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Chapter 4 Chemical Quantities and Aqueous Reactions Part2

Chemistry: A Molecular Approach , 1 st Ed. Nivaldo Tro. Chapter 4 Chemical Quantities and Aqueous Reactions Part2. Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA. 2008, Prentice Hall. Ionic Equations.

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Chapter 4 Chemical Quantities and Aqueous Reactions Part2

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  1. Chemistry: A Molecular Approach, 1st Ed.NivaldoTro Chapter 4Chemical Quantities and Aqueous ReactionsPart2 Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA 2008, Prentice Hall

  2. Ionic Equations • equations which describe the chemicals put into the water and the product molecules are called molecular equations 2 KOH(aq) + Mg(NO3)2(aq) ® 2 KNO3(aq) + Mg(OH)2(s) • equations which describe the actual dissolved species are called completeionic equations • aqueous strong electrolytes are written as ions • soluble salts, strong acids, strong bases • insoluble substances, weak electrolytes, and nonelectrolytes written in molecule form • solids, liquids, and gases are not dissolved, therefore molecule form 2K+1(aq) + 2OH-1(aq) + Mg+2(aq) + 2NO3-1(aq)® 2K+1(aq) + 2NO3-1(aq) + Mg(OH)2(s) Tro, Chemistry: A Molecular Approach

  3. Ionic Equations • ions that are both reactants and products are called spectator ions 2K+1(aq) + 2OH-1(aq) + Mg+2(aq) + 2NO3-1(aq)® 2K+1(aq) + 2NO3-1(aq) + Mg(OH)2(s) • an ionic equation in which the spectator ions are • removed is called a net ionic equation • 2OH-1(aq) + Mg+2(aq)® Mg(OH)2(s) Tro, Chemistry: A Molecular Approach

  4. Acid-Base Reactions • also called neutralization reactions because the acid and base neutralize each other’s properties 2 HNO3(aq) + Ca(OH)2(aq)  Ca(NO3)2(aq) + 2 H2O(l) • the net ionic equation for an acid-base reaction is H+(aq) + OH(aq)  H2O(l) • as long as the salt that forms is soluble in water Tro, Chemistry: A Molecular Approach

  5. Acids and Bases in Solution • acids ionize in water to form H+ ions • more precisely, the H from the acid molecule is donated to a water molecule to form hydronium ion, H3O+ • most chemists use H+ and H3O+ interchangeably • bases dissociate in water to form OH ions • bases, like NH3, that do not contain OH ions, produce OH by pulling H off water molecules • in the reaction of an acid with a base, the H+ from the acid combines with the OH from the base to make water • the cation from the base combines with the anion from the acid to make the salt acid + base salt + water Tro, Chemistry: A Molecular Approach

  6. Common Acids

  7. Common Bases Tro, Chemistry: A Molecular Approach

  8. HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l) Tro, Chemistry: A Molecular Approach

  9. Example - Write the molecular, ionic, and net-ionic equation for the reaction of aqueous nitric acid with aqueous calcium hydroxide • Write the formulas of the reactants HNO3(aq) + Ca(OH)2(aq)  • Determine the possible products • Determine the ions present when each reactant dissociates (H+ + NO3-) + (Ca+2 + OH-)  • Exchange the ions, H+1 combines with OH-1 to make H2O(l) (H+ + NO3-) + (Ca+2 + OH-)  (Ca+2 + NO3-) + H2O(l) • Write the formula of the salt • cross the charges (H+ + NO3-) + (Ca+2 + OH-)  Ca(NO3)2 + H2O(l) Tro, Chemistry: A Molecular Approach

  10. Example - Write the molecular, ionic, and net-ionic equation for the reaction of aqueous nitric acid with aqueous calcium hydroxide • Determine the solubility of the salt Ca(NO3)2 is soluble • Write an (s) after the insoluble products and a (aq) after the soluble products HNO3(aq) + Ca(OH)2(aq)  Ca(NO3)2(aq) + H2O(l) • Balance the equation 2 HNO3(aq) + Ca(OH)2(aq)  Ca(NO3)2(aq) + 2 H2O(l) Tro, Chemistry: A Molecular Approach

  11. Example - Write the molecular, ionic, and net-ionic equation for the reaction of aqueous nitric acid with aqueous calcium hydroxide • Dissociate all aqueous strong electrolytes to get complete ionic equation • not H2O 2 H+(aq) + 2 NO3-(aq) + Ca+2(aq) + 2 OH-(aq)  Ca+2(aq) + 2 NO3-(aq) + H2O(l) • Eliminate spectator ions to get net-ionic equation 2 H+1(aq) + 2 OH-1(aq)  2 H2O(l) H+1(aq) + OH-1(aq)  H2O(l) Tro, Chemistry: A Molecular Approach

  12. Titration • often in the lab, a solution’s concentration is determined by reacting it with another material and using stoichiometry – this process is called titration • in the titration, the unknown solution is added to a known amount of another reactant until the reaction is just completed, at this point, called the endpoint, the reactants are in their stoichiometric ratio • the unknown solution is added slowly from an instrument called a burette • a long glass tube with precise volume markings that allows small additions of solution Tro, Chemistry: A Molecular Approach

  13. Acid-Base Titrations • the difficulty is determining when there has been just enough titrant added to complete the reaction • the titrant is the solution in the burette • in acid-base titrations, because both the reactant and product solutions are colorless, a chemical is added that changes color when the solution undergoes large changes in acidity/alkalinity • the chemical is called an indicator • at the endpoint of an acid-base titration, the number of moles of H+ equals the number of moles of OH • aka the equivalence point Tro, Chemistry: A Molecular Approach

  14. Titration Tro, Chemistry: A Molecular Approach

  15. Titration The base solution is the titrant in the burette. As the base is added to the acid, the H+ reacts with the OH– to form water. But there is still excess acid present so the color does not change. At the titration’s endpoint, just enough base has been added to neutralize all the acid. At this point the indicator changes color. Tro, Chemistry: A Molecular Approach

  16. Write down the given quantity and its units. Given: 10.00 mL HCl 12.54 mL of 0.100 M NaOH Example 4.14:The titration of 10.00 mL of HCl solution of unknown concentration requires 12.54 mL of 0.100 M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution? Tro, Chemistry: A Molecular Approach

  17. Write down the quantity to find, and/or its units. Find: concentration HCl, M Information Given: 10.00 mL HCl 12.54 mL of 0.100 M NaOH Example 4.14:The titration of 10.00 mL of HCl solution of unknown concentration requires 12.54 mL of 0.100 M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution? Tro, Chemistry: A Molecular Approach

  18. Collect Needed Equations and Conversion Factors: HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) 1 mole HCl = 1 mole NaOH 0.100 M NaOH 0.100 mol NaOH  1 L sol’n Information Given: 10.00 mL HCl 12.54 mL of 0.100 M NaOH Find: M HCl Example 4.14:The titration of 10.00 mL of HCl solution of unknown concentration requires 12.54 mL of 0.100 M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution? Tro, Chemistry: A Molecular Approach

  19. Write a Concept Plan: Information Given: 10.00 mL HCl 12.54 mL of 0.100 M NaOH Find: M HCl CF: 1 mol HCl = 1 mol NaOH 0.100 mol NaOH = 1 L M = mol/L Example 4.14:The titration of 10.00 mL of HCl solution of unknown concentration requires 12.54 mL of 0.100 M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution? mL NaOH L NaOH mol NaOH mol HCl mL HCl L HCl Tro, Chemistry: A Molecular Approach

  20. Apply the Solution Map: Information Given: 10.00 mL HCl 12.54 mL of 0.100 M NaOH Find: M HCl CF: 1 mol HCl = 1 mol NaOH 0.100 mol NaOH = 1 L M = mol/L CP: mL NaOH → L NaOH → mol NaOH → mol HCl; mL HCl → L HCl & mol  M Example:The titration of 10.00 mL of HCl solution of unknown concentration requires 12.54 mL of 0.100 M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution? = 1.25 x 10-3 mol HCl Tro, Chemistry: A Molecular Approach

  21. Apply the Concept Plan: Information Given: 10.00 mL HCl 12.54 mL NaOH Find: M HCl CF: 1 mol HCl = 1 mol NaOH 0.100 mol NaOH = 1 L M = mol/L CP: mL NaOH → L NaOH → mol NaOH → mol HCl; mL HCl → L HCl & mol  M Example:The titration of 10.00 mL of HCl solution of unknown concentration requires 12.54 mL of 0.100 M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution? Tro, Chemistry: A Molecular Approach

  22. Check the Solution: Information Given: 10.00 mL HCl 12.54 mL NaOH Find: M HCl CF: 1 mol HCl = 1 mol NaOH 0.100 mol NaOH = 1 L M = mol/L CP: mL NaOH → L NaOH → mol NaOH → mol HCl; mL HCl → L HCl & mol  M Example:The titration of 10.00 mL of HCl solution of unknown concentration requires 12.54 mL of 0.100 M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution? HCl solution = 0.125 M The units of the answer, M, are correct. The magnitude of the answer makes sense since the neutralization takes less HCl solution than NaOH solution, so the HCl should be more concentrated. Tro, Chemistry: A Molecular Approach

  23. Gas Evolving Reactions • Some reactions form a gas directly from the ion exchange K2S(aq) + H2SO4(aq)  K2SO4(aq) + H2S(g) • Other reactions form a gas by the decomposition of one of the ion exchange products into a gas and water K2SO3(aq) + H2SO4(aq)  K2SO4(aq) + H2SO3(aq) H2SO3 H2O(l) + SO2(g) Tro, Chemistry: A Molecular Approach

  24. NaHCO3(aq) + HCl(aq)  NaCl(aq) + CO2(g) + H2O(l) Tro, Chemistry: A Molecular Approach

  25. Compounds that UndergoGas Evolving Reactions Tro, Chemistry: A Molecular Approach

  26. Example 4.15 - When an aqueous solution of sodium carbonate is added to an aqueous solution of nitric acid, a gas evolves • Write the formulas of the reactants Na2CO3(aq) + HNO3(aq)  • Determine the possible products • Determine the ions present when each reactant dissociates (Na+1 + CO3-2) + (H+1 + NO3-1)  • Exchange the anions (Na+1 + CO3-2) + (H+1 + NO3-1)  (Na+1 + NO3-1) + (H+1 + CO3-2) • Write the formula of compounds • cross the charges Na2CO3(aq) + HNO3(aq)  NaNO3 + H2CO3 Tro, Chemistry: A Molecular Approach

  27. Example 4.15 - When an aqueous solution of sodium carbonate is added to an aqueous solution of nitric acid, a gas evolves • Check to see either product H2S - No • Check to see if either product decomposes – Yes • H2CO3 decomposes into CO2(g) + H2O(l) Na2CO3(aq) + HNO3(aq)  NaNO3 + CO2(g) + H2O(l) Tro, Chemistry: A Molecular Approach

  28. Example 4.15 - When an aqueous solution of sodium carbonate is added to an aqueous solution of nitric acid, a gas evolves • Determine the solubility of other product NaNO3 is soluble • Write an (s) after the insoluble products and a (aq) after the soluble products Na2CO3(aq) + 2 HNO3(aq)  2 NaNO3(aq) + CO2(g) + H2O(l) • Balance the equation Na2CO3(aq) + 2 HNO3(aq)  2 NaNO3 + CO2(g) + H2O(l) Tro, Chemistry: A Molecular Approach

  29. Other Patterns in Reactions • the precipitation, acid-base, and gas evolving reactions all involved exchanging the ions in the solution • other kinds of reactions involve transferring electrons from one atom to another – these are called oxidation-reduction reactions • also known as redox reactions • many involve the reaction of a substance with O2(g) 4 Fe(s) + 3 O2(g)  2 Fe2O3(s) Tro, Chemistry: A Molecular Approach

  30. Combustion as Redox2 H2(g) + O2(g) 2 H2O(g) Tro, Chemistry: A Molecular Approach

  31. Redox without Combustion2 Na(s)+ Cl2(g)  2 NaCl(s) 2 Na 2 Na+ + 2 e Cl2+ 2 e  2 Cl Tro, Chemistry: A Molecular Approach

  32. Reactions of Metals with Nonmetals • consider the following reactions: 4 Na(s) + O2(g) → 2 Na2O(s) 2 Na(s) + Cl2(g) → 2 NaCl(s) • the reaction involves a metal reacting with a nonmetal • in addition, both reactions involve the conversion of free elements into ions 4 Na(s) + O2(g) → 2 Na+2O– (s) 2 Na(s) + Cl2(g) → 2 Na+Cl–(s) Tro, Chemistry: A Molecular Approach

  33. Ger Leo Oxidation and Reduction • in order to convert a free element into an ion, the atoms must gain or lose electrons • of course, if one atom loses electrons, another must accept them • reactions where electrons are transferred from one atom to another are redox reactions • atoms that lose electrons are being oxidized, atoms that gain electrons are being reduced 2 Na(s) + Cl2(g) → 2 Na+Cl–(s) Na → Na+ + 1 e– oxidation Cl2+ 2 e– → 2 Cl– reduction Tro, Chemistry: A Molecular Approach

  34. Electron Bookkeeping • for reactions that are not metal + nonmetal, or do not involve O2, we need a method for determining how the electrons are transferred • chemists assign a number to each element in a reaction called an oxidation state that allows them to determine the electron flow in the reaction • even though they look like them, oxidation states are not ion charges! • oxidation states are imaginary charges assigned based on a set of rules • ion charges are real, measurable charges Tro, Chemistry: A Molecular Approach

  35. Rules for Assigning Oxidation States • rules are in order of priority • free elements have an oxidation state = 0 • Na = 0 and Cl2 = 0 in 2 Na(s) + Cl2(g) • monatomic ions have an oxidation state equal to their charge • Na = +1 and Cl = -1 in NaCl • (a) the sum of the oxidation states of all the atoms in a compound is 0 • Na = +1 and Cl = -1 in NaCl, (+1) + (-1) = 0 Tro, Chemistry: A Molecular Approach

  36. Rules for Assigning Oxidation States • (b) the sum of the oxidation states of all the atoms in a polyatomic ion equals the charge on the ion • N = +5 and O = -2 in NO3–, (+5) + 3(-2) = -1 • (a) Group I metals have an oxidation state of +1 in all their compounds • Na = +1 in NaCl • (b) Group II metals have an oxidation state of +2 in all their compounds • Mg = +2 in MgCl2 Tro, Chemistry: A Molecular Approach

  37. Rules for Assigning Oxidation States • in their compounds, nonmetals have oxidation states according to the table below • nonmetals higher on the table take priority Tro, Chemistry: A Molecular Approach

  38. Practice – Assign an Oxidation State to Each Element in the following • Br2 • K+ • LiF • CO2 • SO42- • Na2O2 Tro, Chemistry: A Molecular Approach

  39. Practice – Assign an Oxidation State to Each Element in the following • Br2Br = 0, (Rule 1) • K+ K = +1, (Rule 2) • LiF Li = +1, (Rule 4a) & F = -1, (Rule 5) • CO2 O = -2, (Rule 5) & C = +4, (Rule 3a) • SO42- O = -2, (Rule 5) & S = +6, (Rule 3b) • Na2O2 Na = +1, (Rule 4a) & O = -1, (Rule 3a) Tro, Chemistry: A Molecular Approach

  40. oxidation reduction Oxidation and ReductionAnother Definition • oxidation occurs when an atom’s oxidation state increases during a reaction • reduction occurs when an atom’s oxidation state decreases during a reaction CH4 + 2 O2 → CO2 + 2 H2O -4 +1 0+4 –2 +1 -2 Tro, Chemistry: A Molecular Approach

  41. Oxidation–Reduction • oxidation and reduction must occur simultaneously • if an atom loses electrons another atom must take them • the reactant that reduces an element in another reactant is called the reducing agent • the reducing agent contains the element that is oxidized • the reactant that oxidizes an element in another reactant is called the oxidizing agent • the oxidizing agent contains the element that is reduced 2 Na(s) + Cl2(g) → 2 Na+Cl–(s) Na is oxidized, Cl is reduced Na is the reducing agent, Cl2 is the oxidizing agent Tro, Chemistry: A Molecular Approach

  42. Identify the Oxidizing and Reducing Agents in Each of the Following 3 H2S + 2 NO3– + 2 H+® 3S + 2 NO + 4 H2O MnO2 + 4 HBr ® MnBr2 + Br2 + 2 H2O Tro, Chemistry: A Molecular Approach

  43. oxidation reduction oxidation reduction Identify the Oxidizing and Reducing Agents in Each of the Following 3 H2S + 2 NO3– + 2 H+® 3S + 2 NO + 4 H2O MnO2 + 4 HBr ® MnBr2 + Br2 + 2 H2O red ag ox ag +1 -2 +5 -2 +1 0 +2 -2 +1 -2 ox ag red ag +4 -2 +1 -1 +2 -1 0 +1 -2 Tro, Chemistry: A Molecular Approach

  44. Combustion Reactions • Reactions in which O2(g) is a reactant are called combustion reactions • Combustion reactions release lots of energy • Combustion reactions are a subclass of oxidation-reduction reactions 2 C8H18(g) + 25 O2(g)  16 CO2(g) + 18 H2O(g) Tro, Chemistry: A Molecular Approach

  45. Combustion Products • to predict the products of a combustion reaction, combine each element in the other reactant with oxygen Tro, Chemistry: A Molecular Approach

  46. Practice – Complete the Reactions • combustion of C3H7OH(l) • combustion of CH3NH2(g) Tro, Chemistry: A Molecular Approach

  47. Practice – Complete the Reactions C3H7OH(l) + 5 O2(g)  3 CO2(g) + 4 H2O(g) CH3NH2(g) + 3 O2(g)  CO2(g) + 2 H2O(g) + NO2(g) Tro, Chemistry: A Molecular Approach

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