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There are several key questions to ask. 1. Is the species a strong acid ?

There are several key questions to ask. 1. Is the species a strong acid ? 2. Is the species a strong base ? 3. Is the species ionic or molecular ? 4. Is the species soluble in water? To answer these questions requires some factual information – i.e. a knowledge base. Strong Acids.

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There are several key questions to ask. 1. Is the species a strong acid ?

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  1. There are several key questions to ask. 1. Is the species a strong acid? 2. Is the species a strong base? 3. Is the species ionic or molecular? 4. Is the species soluble in water? To answer these questions requires some factual information – i.e. a knowledge base.

  2. Strong Acids Strong acid: A strong acid dissociates almost 100 % in water. E. g. in water HNO3(aq) H3O+(aq)+ NO3-(aq) ess. 100%

  3. Strong Acids The list of common strong acids is short. It is HCl HNO3 HClO3 HBr H2SO4 HClO4 HI Know this list. There are other more exotic strong acids, but these are not commonly encountered. Any acid not on this list can be regarded as weak.

  4. Strong bases Strong base: A strong base dissociates almost 100 % in water. E. g. in water NaOH(aq) Na+(aq)+ OH-(aq) ess. 100%

  5. The list of common strong bases is short. It is LiOH NaOH KOH RbOH CsOHBa(OH)2* * Solubility limited. Know this list. There are other more exotic strong bases, but these are not commonly encountered. Any base not on this list can be regarded as weak.

  6. Ionic versus molecular

  7. Ionic versus molecular

  8. A. A typical metal combined with a typical nonmetal will yield an ionic compound.

  9. A. A typical metal combined with a typical nonmetal will yield an ionic compound. B. A typical nonmetal combined with a typical nonmetal will yield a molecular compound.

  10. A. A typical metal combined with a typical nonmetal will yield an ionic compound. B. A typical nonmetal combined with a typical nonmetal will yield a molecular compound. Lots of exceptions, particularly further down the periodic table.

  11. Solubility issues

  12. Now we are ready to address the original question!

  13. Now we are ready to address the original question! For the example, all 4 questions will come into play.

  14. Now we are ready to address the original question! For the example, all 4 questions will come into play. 1. Is the species a strong acid?

  15. Now we are ready to address the original question! For the example, all 4 questions will come into play. 1. Is the species a strong acid? 2. Is the species a strong base?

  16. Now we are ready to address the original question! For the example, all 4 questions will come into play. 1. Is the species a strong acid? 2. Is the species a strong base? 3. Is the species ionic or molecular?

  17. Now we are ready to address the original question! For the example, all 4 questions will come into play. 1. Is the species a strong acid? 2. Is the species a strong base? 3. Is the species ionic or molecular? 4. Is the species soluble in water?

  18. NaOH(aq) + HCl(aq)NaCl(aq) + H2O(l) strong strong ionic? ionic? base ? acid ? soluble?

  19. NaOH(aq) + HCl(aq)NaCl(aq) + H2O(l) strong strong ionic? yes ionic? no base ? yes acid ? yes soluble? yes

  20. NaOH(aq) + HCl(aq)NaCl(aq) + H2O(l) strong strong ionic? yes ionic? no base ? yes acid ? yes soluble? yes Na+(aq)+ OH-(aq)+ H+(aq)+ Cl-(aq)Na+(aq) + Cl-(aq) + H2O(l)

  21. NaOH(aq) + HCl(aq)NaCl(aq) + H2O(l) strong strong ionic? yes ionic? no base ? yes acid ? yes soluble? yes Na+(aq)+ OH-(aq)+ H+(aq)+ Cl-(aq)Na+(aq) + Cl-(aq) + H2O(l) Now cancel species that are common to both sides. Note – they must be exactly the same.

  22. NaOH(aq) + HCl(aq)NaCl(aq) + H2O(l) strong strong ionic? yes ionic? no base ? yes acid ? yes soluble? yes Na+(aq)+ OH-(aq)+ H+(aq)+ Cl-(aq)Na+(aq) + Cl-(aq) + H2O(l) Now cancel species that are common to both sides. Note – they must be exactly the same. OH-(aq)+ H+(aq) H2O(l)

  23. NaOH(aq) + HCl(aq)NaCl(aq) + H2O(l) strong strong ionic? yes ionic? no base ? yes acid ? yes soluble? yes Na+(aq)+ OH-(aq)+ H+(aq)+ Cl-(aq)Na+(aq) + Cl-(aq) + H2O(l) Now cancel species that are common to both sides. Note – they must be exactly the same. OH-(aq)+ H+(aq) H2O(l) This is the required net ionic equation for the reaction.

  24. Example 2: Give the net ionic equation for the reaction: K2SO4(aq) + BaCl2(aq) 2 KCl + BaSO4 Note: The question does not tell you if KCl and BaSO4 are dissolved in solution.

  25. For this example, only 2 questions will come into play (since we are not dealing with acids or bases in this example).

  26. For this example, only 2 questions will come into play (since we are not dealing with acids or bases in this example). 3. Is the species ionic or molecular?

  27. For this example, only 2 questions will come into play (since we are not dealing with acids or bases in this example). 3. Is the species ionic or molecular? 4. Is the species soluble in water?

  28. K2SO4(aq) + BaCl2(aq) 2 KCl + BaSO4 ionic? ionic? ionic? ionic? soluble? soluble? soluble? soluble?

  29. K2SO4(aq) + BaCl2(aq) 2 KCl + BaSO4 ionic? yes ionic? yes ionic? yes ionic? yes soluble? yes soluble? yes soluble? yes soluble? no

  30. K2SO4(aq) + BaCl2(aq) 2 KCl + BaSO4 ionic? yes ionic? yes ionic? yes ionic? yes soluble? yes soluble? yes soluble? yes soluble? no Key comment: The subscript label “aq” tells you that the particular compound is soluble.

  31. K2SO4(aq) + BaCl2(aq) 2 KCl + BaSO4 ionic? yes ionic? yes ionic? yes ionic? yes soluble? yes soluble? yes soluble? yes soluble? no Key comment: The subscript label “aq” tells you that the particular compound is soluble. If the answer to both questions is yes, then split the compound into ions.

  32. 2 K+(aq)+ SO42-(aq)+ Ba2+(aq)+ 2 Cl-(aq)2 K+(aq)+ 2 Cl-(aq)+ BaSO4(s)

  33. 2 K+(aq)+ SO42-(aq)+ Ba2+(aq)+ 2 Cl-(aq)2 K+(aq)+ 2 Cl-(aq)+ BaSO4(s) Now cancel the common species that are exactly the same on both sides of the equation.

  34. 2 K+(aq)+ SO42-(aq)+ Ba2+(aq)+ 2 Cl-(aq)2 K+(aq)+ 2 Cl-(aq)+ BaSO4(s) Now cancel the common species that are exactly the same on both sides of the equation. SO42-(aq)+ Ba2+(aq)BaSO4(s)

  35. 2 K+(aq)+ SO42-(aq)+ Ba2+(aq)+ 2 Cl-(aq)2 K+(aq)+ 2 Cl-(aq)+ BaSO4(s) Now cancel the common species that are exactly the same on both sides of the equation. SO42-(aq)+ Ba2+(aq)BaSO4(s) This is the required net ionic equation.

  36. Quantitative Aspects of Chemical Reactions

  37. Quantitative Aspects of Chemical Reactions Objective: The study of the quantitative relations between amounts of reactants and products.

  38. Quantitative Aspects of Chemical Reactions Objective: The study of the quantitative relations between amounts of reactants and products. Stoichiometry: The stoichiometry of a reaction is the description of the relative quantities by moles of the reactants and products as given by the coefficients of the balanced equation for the reaction.

  39. Stoichiometry Calculations Coefficients in a balanced chemical equation give relative quantities of reactants and products by moles as well as by molecules.

  40. Stoichiometry Calculations Coefficients in a balanced chemical equation give relative quantities of reactants and products by moles as well as by molecules. Example: 2 H2 + O2 2H2O This can be read as: 2molecules of H2 react with 1molecule of O2 to give 2molecules of water.

  41. Or multiply both sides by 6.02 x 1023 (Avogadro’s number), so we can read the equation as:

  42. Or multiply both sides by 6.02 x 1023 (Avogadro’s number), so we can read the equation as: 2moles of H2 react with 1mole of O2 to give 2moles of water.

  43. Or multiply both sides by 6.02 x 1023 (Avogadro’s number), so we can read the equation as: 2moles of H2 react with 1mole of O2 to give 2moles of water. This statement in terms of moles will be the most useful one for solving stoichiometry problems.

  44. Or multiply both sides by 6.02 x 1023 (Avogadro’s number), so we can read the equation as: 2moles of H2 react with 1mole of O2 to give 2moles of water. This statement in terms of moles will be the most useful one for solving stoichiometry problems. The essential idea is that the ratio of reactants and products is 2 : 1: 2 in both statements.

  45. Solving Stoichiometry Problems using the mole method

  46. Four steps: (1) Obtain the correct formulas for all reactants and products and balance the chemical equation.

  47. Four steps: (1) Obtain the correct formulas for all reactants and products and balance the chemical equation. (2) Convert all of the known amounts of the substances into moles.

  48. Four steps: (1) Obtain the correct formulas for all reactants and products and balance the chemical equation. (2) Convert all of the known amounts of the substances into moles. (3) Use the coefficients of the substances in the balanced equation to calculate the number of moles of the unknown quantities in the problem.

  49. Four steps: (1) Obtain the correct formulas for all reactants and products and balance the chemical equation. (2) Convert all of the known amounts of the substances into moles. (3) Use the coefficients of the substances in the balanced equation to calculate the number of moles of the unknown quantities in the problem. (4) Convert the number of moles of the unknown quantities to grams (or other mass units).

  50. Example: Nitrogen dioxide is a major air pollutant. It may be formed by the direct combination of dinitrogen and dioxygen.

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