to the OH- from the base. SOLUTION: (a) Total Ionic Equation: (H+ transferred to OH-) 2H3O+ (aq) + 2I− (aq) + Ca2+ (aq) + 2OH− (aq) → 2I− (aq) + Ca2+ (aq) + 4H2O (l) Net Ionic Equation: H3O+ (aq) + OH− (aq) → 2H2O (l) When the water evaporates, the salt remaining is CaI2, calcium iodide. ©McGraw-Hill Education. Sample Problem 4.13A – Plan and Solution (b) PLAN: In (b) the acid is weak; therefore it does not dissociate much and largely exists as intact molecules in solution. SOLUTION: (b) Total Ionic Equation: (H+ transferred to OH-) K+ (aq) + OH− (aq) + CH3CH2COOH (aq) → K+ (aq) + H2O (l) + CH3CH2COO− (aq) Net Ionic Equation: CH3CH2COOH (aq) + OH− (aq) → CH3CH2COO− (aq) + H2O (l) K+ is the only spectator ion in the reaction. ©McGraw-Hill Education. Sample Problem 4.14 – Problem and Plan Calculating the Amounts of Reactants and Products in an Acid-Base Reaction PROBLEM: A 0.10 M HCl solution is used to simulate the acid concentration of the stomach. How many liters of “stomach acid” react with a tablet containing 0.10 g of magnesium hydroxide? PLAN: Write a balanced equation and convert the mass of Mg(OH)2 to moles. Use the mole ratio to determine the moles of HCl, then convert to volume using molarity. ©McGraw-Hill Education. Sample Problem 4.14 - Solution SOLUTION: Mg(OH)2 (s) + 2HCl (aq) → MgCl2 (aq) + 2H2O (l) 1 mol Mg(OH)2 0.10 g Mg(OH)2 x = 1.7 x 10−3 mol Mg(OH)2 58.33 g Mg(OH)2 2 mol HCl 1.7 x 10−3 mol Mg(OH)2 x = 3.4 x 10−3 mol HCl 1 mol Mg(OH)2 1L HCl soln 3.4x10−3 mol HCl x = 3.4 x 10−2 L HCl 0.10 mol HCl ©McGraw-Hill Education. Acid-Base Titrations In a titration, the concentration of one solution is used to determine the concentration of another, based on the stoichiometry of reaction between the reactants present in the two solutions. In an acid-base titration, a standard solution of base is usually added to a sample of acid of unknown molarity (or the other way around). At the equivalence point, the mol of H+ from the acid equals the mol of OH- ion produced by the base. – Amount of H+ ion in flask = amount of OH- ion added An acid-base indicator has different colors in acid and base. It is used as a proxy to detect the equivalence point. The end point occurs when there is a slight excess of base and the indicator changes color permanently. ©McGraw-Hill Education. An Acid-base Titration Figure 4.15. The total ionic equation is shown in caption. ©McGraw-Hill Education. © McGraw-Hill Education/Stephen Frisch, photographer Sample Problem 4.15 – Problem and Plan Finding the Concentration of Acid from a Titration PROBLEM: A 50.00 mL sample of H2SO4 is titrated with 0.1524 M NaOH. The buret reads 0.55 mL at the start and 33.87 mL at the end point. Find the molarity of the H2SO4 solution. PLAN: Write a balanced equation for the reaction. Use the volume of base to find mol OH-, then mol H+ and finally M for the acid. ©McGraw-Hill Education. Sample Problem 4.15 - Solution SOLUTION: 2NaOH (aq) + H2SO4 (aq) → Na2SO4 (aq) + 2H2O (l) volume of base = 33.87 mL – 0.55 mL = 33.32 mL 1L 0.1524 mol NaOH 33.32 mL soln x 3 × = 5.078 x 10−3 mol NaOH 10 mL 1 L soln Since 1 mol of H2SO4 reacts with 2 mol NaOH, the amount of H2SO4 is 1 mol H2SO4 5.078x10−3 mol NaOH× =2.539x10−3 mol H2SO4 2 mol NaOH 2.539x10−3 mol H2SO4 103 mL × = 0.05078 M H2SO4 50.00 mL 1L ©McGraw-Hill Education. 4.4 Oxidation–Reduction (Redox) Reactions The reactions in which electrons are transferred from one reactant to the other are called reduction-oxidation reactions, or redox reactions. Many redox reactions involve the reaction of a substance with oxygen: 4 Fe(s) + 3 O2(g) → 2 Fe2O3(s) (oxidation) 2 C8H18(l) + 25 O2(g) → 16 CO2(g) +18 H2O(g) (combustion) However, redox reactions need not to involve O2: 2 Na(s) + Cl2(g) → 2 NaCl(s) C2H4(g) + Br2(g) → C2H4Br2(l) Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s) © 2017 Pearson Education, Inc. Oxidation–Reduction (Redox) Reactions Redox is a portmanteau for reduction-oxidation. Of what electrons and electron-transfer are we talking about in a redox reaction? In any redox reaction, at least one reactant loses electrons. These electrons are transferred to another reactant which gains exactly the number of electrons lost. There must be perfect electron transfer balance and no electric charge accumulation or loss during a redox reaction. This critical information is used to balance redox reactions. Oxidation = loss of electrons. Reducing species is oxidized. Reduction = gain of electrons. Oxidizing species is reduced. © 2017 Pearson Education, Inc. The Redox Process in Compound Formation Figure 4.16 ©McGraw-Hill Education. Redox Terminology 1 𝑀𝑔 + 𝑂2 → 𝑀𝑔𝑂 2 𝑖𝑜𝑛𝑖𝑐 𝑐𝑜𝑚𝑝𝑜𝑢𝑛𝑑 𝑚𝑎𝑑𝑒 𝑢𝑝 𝑜𝑓 𝑀𝑔2+ 𝑎𝑛𝑑 𝑂2− Oxidation is the loss of electrons: 𝑀𝑔 → 𝑀𝑔2+ + 2𝑒 − 1 2 Reduction is the gain of electrons: 𝑂2 + 2𝑒 − → 𝑂2− The oxidizing agent is the reactant doing the oxidizing: O2. Therefore this reactant gains electrons and is reduced. The reducing agent is the reactant doing the reducing: Mg. Therefore this reactant loses electrons and is oxidized. The transferred electrons are never free, as shown in the breakdown half-reactions above. Oxidation and reduction must occur together. None of the half-reaction occurs in isolation. The electrons are passed on by direct contact between Mg and O2 (collision). ©McGraw-Hill Education. Electron Clouds in Chemical Bonding For ionic compounds such as MgO or NaCl it is easy to understand the electron transfer process. A covalent bond in which the shared electron pair is not shared equally, but remains closer to one atom than the other, is a polar covalent bond. Polar covalent bonds are much more common than either pure ionic or pure covalent bonds. ©McGraw-Hill Education. Na+ Cl- H Cl NaCl HCl H2 H H Summary of Terminology for Redox Reactions Figure 4.17 ©McGraw-Hill Education. Oxidation Numbers (a.k.a. Oxidation States) Chemists assign a number to each element in a reaction called an oxidation number (O.N.) that allows them to determine the electron flow (transfer) in the reaction. In this example, we say that the oxidation state of Carbon changes from 0 to +4, and the oxidation state of Sulfur changes from 0 to -2. We say that Carbon loses electrons and is oxidized. We say that Sulfur gains electrons and is reduced. © 2017 Pearson Education, Inc. Oxidation Numbers (a.k.a. Oxidation States) Even though they look like them, oxidation states (or numbers) are not ion charges! There are no ions in CS2, it is a covalent molecule. – Oxidation numbers (or states) are imaginary charges assigned based on a set of rules. See next slides for rules. – Ion charges are real, measurable charges. For reactions that are NOT simple additions of metal + nonmetal or do not involve O2, we need a method for calculating how the electrons are transferred. This method is based on the oxidation numbers. © 2017 Pearson Education, Inc. Rules for Assigning an Oxidation Number (O.N.) 1. For an element (Na, C, O2, Cl2, P4, etc.): O.N. = 0 2. For a monatomic ion: O.N. = ion charge (with the sign before the numeral) 3. The sum of O.N. values for the atoms in a molecule or formula unit of a compound equals zero. The sum of O.N. values for the atoms in a polyatomic ion equals the ion’s charge. Rules for Specific Atoms or Periodic Table Groups 1. For Group 1A(1): O.N. = +1 in all compounds 2. For Group 2A(2): O.N. = +2 in all compounds 3. For hydrogen: O.N. = +1 in combination with nonmetals O.N. = −1 in combination with metals and boron (e.g. NaH) 4. For fluorine: O.N. = −1 in all compounds 5. For oxygen: O.N. = −1 in peroxides (e.g. H2O2) O.N. = −2 in all other compounds (except with F) 6. For Group 7A(17): O.N. = −1 in combination with metals, nonmetals (except O), and other halogens lower in the group ©McGraw-Hill Education. Sample Problem 4.16 – Problem and Plan Determining the Oxidation Number of Each Element in a Compound (or Ion) PROBLEM: Determine the oxidation number (O.N.) of each element in these species: (a) zinc chloride (b) sulfur trioxide (c) nitric acid (d) dichromate ion PLAN: Use Table 4.4 to assign each atom an O.N. ©McGraw-Hill Education. Sample Problem 4.16 - Solution (a) ZnCl2. The O.N. of each chloride ion is −1 for a total of –2, so the O.N. of Zn must be +2 since the sum of O.N.s must equal zero for a compound. (b) SO3. The O.N. of each oxygen is −2 for a total of −6. The O.N.s must add up to zero, so the O.N. of S is +6. (c) HNO3. The O.N. of H is +1 and the O.N. of each O is −2 for a total of −6. Therefore, the O.N. of N is +5. (d) Cr2O72-. The sum of the O.N. values in a polyatomic ion equals the ion's charge. The O.N. of each O is −2 so the total for seven O atoms is −14. Therefore, each Cr must have an O.N. of +6 in order for the sum of the O.N.s to equal the charge of the ion. ©McGraw-Hill Education. Stoichiometry in redox titration Figure 4.18 ©McGraw-Hill Education. © McGraw-Hill Education/Stephen Frisch, photographer Sample Problem 4.18 - Problem Finding the Amount of Reducing Agent by Titration PROBLEM: To measure the Ca2+ concentration in 1.00 mL of human blood Na2C2O4 solution is added precipitate the Ca2+ as CaC2O4. The solid is dissolved in dilute H2SO4 to release C2O42-, and 2.05mL of 4.88x10-4 M KMnO4 is required to reach the end point. The balanced equation is 2 KMnO4 (aq) + 5CaC2O4 (s) + 8H2SO4 (aq) → 2MnSO4 (aq) + K2SO4 (aq) + 5CaSO4 (s) +10CO2 (g) + 8H2O (l) Calculate the amount (mol) of Ca2+ in 1.00 mL of blood. ©McGraw-Hill Education. Sample Problem 4.19 - Plan ©McGraw-Hill Education. Sample Problem 4.19 - Solution SOLUTION: 1L 4.88x10−4 mol KMnO4 2.05 mL soln x 3 × = 1.00 x 10−6 10 mL 1L soln mol KMnO4 5 mol CaC2O4 1.00 x 10−6 mol KMnO4 x = 2.50 x 10−6 mol 2 mol KMnO4 CaC2O4 1 mol Ca2+ 2.50 x 10−6 mol CaC2O4 x = 2.50 x 10−6 mol Ca2+ 1 mol CaC2O4 ©McGraw-Hill Education. 4.5 Chemical Elements in Redox Reactions Combination Reactions – Two or more reactants combine to form a new compound: X+Y→Z Decomposition Reactions (thermal or electrolytic) – A single compound Z decomposes to form two or more products: Z→X+Y Single Displacement Reactions X + YZ → XZ + Y Combustion: reaction with O2 to produce heat and oxides ©McGraw-Hill Education. Combining Elements to Form an Ionic Compound Figure 4.19 ©McGraw-Hill Education. © McGraw-Hill Education/Stephen Frisch, photographer Decomposition of Mercury(II) oxide Figure 4.20 ©McGraw-Hill Education. © McGraw-Hill Education/Stephen Frisch, photographer Lithium Displaces H2 from Water Figure 4.21 ©McGraw-Hill Education. © McGraw-Hill Education/Stephen Frisch, photographer Displacement of H2 from Acid by Nickel Figure 4.22 ©McGraw-Hill Education. © McGraw-Hill Education/Stephen Frisch, photographer Copper Displacement of Silver Figure 4.23 ©McGraw-Hill Education. © McGraw-Hill Education/Stephen Frisch, photographer The Activity Series of the Metals Figure 4.24 ©McGraw-Hill Education. The Activity Series of the Halogens Figure 4.24 © McGraw Hill Combustion Reaction Combustion: reaction with O2 to produce heat and oxides – Must have O2(g) as reactant – Typically refers to Carbon and Hydrogen rich compounds as the other reactant, yielding CO2 and H2O – Example: stovetop burner 𝐶𝐻4 𝑔 + 2𝑂2 𝑔 → 𝐶𝑂2 𝑔 + 2𝐻2 𝑂(𝑔) ©McGraw-Hill Education. Sample Problem 4.19 – Problem and Plan Identifying the Type of Redox Reaction PROBLEM: Classify each of the following redox reactions as a combination, decomposition, or displacement reaction. Write a balanced molecular equation for each, as well as total and net ionic equations for part (c), and identify the oxidizing and reducing agents: (a) magnesium (s) + nitrogen (g) → magnesium nitride (s) (b) hydrogen peroxide (l) → water (l) + oxygen gas (c) aluminum (s) + lead(II) nitrate (aq) → aluminum nitrate (aq) + lead (s) PLAN: Combination reactions combine reactants, decomposition reactions involve more products than reactants, and displacement reactions have the same number of reactants and products. ©McGraw-Hill Education. Sample Problem 4.19 - Solution SOLUTION: (a) This is a combination reaction, since Mg and N2 combine. Mg is the reducing agent; N2 is the oxidizing agent. (b) This is a decomposition reaction, since H2O2 breaks down. H2O2 is both the reducing and the oxidizing agent. (c) This is a displacement reaction, since Al displaces Pb2+ from solution. Al is the reducing agent; Pb(NO3)2 is the oxidizing agent. The total ionic equation is: 2Al (s) + 3Pb2+ (aq) + 6NO3− (aq) → 2Al3+ (aq) + 6NO3− (aq) + 3Pb (s) The net ionic equation is: 2Al (s) + 3Pb2+ (aq) → 2Al3+ (aq) + 3Pb (s) ©McGraw-Hill Education.