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Unit B Chapter 4 Electrolytes Precipitation (aka Double Replacement or Metathesis) Reactions Acids & Bases Neutralization Reactions Oxidation Reduction Concentration of Solutions Solution Stoichiometry Sucrose with Sulfuric Acid 1. put 2-3 pellets NaOH in water with universal indicator 1) Why this Color? 2) What was the temperature? the pH? 2 Just what was that sugar reaction? • • • Sucrose or common table sugar was mixed with concentrated sulfuric acid. Soon an exothermic reaction takes places during which a column of carbon rises from the beaker and a cloud of steam is produced. Concentrated sulfuric acid acts as a catalyst to dehydrate sucrose to produce carbon and water. The heat of the reaction vaporizes the water, and the gas causing the column of carbon puff up, just like gases during cooking cause a cake to rise. The name carbohydrate derives from the formula of sugars such as sucrose, C12H22O11 (notice the 2:1 H:O ratio C12(H2O)11) in which the formula appears to be a hydrate of carbon. C12H22O11(s) → 11H2O(g) + 12C(s) 3 Reactions in Aqueous Solutions Unit B Chapter 4 slide view 4 133 g of aluminum chloride in 1 L of water would be considered (select all that apply) 1. 2. 3. 4. 5. 6. soluble ionized dissociated aqueous concentrated dilute 7. weak electrolyte 8. strong electrolyte 9. acidic 5 133 g of AlCl3 in 1 L of water would be 3+ − considered AlCl3 → Al + 3Cl 1. 2. 3. 4. 5. 6. soluble ionized dissociated aqueous concentrated dilute generally <0.1 M is considered dilute 7. weak electrolyte 8. strong electrolyte 9. acidic (for reasons you don’t know yet…) slide show 6 Electrolytes • Compounds that dissolve and dissociate into ions Strong or weak = amount ionized ✓ Concentrated or dilute = amount dissolved ✓ • Electrolytes are Soluble ionic compounds ✓ Acids or bases ✓ • Molecular compounds tend to be non-electrolytes 7 Dissolving & Dissociating • When ionic compounds dissolve they dissociate ✓ Observe the orientation of the water molecules as the solvate around the dissociated ions • When molecular compounds dissolve, they do not dissociate Slide View 8 Which solution represented below is a nonelectrolyte? 1 2 3 9 Which solution represented below is a nonelectrolyte? 1. • • 1 2 3 there are no ions in the solution Which representation would conduct electricity the best? 10 Rank the following in order of increasing conductivity in water: (equal moles of each substance put in the water) C6H12O6, KBr, HF, Zn(OH)2 Least conductive → more conductive 1. C6H12O6 < KBr < HF < Zn(OH)2 2. Zn(OH)2 < C6H12O6 < HF < KBr 3. C6H12O6 < Zn(OH)2 < HF < KBr 4. C6H12O6 < HF < Zn(OH)2 < KBr 5. C6H12O6 < Zn(OH)2 < KBr < HF 6. KBr < Zn(OH)2 < HF < C6H12O6 11 Rank the following in order of increasing conductivity in water: (equal moles of each substance put in the water) C6H12O6, KBr, HF, Zn(OH)2 Remember that conductivity is related to the amount of ions in solution 1. C6H12O6 < KBr < HF < Zn(OH)2 2. Zn(OH)2 < C6H12O6 < HF < KBr 3. C6H12O6 < Zn(OH)2 < HF < KBr • Molecular sugar is least conductive, next is zinc hydroxide, because although it is a base (and actually a strong base), it is very insoluble, hydrofluoric acid is weak, and KBr is a soluble salt. • Time to memorize the 7 strong acids: HCl, HBr, HI, HNO3, H2SO4, HClO3, HClO4 4. C6H12O6 < Zn(OH)2 < KBr < HF 5. KBr < Zn(OH)2 < HF < C6H12O6 12 Rank the following in order of increasing conductivity in water: (equal moles of each substance put in the same amount of water) C3H7OH, KCl, HClO, AlCl3 1. C3H7OH < HClO < KCl < AlCl3 2. AlCl3 < C3H7OH < KCl <HClO 3. HClO < C3H7OH < KCl < AlCl3 4. C3H7OH < HClO < KCl = AlCl3 5. C3H7OH < KCl < AlCl3 < HClO 6. C3H7OH < HClO < AlCl3 < KCl 13 Rank the following in order of increasing conductivity in water: (equal moles of each substance put in the water) C3H7OH, KCl, HClO, AlCl3 1. C3H7OH < HClO < KCl < AlCl3 • 2. C3H7OH is an alcohol, it is molecular and therefore least conductive. HClO, perchlorous acid is a weak acid. KCl, AlCl3 are salts and you might think are equally conductive, but because AlCl3 dissociates into 4 ions (Al+3 and 3 Cl−) it produces more ions per equal concentration and thus is a more conductive solution. AlCl3 < C3H7OH < KCl <HClO 3. HClO < C3H7OH < KCl < AlCl3 4. C3H7OH < HClO < KCl = AlCl3 5. C3H7OH < KCl < AlCl3 < HClO • • 14 Calculate the concentration of chloride ion when 5.8 g of sodium chloride and 4.8 g of magnesium chloride is dissolved to produce 400 ml of solution. NaCl 58.5g/mol No Calculator MgCl2 95.21 1. 2. 3. 4. 0.05 M 0.10 M 0.15 M 0.20 M 5. 6. 7. 8. 0.50 M 2.0 M 0.40 M 0.75 M 15 Calculate the concentration of chloride ion when 5.8 g of sodium chloride and 4.8 g of magnesium chloride is dissolved to produce 400 ml of solution. 1. 0.05 M No Calculator 5. 0.50 M 2. 0.10 M 6. 2.0 M 3. 0.15 M 7. 0.40 M 4. 0.20 M 8. 0.75 M ⎛ 1mol ⎞ − 5.8g ⎜ = 0.1molNaCl = 0.1molCl ⎟ ⎝ 58g ⎠ ⎛ 2Cl ⎞ ⎛ 1mol ⎞ − 4.8g ⎜ = 0.05molMgCl2 × ⎜ = 0.1molCl ⎟ ⎝ 95g ⎠ ⎝ 1MgCl2 ⎟⎠ − ⎛ 0.2molCl ⎞ − = 0.5MforCl ⎜⎝ 0.4 L ⎟⎠ − 16 When 30. ml of 0.50 M K2SO4 are added to 70. mL of 0.50 M of KMnO4, the resulting concentration of potassium is 1. 1.0 2. 0.85 3. 0.65 4. 0.50 5. 0.33 No Calculator K2SO4 174 g/mol KMnO4 158 g/mol 17 When 30. ml of 0.50 M K2SO4 are added to 70. mL of 0.50 M of KMnO4, the resulting concentration of potassium is 1. 2. 3. 4. 5. 1.0 0.85 0.65 0.50 0.33 No Calculator ⎛ 2K ⎞ + (30ml)(0.5M ) = 15mmolK 2 SO4 × ⎜ = 30mmolK ⎟ ⎝ 1K 2 SO4 ⎠ + (70ml)(0.5M ) = 35mmolKMnO4 = 35mmolK + ⎛ 65mmol ⎞ + ⎜⎝ ⎟⎠ = 0.65molarK 100ml 18 Write the reaction that best represents the dissolving of magnesium chloride in water 1. MgCl → Mg(aq) + Cl(aq) + − 2. MgCl → Mg + Cl 2+ − 3. MgCl2 → Mg + (Cl )2 2+ − 4. MgCl2 → Mg + Cl 2 5. MgCl2 → Mg + Cl2 2+ − 6. MgCl2 → Mg + 2Cl 19 Write the reaction that best represents the dissolving of magnesium chloride in water 1. 2. 3. 4. 5. 6. • • MgCl → Mg(aq) + Cl(aq) + − MgCl → Mg + Cl MgCl2 → Mg2+ + (Cl−)2 2+ − MgCl2 → Mg + Cl 2 MgCl2 → Mg + Cl2 MgCl2 → Mg2+ + 2Cl− Be sure you know the correct charges, and show the charge on the dissociated ions. Remember that multiple amounts of the same ion separate from each other, 2Cl− not Cl2− 20 Write the reaction that best represents the dissolving of perchloric acid in water + − 1. HClO4 → H + ClO4 + − 2. HClO4 → H + 4ClO + − 2− 3. HClO4 → H + Cl + 4O 21 Write the reaction that best represents the dissolving of perchloric acid in water + − 1. HClO4 → H + ClO4 + − 2. HClO4 → H + 4ClO + − 2− 3. HClO4 → H + Cl + 4O 22 Write the reaction that best represents the dissolving of chlorous acid in water 1. 2. 3. 4. 5. + H HClO2 → + HClO2 → H + HClO2 → H + HClO2 ! H + HClO2 ! H + + + + + ClO2 − 2ClO − 2− Cl + 2O − ClO2 − 2ClO − 23 Write the reaction that best represents the dissolving of chlorous acid in water 1. 2. 3. • • 4. • • 5. HClO2 → H+ + ClO2− HClO2 → H+ + 2ClO− HClO2 → H+ + Cl− + 2O2− Polyatomic ions do NOT break up in aqueous solutions. The polyatomic ion, chlorite ion does not break apart in solution. HClO2 ! H+ + ClO2− This is even better HClO2 ← This is a weak acid so you should use the double arrows to indicate only some of the acid ionizes. HClO2 ! H+ + 2ClO− → H+ + ClO − 2 24 In the following reaction, which ions are spectator ions: AgNO3 + NaCl → NaNO3 + AgCl ? 1. Ag+ and NO3− 2. Na+ and Cl− 3. − Cl and NO3 4. + Na and NO3 − Can you do the problem without your solubility table? − 5. Ag+ and Cl− 25 In the following reaction, which ions are spectator ions: AgNO3 + NaCl → NaNO3 + AgCl ? 1. Ag+ and NO3− 2. Na+ and Cl− 3. − Cl 4. + Na • and NO3 − Can you do the problem without your solubility table? and NO3 Since silver chloride is the precipitate, the nitrate and alkali salts are the spectator ions. − 5. Ag+ and Cl− 26 Identify the precipitate when sodium hydroxide is combined with aluminum nitrate. 1. NaNO3 2. AlOH3 Do the problem without your solubility table. 3. NaOH 4. AlOH 5. Al(OH)3 27 Identify the precipitate when sodium hydroxide is combined with aluminum nitrate. 1. 2. 3. 4. 5. • • NaNO3 AlOH3 Do the problem without NaOH your solubility table. AlOH Al(OH)3 Alkali nitrates will always be soluble. By process of elimination, the precipitate must be the aluminum hydroxide. 28 Which of the compounds listed below are insoluble: 1. NiCO3 5. ZnS 2. Ba(NO3)2 6. K2CrO4 3. C6H12O6 7. Al2(SO4)3 4. Na3PO3 8. SnF2 Do the problem without your solubility table. 29 Which of the compounds listed below are insoluble: 1. NiCO3 2. Ba(NO3)2 3. C6H12O6 4. Na3PO3 5. ZnS Do the problem without your solubility table. 6. K2CrO4 7. Al2(SO4)3 8. SnF2 • • • • Remember that nitrates and alkali salts are always soluble. Sugar is also of course soluble. Sulfates are usually soluble Carbonates and sulfides are usually NOT soluble 30 In the reaction done in water, shown below, which element is oxidized? NiCl2 + Al → AlCl3 + Ni 1. 2. 3. 4. 5. Ni Balance this equation into a balanced net Cl ionic equation. O Al None of them are since there is no oxygen in the reaction. 31 In the reaction below, which element is oxidized? NiCl2 + Al → AlCl3 + Ni Turn this equation into a balanced net ionic equation. 1. 2. 3. 4. 5. Ni 3 NiCl2 + 2Al → 2AlCl3 + 3Ni Cl or Ni2+ + Al → Al3+ + Ni O 3Ni2+ + 2Al → 2Al3+ + 3Ni Al None of them are since there is no oxygen in the reaction. • Aluminum loses three electrons per atom as it changes from atom form to Al3+ ion form. 32 For the reaction below, select the oxidizing agent (the substance that causes oxidation of some other substance): Fe2O3(s) + 3H2(g) → Fe(s) + H2O(L) 1. O 2. H 3. Fe 4. H2O 5. Fe2O3 33 For the reaction below, select the oxidizing agent: Fe2O3(s) + 3H2(g) → Fe(s) + H2O(L) 1. 2. 3. 4. 5. • • O H Fe H 2O Fe2O3 The oxidation number of iron changes from 3+ as a reactant to 0 as a product, gaining electrons, and is therefore reduced. Thus the substance that the iron is part of is the oxidizing agent. 34 Which reaction below is not an oxidation reduction reaction? 1. 2+ 3Co + 2Al → 3Co + 3+ 2Al 2. 2Na + 2H2O → 2NaOH + H2 3. 2HCl + Zn → ZnCl2 + H2 4. 2HNO3 + Mg(OH)2 → 2H2O + Mg(NO3)2 5. CH4 + 2O2 → CO2 + 2H2O 35 Which reaction below is not an oxidation reduction reaction? 1. 3Co2+ + 2Al → 3Co + 2Al3+ 2. 2Na + 2H2O → 2NaOH + H2 3. 2HCl + Zn → ZnCl2 + H2 4. 2HNO3 + Mg(OH)2 → 2H2O + Mg(NO3)2 5. CH4 + 2O2 → CO2 + 2H2O • Double replacement reactions are never oxidation reduction reactions since none of the atoms change oxidation numbers. 36 LAD B3 Redox Titration 37 Reading the label of a cobalt supplement, the bottle tells us that each there are 35 mg of elemental cobalt delivered in 150 mg of cobalt(II) acetate in a 457 mg tablet. Is the cobalt(II) acetate in the pill a hydrate (if so, with how many water?) or anhydrate? 38 The iron pill bottle tells us that each there are 68 mg of elemental iron delivered in 338 mg of iron(II) sulfate. Is the iron sulfate in the pill a hydrate (if so,with how many water?) or is it an anhydrate? Two ways to think about this…. ✓ Yes it is a hydrate, 7 water 1. Set up a ratio to find the mass of of the iron compound in the tablet ✓ 68g/338g = 55.85g/x solve for x = 277.6g then 2. Subtract the mass caused by the anhydrate, FeSO4 ✓ 277.6g - 151.92g = 126g water 3. Determine the moles of water ✓ 126 / 18 = 7 H2O ✓ Yes it is a hydrate, 7 water 1. Set up a ratio to find the mole of the iron in 68 mg ✓ 68/55.85 = 1.218 mole Fe in FeSO4 = mole of FeSO4 2. Find the MM of the FeSO4 compound in the tablet ✓ 338 g / 1.218 mol = 277 g/mol 3. Determine the mass, then moles of this compound that must be water ✓ 277 g - 152 g = 125 g ✓ 125 / 18 = ~7 H2O 39 Solution Stoichiometry If you’re not part of the solution, you’re part of the precipitate 40 24 ml of a 0.10 M solution of Al(NO3)3 is combined with 24 ml of 0.10 M solution of Na2CO3. • Write out balanced “overall” equation. • Formulas in black, balancing in color. • Be sure and indicate the precipitate by circling in color. 41 24 ml of a 0.10 M solution of Al(NO3)3 is combined with 24 ml of 0.10 M Na2CO3. • Write out balanced “overall” equation on your whiteboard. • 3Na2CO3 + 2Al(NO3)3 → Al2(CO3)3(ppt) + 6NaNO3 • Convert this to a net ionic equation 42 24 ml of a 0.10 M solution of Al(NO3)3 is combined with 24 ml of 0.10 M Na2CO3. • 3Na2CO3 + 2Al(NO3)3 → Al2(CO3)3(ppt) + 6NaNO3 2− + 3+ 2Al • 3CO3 → Al2(CO3)3(ppt) • sodium and nitrate ions are spectator ions 43 24 ml of a 0.10 M solution of Al(NO3)3 is combined with 24 ml of 0.10 M Na2CO3. 3Na2CO3 + 2Al(NO3)3 → Al2(CO3)3(ppt) + 6NaNO3 What is the mass of the dried precipitate? 1. 7.8 g 2. 70.2 g 3. 0.19 g 4. 0.28 g 5. 0.56 g MM (g/mol) 3Na2CO3 = 106 Al(NO3)3 = 213 Al2(CO3)3 = 234 NaNO3 = 85 6. 187 g 7. 842 g 8. 1123 g 9. 1685 g 44 24 ml of a 0.10 M solution of Al(NO3)3 is combined with 24 ml of 0.10 M Na2CO3. 3Na2CO3 + 2Al(NO3)3 → Al2(CO3)3(ppt) + 6NaNO3 What is the mass of the dried precipitate? #3 0.19 g • 2.4 mmol each, and because of the 3:2 ratio between the reactants, the 3Na2CO3 limits 1Al2 (CO3 )3 234g 2.4mmol × = 0.8mmol × = 187mg 3Na2CO3 1mol • which equals 0.19 g Al2(CO3)3 MM (g/mol) 3Na2CO3 = 106 Al(NO3)3 = 213 Al2(CO3)3 = 234 NaNO3 = 85 45 24 ml of a 0.10 M solution of Al(NO3)3 is combined with 24 ml of 0.10 M Na2CO3. 3Na2CO3 + 2Al(NO3)3 → Al2(CO3)3(ppt) + 6NaNO3 What is the molar concentration of each ion in solution after the reaction? 46 24 ml of a 0.10 M solution of Al(NO3)3 is combined with 24 ml of 0.10 M Na2CO3. 3Na2CO3 + 2Al(NO3)3 → Al2(CO3)3(ppt) + 6NaNO3 What is the molar concentration of each ion in solution after the reaction? • • • • • Remember - the total volume of the solution has doubled which affects the concentrations Treat the spectator ion concentrations as completely unreacted in twice the volume. Don’t forget they are sometimes “buy 1 get many” ions Assume the limiting ion in the precipitate is effectively gone from solution. Calculate the mmoles left over (this means a subtraction) of the excess ion that forms the precipitate and calculate its molarity in the total volume. 47 24 ml of a 0.10 M solution of Al(NO3)3 is combined with 24 ml of 0.10 M Na2CO3. 3Na2CO3 + 2Al(NO3)3 → Al2(CO3)3(ppt) + 6NaNO3 What is the molar concentration of each ion in solution after the reaction? • Assume the CO32− ≃ 0 + 2Na 4.8mmol 2.4mmolNa2CO3 × = = 0.10MNa + Still in Solution 1Na2CO3 48ml − 3 − 3 3NO 7.2mmolNO − 2.4mmolAl(NO3 )3 × = = 0.15M NO3 Still in Solution Al(NO3 )3 48ml 48 24 ml of a 0.10 M solution of Al(NO3)3 is combined with 24 ml of 0.10 M Na2CO3. 3Na2CO3 + 2Al(NO3)3 → Al2(CO3)3(ppt) + 6NaNO3 What is the molar concentration of each ion in solution after the reaction? • Assume the CO32− ≃ 0 2Na + 4.8mmol 2.4mmolNa2CO3 × = = 0.10MNa + Still in Solution 1Na2CO3 48ml 3NO3− 7.2mmolNO3− 2.4mmolAl(NO3 )3 × = = 0.15M NO3− Still in Solution Al(NO3 )3 48ml 3+ 1Al 3+ 2.4mmol1Al(NO3 )3 × = 2.4mmolAl inSolution 1Al(NO3 )3 2Al(NO3 )3 2.4mmolNa2CO3 × = 1.6mmolAl(NO3 )3 needed 3Na2CO3 2.4mmolAl 3+ − 1.6mmolAl 3+ 0.8mmolAl 3+ Left = = 0.017MAl 3+ Still in Solution 48ml 49 − NO3 Calculate the Molarity of ions in 40 ml of a 0.20 M solution of Mg(NO3)2 after mixing with 60 ml of a 0.10 M solution of Al(NO3)3 No Calculator • Type in a numerical answer. 50 − NO3 ions Calculate the Molarity of in 40 ml of a 0.20 M solution of Mg(NO3)2 after reacting with 60 ml of a 0.10 M solution of Al(NO3)3 − 3 2NO − 40ml × 0.2Μ × = 16mmol NO3 1Mg(NO3 )2 No Calculator − 3 3NO − 60ml × 0.1Μ × = 18mmol NO3 1Al(NO3 )3 − 3 34NO total − = 0.34M NO3 100ml Total 51 25 ml of 0.40-molar iron(III) bromide solution is combined with 35 ml of 0.30-molar lead(II) nitrate solution. Write out the balanced overall (aka molecular) equation for this reaction molar mass iron(III) bromide lead(II) nitrate 52 25 ml of 0.40-molar iron(III) bromide solution is combined with 35 ml of 0.30-molar lead(II) nitrate solution. 2FeBr3 + 3Pb(NO3)2 → 3PbBr2 + 2Fe(NO3)3 molar mass iron(III) bromide 295.55 lead(II) nitrate 331.22 lead(II) bromide 367 iron(III) nitrate 248.88 1. What mass of precipitate forms. 2. What is the concentration of spectator ions after the precipitation is complete? 3. What is the concentration of the other ions after the precipitation is complete 53 25 ml of 0.40-molar iron(III) bromide solution is combined with 35 ml of 0.30-molar lead(II) nitrate solution. 2FeBr3 + 3Pb(NO3)2 → 3PbBr2 + 2Fe(NO3)3 molar mass lead(II) bromide 367 25ml × 0.4M = 10mmolFeBr2 35ml × 0.3M = 10.5mmolPb ( NO3 )2 lead(II) nitrate limits 3PbBr2 10.5mmolPb ( NO3 )2 × = 10.5mmolPb ( NO3 )2 × 367g / mol = 3853mg = 3.85g 3Pb ( NO3 )2 54 25 ml of 0.40-molar iron(III) bromide solution is combined with 35 ml of 0.30-molar lead(II) nitrate solution. 2FeBr3 + 3Pb(NO3)2 → 3PbBr2 + 2Fe(NO3)3 molar mass lead(II) bromide 367 25ml × 0.4M = 10mmolFeBr2 35ml × 0.3M = 10.5mmolPb ( NO3 )2 lead(II) nitrate limits 2NO3− 10.5mmolPb ( NO3 )2 × = 21mmolNO3− 1Pb ( NO3 )2 3+ Fe 10mmolFe 10mmolFeBr2 × = FeBr2 60ml 3+ 21mmolNO3− = 0.35M NO3− 60ml = 0.17M Fe Still in Solution 3+ 55 25 ml of 0.40-molar iron(III) bromide solution is combined with 35 ml of 0.30-molar lead(II) nitrate solution. 2FeBr3 + 3Pb(NO3)2 → 3PbBr2 + 2Fe(NO3)3 molar mass lead(II) bromide 367 25ml × 0.4M = 10mmolFeBr2 35ml × 0.3M = 10.5mmolPb ( NO3 )2 lead(II) nitrate limits 2FeBr3 10.5mmolPb ( NO3 )2 × = 7mmolFeBr3used 3Pb ( NO3 )2 − 3Br 10FeBr3started − 7FeBr3used =3FeBr3leftover × = 9Br − leftover 1FeBr3 9mmolBr − = 0.15M Br − 60ml No lead ions left in solution...all in the precipitate 56 100 ml of 0.050 M lead(II) nitrate solution is combined with 100 ml of 0.080-molar sodium chloride solution. Here’s a series of questions.... Write out the balanced overall (aka molecular) and net ionic equations for this reaction 1. On your notebook paper. 57 100 ml of 0.050 M lead(II) nitrate solution is combined with 100 ml of 0.080-molar sodium chloride solution. Write out the balanced overall (aka molecular) equation for this reaction 1. What mass of precipitate forms 2. What is the concentration of sodium ions after the precipitation is complete? 3. What is the concentration of lead ions after the precipitation is complete 58 100 ml of 0.050 M lead(II) nitrate is combined with 100 ml of 0.080-molar sodium chloride. Here’s a series of questions.... Write out the overall and net ionic equations for this reaction 1. Pb(NO3)2(aq) + 2NaCl(aq) → 2NaNO3(aq) + PbCl2(ppt) 2+ − 2. Pb + 2Cl → PbCl2 59 100.0 ml of 0.050 M lead(II) nitrate solution is combined with 100.0 ml of 0.080-molar sodium chloride solution. Calculate the mass of precipitate formed (MM PbCl2 = 278 g/mol) 1. Pb(NO3)2 + 2NaCl → 2NaNO3 + PbCl2 2. +2 Pb + − 2Cl → PbCl2 60 100.0 ml of 0.050 M lead(II) nitrate is combined with 100.0 ml of 0.080-molar sodium chloride. Calculate the mass of precipitate formed. (MM PbCl2 = 278 g/mol) 1. Pb(NO3)2 + 2NaCl → 2NaNO3 + PbCl2 2. (0.05 M)(0.1L) = 0.005 mol Pb(NO3)2 3. (0.08 M)(0.1L) = 0.008 mol NaCl 4. use the “limiting trick” to determine that NaCl limits 5. ⎛ 1PbCl2 ⎞ ⎛ 278g ⎞ 0.008molNaCl ⎜ = 1.1gPbCl2 precipitate ⎟ ⎜ ⎟ ⎝ 2NaCl ⎠ ⎝ 1mol ⎠ 61 100 ml of 0.050 M lead(II) nitrate is combined with 100 ml of 0.080-molar sodium chloride. What are the concentrations of the sodium ions and nitrate ions in solution after the reaction is complete? 1. Pb(NO3)2 + 2NaCl → 2NaNO3 + PbCl2 62 100 ml of 0.050 M lead(II) nitrate is combined with 100 ml of 0.080-molar sodium chloride. What is the concentration of the sodium and nitrate ions in solution after the reaction is complete? 1. Pb(NO3)2 + 2NaCl → 2NaNO3 + PbCl2 2. All of the starting sodium and nitrate ions are still present, since they are spectator ions, swimming in the 200 ml total volume. 3. (0.05 M)(100 ml) = 5 mmol Pb(NO3)2 4. ⎛ 2NO3− ⎞ − 5mmolPb(NO3 )2 ⎜ = 10mmolNO 3 ⎟ 1Pb(NO ) ⎝ 3 2⎠ 10mmolNO3− = 0.050MforNO3− 200mltotalvolume 5. (0.08 M)(100 ml) = 8 mmol NaCl 6. ⎛ 1Na + ⎞ + 8mmolNaCl ⎜ = 8mmolNa ⎝ 1NaCl ⎟⎠ 8mmolNa + = 0.040MforNa + 200mltotalvolume 63 100 ml of 0.050 M lead(II) nitrate is combined with 100 ml of 0.080-molar sodium chloride. What is the concentration of the lead ions and chloride ions in solution after the reaction is complete? 1. Pb(NO3)2 + 2NaCl → 2NaNO3 + PbCl2 64 100 ml of 0.050 M lead(II) nitrate is combined with 100 ml of 0.080-molar sodium chloride. What is the concentration of the lead and chloride ions in solution after the reaction is complete? 1. Pb(NO3)2 + 2NaCl → 2NaNO3 + PbCl2 2. Since the NaCl limited, we can assume all the Cl− ion are in the precipitate not the solution, thus Cl− = 0 M 3. Some of the lead ions are in the precipitate, some are still in solution. 4. (0.05 M)(100 ml) = 5 mmol Pb2+ to start 5. ⎛ 1Pb(NO3 )2 ⎞ 2+ 8mmolNaCl ⎜ = 4mmolPb reactedtomakeprecipitate ⎟ ⎝ 2NaCl ⎠ 6. 5mmol Pb2+ start − 4 mmol Pb2+ reacted = 1 mmol Pb2+ remain in soln 7. 1mmolPb 2+ = 0.0050MPb 2+ RemainInSolution 200mlTotalVolume 65 100. ml of 0.050 M (mol/L) cobalt(III) nitrate is combined with 200. ml of 0.040-molar potassium chromate. (molar mass of cobalt chromate = 465.86 g/mol) Better write a balanced equation. 1. 2. 3. 4. Is this a redox reaction? Which ion limits the reaction? What is the mass of the precipitate? What is the concentration of each of the ions left in the solution? 66 100. ml of 0.050 M (mol/L) cobalt(III) nitrate is combined with 200. ml of 0.040-molar potassium chromate. (molar mass of cobalt chromate = 465.86 g/mol) Better write a balanced equation. 1. Is this a redox reaction? • No, DR reactions are never redox 2. Which ion limits the reaction? • Co2+ 3. What is the mass of the precipitate? • 1.2 g (rounded to 2 SF) 4. What is the concentration of each of the ions left in the solution? 67 100. ml of 0.050 M (mol/L) cobalt(III) nitrate is combined with 200. ml of 0.040-molar potassium chromate. (molar mass of cobalt chromate = 465.86 g/mol) 2Co(NO3)3 + 3 K2CrO4 → Co2(CrO4)3(ppt) + 6KNO3 100mlCo(NO3 )3 × 0.05M = 5mmolCo(NO3 )3 200mlK 2CrO4 × 0.04Μ = 8mmK 2CrO4 3NO3− 15mmol 5mmolCo(NO3 )3 × = = 0.050M NO3− Still in Solution 1Co(NO3 )3 300ml 2K + 16mmol 8mmolK 2CrO4 × = = 0.053M K + Still in Solution K 2CrO4 300ml • Assume the Co3+ ≃ 0 3K 2CrO4 5mmolCo(NO3 )3 × = 7.5mmolCrO42− needed 1Co(NO3 )3 2− 0.05mmolCrO 4 8mmolCrO42− − 7.5mmolCrO42− = = 0.0017MCrO42− Still in Solution 300ml 68 A 1.0 L sample of an aqueous solution contains 0.030 mol of KCl and 0.020 mol of AlCl3. What is the minimum number of moles of AgNO3 that must be added to the solution in order to precipitate all of − the Cl as AgCl(s)? No calculator 69 A 1.0 L sample of an aqueous solution contains 0.030 mol of KCl and 0.020 mol of AlCl3. What is the minimum number of moles of AgNO3 that must be added to the solution in order to precipitate all of − the Cl as AgCl(s)? No calculator 0.09 mol AgNO3 must be added 70 A 1.0 L sample of an aqueous solution contains 0.030 mol of KCl and 0.020 mol of AlCl3. What is the minimum number of moles of Pb(NO3)2 that must be added to the solution in order to precipitate − all of the Cl as PbCl2(s)? No calculator 71 A 1.0 L sample of an aqueous solution contains 0.030 mol of KCl and 0.020 mol of AlCl3. What is the minimum number of moles of Pb(NO3)2 that must be added to the solution in order to precipitate − all of the Cl as PbCl2(s)? No calculator 0.045 mole Pb(NO3)2 must be added 72 When 10. milliliter of 0.4-molar sodium sulfate is added to 20. milliliters of 0.10-molar aluminum sulfate the number of moles of 2+ Pb that must be added to 2− precipitate out all of the SO4 would be No calculator 1. 0.020 mol 2. 0.0060 mol 3. 0.010 mol 4. 10. mol 5. 0.0010 mol 6. 0.10 mol 7. 0.005 mol 8. 1.0 mol 73 When 10. milliliter of 0.4-molar sodium sulfate is added to 20. milliliters of 0.10-molar aluminum sulfate the number of moles of 2+ Pb that must be added to 2− precipitate out all of the SO4 No calculator would be 1. 0.020 mol 2. 0.0060 mol 3. 0.010 mol 4. 10. mol 5. 0.0010 mol 6. 0.10 mol 7. 0.005 mol 8. 1.0 mol 74 A 10.0-milliliter sample of 0.200molar K3PO4 solution is added to 20.0 milliliters of 0.450-molar Ba(NO3)2 solution. Barium phosphate precipitates. The 2+ concentration of barium ion, Ba , in solution after reaction is No calculator 75 A 10.0-milliliter sample of 0.200molar K3PO4 solution is added to 20.0 milliliters of 0.450-molar Ba(NO3)2 solution. Barium phosphate precipitates. The 2+ concentration of barium ion, Ba , in solution after reaction is No calculator 76 A 10.0-milliliter sample of 0.200molar K3PO4 solution is added to 20.0 milliliters of 0.450-molar Ba(NO3)2 solution. Barium phosphate precipitates. The 2+ concentration of barium ion, Ba , in solution after reaction is No calculator 3−, 2+ Ba , 2+ Ba 2 mmol PO4 9 mmol 3 mmol used 6 mmol Ba2+ left over/30 ml = 0.2 M Ba2+ after ppt 77 If 40.0 ml of a 0.30 M solution of barium nitrate is combined with 25.0 ml of 0.35 M sodium phosphate solution, 1. calculate the mass of the precipitate that should form. 2. If 1.32 g of precipitate did form in the laboratory, calculate the % yield. 3. Determine the concentration (molarity) of any ions still floating in the solution. Acid Base Review from First Year and Preview of What’s New This Year weak? strong? concentrated? dilute? What do these words mean? 79 concentrated vs dilute describes the amount dissolved. • Concentration refers to the quantity of solute that is dissolved in the solvent. in other words, how much stuff did you put in the water…. ✓ concentrated = lots ✓ dilute = little ✓ 80 Reminder: When soluble ionic compounds dissolve, they dissociate into separated ions. • Generic Reaction for the dissolving of any ionic compound: + + X− AX → A ‣ (s) The break up of the salt and subsequent surrounding of ions by water molecules is called hydrolysis. Notice how the water molecules arrange differently around the + compared to the − ions. 81 Reminder: When sugar (or other molecular compounds) dissolves in water, the molecules do NOT dissociate into ions. Sugar We represent this in an equation as: H 2O C6H12O6(s) → C6H12O6(aq) Salt We represent this in an equation as: H 2O NaCl(s) → Na+ + Cl− 82 Acids • Molecular compounds made of H&(−ion) ✓ HCl, HC2H3O2, H2SO4, HClO3, HBr, HNO2, etc • Acids are molecular compounds that “stand on the line.” The line that differentiates ionic from molecular. • An acid molecule is a covalently bonded compound, that when dissolved in water, dissociates (all or partly) into ions. • However, the degree to which acids ionize when placed in water, varies from acid to acid. ✓ We call this ability to dissociate, the strength or weakness of the acid. 83 Particle View: • Use this as HCl • Use this as HF • Sketch two separate beakers (side view) depicting no more than 5 molecules in each beaker. Represent the two substances as they would appear when dissolved in water 84 Particle View: strong vs weak HCl is a strong acid. Plenty of HCl molecules dissolve, and virtually all of them ionize into H+ and Cl− We represent this in an equation as: HCl(aq) → H+ + Cl− + + + + H+ Both strong and weak acids will be able to neutralize a base equally well. − + − + H+ HF is a weak acid. While plenty of HF molecules dissolve, only a fraction of them ionize into H+ and F− We represent this in an equation as: HF(aq) H + + F− 85 concentrated vs dilute and strong vs weak • You can have a concentrated strong acid • You can have a dilute strong acid • You can have a concentrated weak acid • You can have a dilute weak acid 86 Select the reaction below that best represents the net ionic equation for sulfuric acid with potassium hydroxide. 1. H2SO4 + KOH → K2SO4 + H2O 2. H2SO4 + 2KOH → K2SO4 + 2H2O 3. 2H+ + OH− → 2H2O + − 4. 2H + 2OH → 2H2O + − 5. H + OH → H2O 87 Select the reaction below that best represents the net ionic equation for sulfuric acid with potassium hydroxide. 1. H2SO4 + KOH → K2SO4 + H2O 2. H2SO4 + 2KOH → K2SO4 + 2H2O 3. 2H+ + OH− → 2H2O + − 4. 2H + 2OH → 2H2O • this would be ok too, but it is probably best to reduce it. 5. + H + − OH → H 2O 88 Acids • Arrhenius defined acids as substances that produce + H ions in an aqueous solution. HBr → H+ + Br− ✓ HBr + H2O → H3O+ + Br− ✓ • Bronstead & Lowrey defined acids as proton donors. Monoprotic acids produce 1 + H • Di- or Tri- protic acids produce 2 or 3 H+ H2SO4 can produce 2 H+ per acid molecule. ✓ We can think of the dissociation occurring in steps: ✓ H2SO4 → H+ + HSO4− ‣ HSO4− → H+ + SO42− ‣ ✓ or all in one step: H2SO4 → H+ + SO42− 89 Strong vs Weak • Strong electrolytes dissociate completely ✓ HNO3 → H+ + NO3− • Weak electrolytes only partly dissociate. Some molecules remain “whole” while some dissociate. ✓ HF H+ + F− 90 Strong vs Weak • HF H+ + F− ✓ The double arrow indicates that both the forward and reverse reactions are occurring at the same rate, so the reaction seems to “stop”. ✓ It does not mean that the reaction occurs only as far as the half way point. ✓ This balance is called chemical equilibrium. 91 Bases • Substances that react with H+ ions • Arrhenius defined bases as substances that produce OH- ions NaOH → Na+ + OH− ✓ Ba(OH)2 → Ba2+ + 2 OH− ✓ • Bronstead & Lowrey defined bases as proton acceptors because some common weak bases appear not to contain OH ions. • Ammonia is a common weak base. It reacts with water, accepts an H+, and produces OH− NH4+ + OH− ✓ NH3 + H2O ✓ When you see the double arrow, it tells us the reaction “comes to equilibrium” and it is a weak base. 92 Memorize the Strong Acids & Strong Bases • Strong acids ✓ ✓ ✓ ✓ (All the rest we will encounter in AP are weak) HCl, HBr, HI HClO3, HClO4 HNO3 H2SO4 • Strong bases - metal hydroxides that dissolve (most metal hydroxides do not dissolve very well) ✓ Alkali hydroxides ‣ ✓ LiOH, NaOH, KOH, RbOH, CsOH Heavy alkaline earth hydroxides ‣ Ca(OH)2, Sr(OH)2, Ba(OH)2 93 Acids React with Bases • Neutralization reaction ✓ aka Double Replacement Rx • HNO3 + KOH → H2O + KNO3 • Acid + base water + salt • Net Ionic: + H + OH → H 2O 94 Acids • Arrhenius defined acids as substances that produce + H ✓ ✓ ions in an aqueous solution. HBr → H+ + Br− HBr + H2O → H3O+ + Br− • Bronstead & Lowrey defined acids as proton donors. Monoprotic acids produce 1 H+ • Di- or Tri- protic acids produce 2 or 3 H+ ✓ ✓ ✓ ✓ ✓ H2SO4 can produce 2 H+ per acid molecule. We can think of the dissociation occuring in steps: H2SO4 → H+ + HSO4− HSO4− H+ + SO42− In this case, only the first ionization is complete, thus a solution of sulfuric acid will actualy contain a mixture of H+, HSO4−, and SO42− 95 How do you make up 500 ml of a 2 M solution of sodium hydroxide? 96 How do you make 500 ml of a 2 M solution of sodium hydroxide? 1. Use the formula M = moles/Liter to determine that you need 1 mole in the 500 ml of solution. 2. NaOH, MM = 40g/mol 3. Mass 40 g of NaOH into a volumetric flask. 4. Add ⅔ of the water and swirl to dissolve 5. Add the remaining water to bring the solution to 500 ml total. 97 How to make 500 ml of a 2.0-molar solution of hydrochloric acid starting with concentrated HCl, 12 M? slide show 98 How to make 500 ml of a 2 M solution of hydrochloric acid starting with concentrated HCl, 12 M? 1. This is a dilution problem, and during a dilution the number of moles in the flask does not change while adding water to dilute the more concentrated solution. 2. Use the formula M×V = M×V 3. (2M)(500 ml) = (12M)(V) V = 83.3 ml 4. Add water to the flask, ⅔ full. Measure out the 83.3ml of concentrated HCl into a graduated cylinder and add to the water in the volumetric flask. 5. Add the remaining water to bring the solution to 500ml total. 99 On paper write an overall and a net ionic equation to describe the reaction between a 500 ml of 2M solution of NaOH and 500 ml of a 2M solution of HCl. 100 On your white boards write an overall and a net ionic equation to describe the reaction between a 500 ml of 2M solution of NaOH and 500 ml of a 2M solution of HCl. Calculate the moles of water that would form during this reaction. • • HCl + NaOH → H2O + NaCl + − H + OH → H2O 101 On your white boards write an overall and a net ionic equation to describe the reaction between a 500 ml of 2M solution of NaOH and 500 ml of a 2M solution of HCl. Calculate the moles of water that would form during this reaction. What volume of water is this? • • • + H + − OH → H 2O + H, (2M)(.5L) = 1mol combined with − (2M)(.5L) = 1 mol OH = 1 mole H2O 102 On your white boards write an overall and a net ionic equation to describe the reaction between a 500 ml of 2M solution of NaOH and 500 ml of a 2M solution of HCl. Calculate the moles of water that would form during this reaction. What volume of water is this? • • • • + H + − OH → H 2O + H, (2M)(.5L) = 1mol combined with (2M)(.5L) = 1 mol OH = 1 mol H2O 1ml 1molH 2O = 18g × = 18mlH 2O 1g Density of water slide show 103 In class, 500 ml of 2M solution of NaOH and 500 ml of a 2M solution of HCl were poured together into a 1000 ml flask. At first, you may have been surprised to see ~18 ml more than 1000 ml of solution, however, after the calculations shown below, you could see why this must occur. • • • • H+ + OH− → H2O (2M)(.5L) = 1mol H+, combined with (2M)(.5L) = 1 mol OH- = 1 mol H2O of water will form 1mol H2O = (18g)(1ml/1g)= 18 ml In the demonstration done in class, the extra water above the line was measured and turned out to be ~18 ml 1000 ml mark 500 ml mark 500 ml + 500 ml = 1000 ml slide show104 Insoluble (Dilute) Strong Base Mg(OH)2 slide view 105 Insoluble (Dilute) Strong Bases • Mg(OH)2 is an insoluble ionic compound in H2O • In an acid solution, Mg(OH)2 does dissolve • Demonstrate • The acid ions, H3O+ (or H+) attack the OH − ions that are in the solid crystal structure creating water molecules H+ + OH− → H2O ✓ H3O+ + OH− → 2H2O ✓ 106 Adding acid to “Milk of Magnesia” a suspension of Mg(OH)2 • Write an overall equation for the reaction between a suspension of magnesium hydroxide and hydrochloric acid. ✓ Mg(OH)2 + HCl → H2O + MgCl2 • Convert to a net ionic equation. ✓ ✓ ✓ Mg(OH)2 + H+ → H2O + Mg2+ Mg(OH)2 must be written as a compound because it is not soluble in water (to any appreciable degree, and it is simply a suspension in the water.) HCl is soluble, thus the Cl− ions are spectator ions. 107 Oxidation Numbers more practice 108 Write a balanced equation for the reaction that represents the combustion of magnesium in air. 1. This is a redox reaction in which magnesium is oxidized. 2. This is a redox reaction in which oxygen is oxidized. 3. This is a redox reaction in which magnesium is reduced. 4. This is a redox reaction in which oxygen is oxidized. 5. This is not a redox reaction. 109 Write a balanced equation for the reaction that represents the combustion of magnesium in air. 2Mg + O2 → 2MgO or Mg + ½O2 → MgO 1. 2. 3. 4. 5. • • This is a redox reaction in which magnesium is oxidized. This is a redox reaction in which oxygen is oxidized. This is a redox reaction in which magnesium is reduced. This is a redox reaction in which oxygen is reduced. This is not a redox reaction. Combustion reactions are always redox reactions. In redox reactions, particles must be both oxidized and reduced. 110 slide view Oxidation - Reduction • Oxidation (LEO) [OIL] ✓ When an element loses electrons • Reduction (GER) [RIG] ✓ When an element gains electrons • The name oxidation is used because this type of reaction was first studied by reacting metals with oxygen. ✓ Rusting: Fe + O2 → Fe2O3 • It was called reduction because when elements are isolated from minerals, we say they are “reduced” to their elemental form. ✓ AlCl3 → Al(s) + Cl2(g) 111 Select the true statement(s) about the reaction below: Mg(OH)2 + HCl → H2O + MgCl2 1. 2. 3. 4. 5. 6. 7. 8. 9. This is a redox reaction in which chlorine is reduced. This is a redox reaction in which magnesium is reduced. This is a redox reaction in which hydrogen is reduced. This is a redox reaction in which oxygen is reduced. This is a redox reaction in which chlorine is oxidized. This is a redox reaction in which magnesium is oxidized. This is a redox reaction in which hydrogen is oxidized. This is a redox reaction in which oxygen is oxidized. This is not a redox reaction. 112 Select the true statement(s) about the reaction below: Mg(OH)2 + HCl → H2O + MgCl2 1. 2. 3. 4. 5. 6. 7. 8. 9. • This is a redox reaction in which chlorine is reduced. This is a redox reaction in which magnesium is reduced. This is a redox reaction in which hydrogen is reduced. This is a redox reaction in which oxygen is reduced. This is a redox reaction in which chlorine is oxidized. This is a redox reaction in which magnesium is oxidized. This is a redox reaction in which hydrogen is oxidized. This is a redox reaction in which oxygen is oxidized. This is not a redox reaction. Acid base neutralization is never redox. 113 What is the oxidation number of sulfur in SF2? 1. 0 2. +2 3. -2 4. +4 5. -4 114 What is the oxidation number of sulfur in SF2? 1. 0 2. +2 3. -2 4. +4 5. -4 • Since fluorine is -1 × 2 = −2 the sulfur must be +2 to balance. 115 What is the oxidation state of chlorine in KClO4 ? 1. 2. 3. 4. 5. −2 −4 +4 +7 +8 116 What is the oxidation state of chlorine in KClO4 ? 1. 2. 3. 4. 5. • −2 −4 +4 +7 +8 O is −2 × 4 = −8, the K is +1 thus the Cl must be the remaining +7. 117 What is the oxidation state of nitrogen in N2O5 ? 1. 2. 3. 4. 5. +5 +2 -10 +10 -5 118 What is the oxidation state of nitrogen in N2O5 ? 1. 2. 3. 4. 5. • +5 +2 -10 +10 −5 Since oxygen is −2 × 5 = -10, the nitrogens must total +10, but 2 nitrogens cause the +10, so each one individually is +5. slide show 119 Balancing Aqueous Redox Equations Chapter 4 and 20 (section 1 & 2) 120 Redox Reactions • NOT double replacement ✓ ✓ precipitation acid base • Always combustion • Always single replacement • Sometimes synthesis • Sometimes decomposition • Complex redox reactions in solution 121 Redox Reactions − NO3 + Al → 3+ Al in acidic solution + NO • List the Oxidation numbers of all elements above • Determine which element is oxidized, • Determine which element is reduced • How many total number of electrons are transferred? 122 Redox Reactions +5 -2 − 0 NO3 + Al → +3 3+ Al in acidic solution +2 -2 + NO • List the Oxidation numbers of all elements above • Determine which element is oxidized, Al • Determine which element is reduced N • How many total number of electrons are transferred? 3e 123 Redox Reactions this Rx occurs in acid solution SO3 2− + − MnO4 → SO4 2− + 2+ Mn • Write out each element’s oxidation number • Determine which element is oxidized • Determine which element is reduced • How many electrons are transferred? 124 In an acidic solution SO3 2− + − MnO4 → SO4 2− + 2+ Mn 1. Identify which elements are oxidized and reduced 2. Balance the number of redox atoms by inspection 3. Adjust the number of redox atoms so the number of electrons lost equals the number of electrons gained. (This is the number of electrons transferred.) 4. balance any other elements (other than H and O) 5. balance oxygen by adding water as necessary 6. balance hydrogen by adding H+ as necessary 7. recheck by confirming that ion charge is balanced 125 Redox Reactions this Rx occurs in acid solution MnO4− + H2O2 → Mn2+ + O2 + H2O 1. Identify which elements are oxidized and reduced 2. Balance the number of redox atoms by inspection 3. Adjust the number of redox atoms so the number of electrons lost equals the number of electrons gained. (This is the number of electrons transferred.) 4. balance any other elements (other than H and O) 5. balance oxygen by adding water as necessary 6. balance hydrogen by adding H+ as necessary 7. recheck by confirming that ion charge is balanced 126 Redox Reactions this Rx occurs in acid solution 6H+ + 2MnO4− + 5H2O2 → 2Mn2+ + 5O2 + 8H2O • Identify the element that is oxidized. • Identify the element that is reduced. +1 + 6H + +7 −2 +1 −1 − 2MnO4 + 5H2O2 → +2 2+ 2Mn 0 +1 −2 + 5O2 + 8H2O 127 • A solution of hypochlorous acid and hydrochloric acid is formed when chlorine gas is dissolved in cold water. • Write out this reaction and determine the oxidation numbers for all elements in this reaction. • The balanced reaction is: + H − Cl ✓ Cl2 + H2O → HOCl + + • In this reaction, chlorine is both oxidized and reduced. This is called a... Disproportionation Reaction 128 In an acidic solution AuCl4 −+ Cu → Au + − Cl + 2+ Cu 1. Identify which elements are oxidized and reduced 2. Balance the number of redox atoms by inspection 3. Adjust the number of redox atoms so the number of electrons lost equals the number of electrons gained. (This is the number of electrons transferred.) 4. balance any other elements (other than H and O) 5. balance oxygen by adding water as necessary 6. balance hydrogen by adding H+ as necessary 7. recheck by confirming that ion charge is balanced 129 AuCl4 −+ Cu → Au + − Cl + 2+ Cu 1. Identify which elements are oxidized and reduced + 0 → 0 + -1 + +2 • +3, -1 • Au reduced 3e- gained, Cu oxidized 2e- lost 2. Balance the number of redox atoms by inspection − they are already 3. Adjust the number of redox atoms so the number of electrons lost equals the number of electrons gained. (This is the number of electrons transferred.) • 2AuCl4− + 3Cu → 2Au + Cl− + 3Cu2+ (a total of 6e− transferred) 4. balance any other elements (other than H and O) • 2AuCl4− + 3Cu → 2Au + 8Cl− + 3Cu2+ 5. balance oxygen by adding water as necessary − not necessary 6. balance hydrogen by adding H+ as necessary − not necessary 7. recheck by confirming that charge is balanced − 2− = 2− 130 2AuCl4 −+ 3Cu → 2Au + − 8Cl + 2+ 3Cu 1. It will be important later, for us to identify the number of electrons transferred. • in this reaction, there are 6 electrons lost • and the same 6 electrons are gained • thus, a total of 6 (not 12) electrons are transferred 2. In every redox reaction there are two halves. • the oxidation • the reduction 3. We call these the half reactions. 131 2AuCl4 −+ 3Cu → 2Au + − 8Cl + 2+ 3Cu From time to time, AP will ask you to write out the half reactions. 1. Pull out the two substances involved in oxidation, then balance, first the oxidation atom, then the other atoms, then oxygen, then hydrogen − then include the number of electrons 2. The oxidation half reaction is 2+ + 2e− Cu → Cu • 3. reduction − + 3e− → Au + 4Cl− AuCl 4 • 4. this corresponds with what we had determined by the previous method • Au reduced 3e− gained, Cu oxidized 2e− lost 132 In an acidic solution − NO3 + Al → 3+ Al + NO 1. Identify which elements are oxidized and reduced 2. Balance the number of redox atoms by inspection 3. Adjust the number of redox atoms so the number of electrons lost equals the number of electrons gained. (This is the number of electrons transferred.) 4. balance any other elements (other than H and O) 5. balance oxygen by adding water as necessary 6. balance hydrogen by adding H+ as necessary 7. recheck by confirming that ion charge is balanced 133 In an acidic sol’n Zn + NO3 −→ 2+ Zn + N2 1. Identify which elements are oxidized and reduced Alert, alert...don’t miss this step 2. Balance the number of redox atoms by inspection 3. Adjust the number of redox atoms so the number of electrons lost equals the number of electrons gained. (This is the number of electrons transferred.) 4. balance any other elements (other than H and O) 5. balance oxygen by adding water as necessary 6. balance hydrogen by adding H+ as necessary 7. recheck by confirming that ion charge is balanced 134 (in acidic solution) Zn + NO3 −→ 2+ Zn + N2 1. Identify which elements are oxidized and reduced 0 + +5,-2 → +2 + 0 • • N reduced 5e-/atom, Zn oxidized 2e2. Balance the number of redox atoms by inspection • Zn + 2NO3− → Zn2+ + N2 3. Adjust the number of redox atoms so the number of electrons lost equals the number of electrons gained. (This is the number of electrons transferred.) • 5Zn + 2NO3− → 5Zn2+ + N2 4. balance any other elements (other than H and O) - none 5. balance oxygen by adding water as necessary • 5Zn + 2NO3− → 5Zn2+ + N2 + 6H2O 6. balance hydrogen by adding H+ as necessary • 12H+ + 5Zn + 2NO3− → 5Zn2+ + N2 + 6H2O 7. recheck by confirming that ion charge is balanced − +10 = +10 135 Write out the two half reactions. + 12H − + 5Zn + 2NO3 → 2+ 5Zn + N2 + 6H2O 136 − Zn + NO3 → 2+ Zn + N2 (in acidic) 1. oxidation 2+ + 2e− Zn → Zn • • note that charge balances in half reactions as well 2. reduction − → N 2NO 3 2 • + ions this may involve using water and H • • 12H+ + 2NO3− + 10e− → N2 + 6H2O 3. this corresponds with what we had determined by the previous method • N reduced 5e- gained per N, but the N2 requires 10etotal, Zn oxidized 2e- lost 137 In a basic solution Br2 + AsO2 −→ − Br + AsO4 3− 1. 2. 3. 4. 5. 6. 7. Identify which elements are oxidized and reduced Balance the redox atoms by inspection balance the redox atoms based on total electrons transferred balance any other elements (other than H and O) balance O by adding water as necessary balance H by adding H+ as necessary if the reaction occurs in basic solution, convert to basic by adding an equal number of OH− ions to both sides to cancel out the H+ ions 8. recheck by confirming that charge is balanced 138 -1 Br2 + AsO2 → -1 Br + AsO4 -3 “pretend” in acid, then convert to basic 1. 2. 3. 4. 5. 6. 7. Identify which elements are oxidized and reduced 0 + +3, -2 → -1 + +5, -2 • • Br reduced 1e-/atom, As oxidized 2ebalance those atoms based on total electrons transferred • Br2 + AsO2-1 → 2Br-1 + AsO4-3 balance any other elements (other than H and O) - none balance O by adding water as necessary • 2H2O + Br2 + AsO2− → 2Br− + AsO43− balance H by adding H+1 as necessary • 2H2O + Br2 + AsO2− → 2Br− + AsO43− + 4H+ if the reaction occurs in basic solution, convert to by adding an equal number of OH-1 ions to both sides to cancel out the H+1 ions • 4OH− + 2H2O + Br2 + AsO2− → 2Br− + AsO43− + 4H+ + 4OH− • 4OH− + 2H2O + Br2 + AsO2− → 2Br− + AsO43− + 4H2O • 4OH− + Br2 + AsO2− → 2Br− + AsO43− + 2H2O recheck by confirming that charge is balanced −5 = −5 139 In basic solution − Br2 → BrO3 + − Br 1. Identify which elements are oxidized and reduced • This is a disproportionation reaction - the same element is both oxidized and reduced. 2. Balance the number of redox atoms by inspection 3. Adjust the number of redox atoms so the number of electrons lost equals the number of electrons gained. (This is the number of electrons transferred.) 4. balance any other elements (other than H and O) 5. balance oxygen by adding water as necessary 6. balance hydrogen by adding H+ as necessary 7. since the reaction occurs in basic solution, convert to by adding an equal number of OH− ions to both sides to cancel out the H+ ions 8. recheck by confirming that ion charge is balanced 140 − Br2 → BrO3 + − Br “pretend” in acid, then convert to basic 1. • • • 2. 3. • 4. 5. • 6. • 7. 8. • • • 9. Identify which elements are oxidized and reduced 0 → +5, -2 + -1 This is a disproportionation reaction - the same element is both oxidized and reduced. Br reduced 1e-/atom, and Br oxidized 5e-/atom Balance the number of redox atoms by inspection − they are already Adjust the number of redox atoms so the number of electrons lost equals the number of electrons gained. (This is the number of electrons transferred.) 6Br2 → 2BrO3− + 10Br− balance any other elements (other than H and O) − none balance oxygen by adding water as necessary 6 Br2 + 6 H2O → 2 BrO3− + 10Br− balance hydrogen by adding H+ as necessary 6 Br2 + 6 H2O → 2 BrO3− + 10 Br− + 12 H+ recheck by confirming that ion charge is balanced since the reaction occurs in basic solution, convert to by adding an equal number of OH− ions to both sides to cancel out the H+ ions 12OH− + 6 Br2 + 6H2O → 2 BrO3− + 10Br− + 12H+ + 12OH− 12OH− + 6 Br2 + 6H2O → 2 BrO3− + 10Br− + 12H2O 12OH− + 6 Br2 → 2 BrO3− + 10Br− + 6H2O recheck by confirming that charge is balanced −12 = −12 141 In basic solution − Br2 → BrO3 + − Br Write out the half reactions which also should be balnced as in a basic solution. 142 − Br2 → BrO3 + − Br (in basic) Write out the half reactions. 1. oxidation − + 10e− Br → 2 BrO 2 3 • − + 12 H+ + 10e− (pretending acidic) Br + 6 H O → 2 BrO 2 2 3 • • Br2 + 12OH− → 2 BrO3− + 6 H2O + 10e− 2. reduction − Br + 2e− → 2Br 2 • 3. this corresponds with what we had determined by the previous method • N reduced 5e− gained per N, but the N2 requires 10e− total, Zn oxidized 2e− lost 143 − CN + − MnO4 → basic sol’n − CNO + MnO2 1. Identify which elements are oxidized and reduced 2. Balance the number of redox atoms by inspection 3. Adjust the number of redox atoms so the number of electrons lost equals the number of electrons gained. (This is the number of electrons transferred.) 4. balance any other elements (other than H and O) 5. balance oxygen by adding water as necessary 6. balance hydrogen by adding H+ as necessary 7. since the reaction occurs in basic solution, convert to by adding an equal number of OH− ions to both sides to cancel out the H+ ions 8. recheck by confirming that ion charge is balanced 144 − CN + 1. • • • 2. 3. • 4. 5. • 6. • 7. • • • 8. − basic sol’n MnO4 → − CNO + MnO2 Identify which elements are oxidized and reduced You may worry about what to do with the oxidation numbers - consider N to be -3 and watch the C change +2, -3 + +7, -2 → +4, -3, -2 + +4, -2 Mn reduced 3e-, and C oxidized 2eBalance the number of redox atoms by inspection − they are already Adjust the number of redox atoms so the number of electrons lost equals the number of electrons gained. (This is the number of electrons transferred.) 3CN− + 2MnO4− → 3CNO− + 2MnO2 balance any other elements (other than H and O) − already done balance oxygen by adding water as necessary 3CN− + 2MnO4− → 3CNO− + 2MnO2 + H2O balance hydrogen by adding H+ as necessary 3CN− + 2MnO4− + 2H→ 3CNO− + 2MnO2 + H2O since the reaction occurs in basic solution, convert to by adding an equal number of OH− ions to both sides to cancel out the H+ ions 2OH− + 3CN− + 2MnO4− + 2H+ → 3CNO− + 2MnO2 + H2O + 2OH− 3CN− + 2MnO4− + 2H2O → 3CNO− + 2MnO2 + H2O + 2OH− 3CN− + 2MnO4− + H2O → 3CNO− + 2MnO2 + 2OH− recheck by confirming that ion charge is balanced −5 = − 5 145 − CN + − basic sol’n MnO4 → − CNO + MnO2 1. oxidation − → CNO− + 2e− CN • • CN− + H2O → CNO− + 2e− + 2H+ (pretending acidic) − + 2OH− → CNO− + 2e− + H O CN 2 • 2. reduction − + 3e− → MnO MnO 4 2 • • MnO4− + 4H+ + 3e− → MnO2 + 2H2O (pretending acidic) − + 2H O + 3e− → MnO + + 4OH− MnO 4 2 2 • 146 Pull out the two half reactions, and balance (basic) 2− − − Pb(OH)4 + ClO → PbO2 + Cl 147 Pull out the two half reactions, and balance (basic) 2− − − Pb(OH)4 + ClO → PbO2 + Cl oxidation Pb(OH)4 2− → PbO2 + 2H2O + 2e− reduction H 2O + − ClO + 2e− → − Cl + − 2OH 148 Pull out the two half reactions, and balance (acidic) each one individually. Label which one is oxidation and which is reduction. I2 + − OCl − → IO3 + − Cl 149 Pull out the two half reactions, and balance (acidic) each one individually. Label which one is oxidation and which is reduction. I2 + − OCl − → IO3 + − Cl oxidation − I2 + 6H2O → 2IO3 + + 12H + 10e− reduction + 2H + − OCl + 2e− → − Cl + H 2O 150 Pull out the two half reactions, and balance (basic) H2O2 + Cl2O7 → O2 + ClO2 − 151 Pull out the two half reactions, and balance (basic) H2O2 + Cl2O7 → O2 + ClO2 − redution − 3H2O + Cl2O7 + 8e− → 2ClO2 + − 6OH oxidation − 2OH + H2O2 → O2 + H2O + 2e− 152 When a basic solution of KMnO4 is added to an SnCl2 solution, a brown precipitate of MnO2 4+ forms and Sn remains in solution. When the same basic solution of KMnO4 is added to an MgCl2 solution, no reaction occurs. Which of the substances involved in these reactions serves as the best reducing agent (i.e. the substance that is best at causing reduction in some other substance) ? 1. 2. 3. 4. 5. KMnO4 SnCl2 MgCl2 Sn4+ Sn2+ 6. 7. 8. 9. K+ MnO2 Mg2+ Cl− 153 When a basic solution of KMnO4 is added to an SnCl2 solution, a brown precipitate of MnO2 forms and Sn4+ remains in solution. When the same basic solution of KMnO4 is added to an MgCl2 solution, no reaction occurs. Which of the substances involved in these reactions serves as the best reducing agent (e.g. the substance that is best at causing reduction in some other substance) ? The Mn in the KMnO4 was reduced, and by SnCl2, not MgCl2. 1. 2. 3. 4. KMnO4 SnCl2 MgCl2 4+ Sn + K 5. 6. MnO2 2+ 7. Mg − 8. Cl 154