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Types of Chemical Reactions & Solution Stoichiometry

Types of Chemical Reactions & Solution Stoichiometry. Chapter 4. Aqueous Solutions. Water is the dissolving medium, or solvent. Some Properties of Water. Water is able to dissolve so many substances because: Water is “bent” or V-shaped . The O-H bonds are covalent .

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Types of Chemical Reactions & Solution Stoichiometry

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  1. Types of Chemical Reactions & Solution Stoichiometry • Chapter 4

  2. Aqueous Solutions Water is the dissolving medium, or solvent.

  3. Some Properties of Water • Water is able to dissolve so many substances because: • Water is “bent” or V-shaped. • The O-H bonds are covalent. • Water is a polar molecule. • Hydration occurs when salts dissolve in water.

  4. Water is a polar molecule because it is a bent molecule. The hydrogen end is + while the oxygen end is -, Delta () is a partial charge--less than 1.

  5. A Solute • dissolves in water (or other “solvent”) • changes phase (if different from the solvent) • is present in lesser amount (if the same phase as the solvent)

  6. A Solvent • retains its phase (if different from the solute) • is present in greater amount (if the same phase as the solute)

  7. Polar water molecules interact with the positive and negative ions of a salt, assisting in the dissolving process. This process is called hydration.

  8. Solubility • The general rule for solubility is: • “Like dissolves like.” • Polar water molecules can dissolve other polar molecules such as alcohol and, also, ionic substances such as NaCl. • Nonpolar molecules can dissolve other nonpolar molecules but not polar or ionic substances. Gasoline can dissolve grease.

  9. Miscibility • Miscible -- two substances that will mix together in any proportion to make a solution. Alcohol and water are miscible because they are both polar and form hydrogen bonds. • Immiscible -- two substances that will not dissolve in each other. Oil and vinegar are immiscible because oil is nonpolar and vinegar is polar.

  10. Solubility • How does the rule “Like dissolves like.” apply to cleaning paint brushes used for latex paint as opposed to those used with oil-based paint?

  11. Simple Rules for Solubility • 1. Most nitrate (NO3) salts are soluble. • 2. Most alkali (group 1A) salts and NH4+are soluble. • 3. Most Cl, Br, and I salts are soluble (NOTAg+, Pb2+, Hg22+) • 4. Most sulfate salts are soluble (NOTBaSO4, PbSO4, HgSO4, CaSO4) • 5. Most OH salts are only slightly soluble (NaOH, KOH are soluble, Ba(OH)2, Ca(OH)2 are marginally soluble) • 6. Most S2, CO32, CrO42, PO43 salts are only slightly soluble.

  12. Electrolytes & Nonelectrolytes • An electrolyte is a material that dissolves in water to give a solution that conducts an electric current. • A nonelectrolyte is a substance which, when dissolved in water, gives a nonconducting solution.

  13. Electrolytes • Strong - conduct current efficiently and are soluble salts, strong acids, and strong bases. • NaCl, KNO3, HNO3, NaOH • Weak - conduct only a small current and are weak acids and weak bases. • HC2H3O2, aq. NH3, tap H2O • Non - no current flows and are molecular substances • pure H2O, sugar solution, glycerol

  14. Electrical conductivity of aqueous solutions. a) strong electrolyte b) weak electrolyte c) nonelectrolyte in solution. Svante Arrhenius first identified these electrical properties.

  15. When BaCl2 dissolves, the Ba2+ and Cl- ions are randomly dispersed in the water. BaCl2 is a strong electrolyte.

  16. Acids • Strong acids - dissociate completely (~100 %) to produce H+ in solution • HCl, H2SO4, HNO3, HBr, HI, & HClO4 • Weak acids - dissociate to a slight extent (~ 1 %) to give H+ in solution • HC2H3O2, HCOOH, HNO2, & H2SO3

  17. HCl is completely ionized and is a strong electrolyte.

  18. Bases • Strong bases - react completely with water to give OH ions. sodium hydroxide • NaOH(s) ---> Na+(aq) + OH-(aq) • Weak bases - react only slightly with water to give OH ions. ammonia • NH3(aq) + HOH(l) <---> NH4+(aq) +OH-(aq)

  19. An aqueous solution of sodium hydroxide which is a strong bases dissociating almost 100 %.

  20. Acetic acid(CH3COOH) exists in water mostly as undissociated molecules. Only a small percent of the molecules are ionized.

  21. Molarity • Molarity (M) = moles of solute per volume of solution in liters:

  22. Molarity Calculations • Calculate the molarity of a solution prepared by dissolving 11.5 g of solid NaOH in enough water to make 1.50 L of solution. • (11.5g NaOH/1.50L)(1 mol NaOH/40.00g NaOH) • = 0.192M NaOH

  23. Molarity Calculations • Calculate the molarity of a solution prepared by dissolving 1.56 g of gaseous HCl in enough water to make 26.8 mL of solution. • (1.56g HCl/26.8mL)(1 mol HCl/36.46g HCl) (1000mL/1L) = 1.60M HCl

  24. Molarity Calculations • How many moles of nitrate ions are present in 25.00 mL of a 0.75 M Co(NO3)2 solution? • (25.00mL)(1L/1000mL)(0.75mol Co(NO3)2/1L) • (2 mol NO3-/1 mol Co(NO3)2) = 3.8 x 10-2 mol NO3-

  25. Molarity Calculations • Typical blood serum is about 0.14M NaCl. What volume of blood contains 1.0 mg of NaCl? • (1.0mg NaCl)(1g/1000mg)(1mol/58.45g)(1L/0.14mol) = 1.2 x 10-4 L blood serum

  26. Standard Solution • A standard solution is a solution whose concentration is accurately known. • Standard solutions are made using a volumetric flask as follows: • mass the solute accurately and add it to the volumetric flask • add a small quantity of distilled HOH • dissolve the solute by gently swirling the flask • add more distilled HOH until the level of the solution reaches the mark on the neck • invert the capped volumetric 25X to thoroughly mix the solution.

  27. Steps involved in making a standard solution.

  28. Common Terms of Solution Concentration • Stock - routinely used solutions prepared in concentrated form. • Concentrated - relatively large ratio of solute to solvent. (5.0 M NaCl) • Dilute - relatively small ratio of solute to solvent. (0.01 M NaCl)

  29. Dilution of Stock Solutions • When diluting stock solutions, the moles of solute after dilution must equal the moles of solute before dilution. • Stock solutions are diluted using either a measuring or a delivery pipet and a volumetric flask.

  30. Steps to dilute a stock solution.

  31. Dilution Calculations • What volume of 16 M sulfuric acid must be used to prepare 1.5 L of a 0.10 M H2SO4 solution? • (0.10mol H2SO4/1L)(1.5L)(1L/16mol)(1000mL/1L) • = 9.4 mL conc H2SO4

  32. Types of Solution Reactions • Precipitation reactions • AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq) • Acid-base reactions • NaOH(aq) + HCl(aq)  NaCl(aq) + H2O(l) • Oxidation-reduction reactions • Fe2O3(s) + 2Al(s)  2Fe(l) + Al2O3(s)

  33. Describing Reactions in Solution • 1. Molecular equation (reactants and products as compounds) • AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq) • 2.Complete ionic equation (all strong electrolytes shown as ions) • Ag+(aq) + NO3(aq) + Na+(aq) + Cl(aq)  • AgCl(s) + Na+(aq) + NO3(aq)

  34. Describing Reactions in Solution (continued) • 3. Net ionic equation (show only components that actually react) • Ag+(aq) + Cl(aq)  AgCl(s) • Na+ and NO3 are spectator ions.

  35. Simple Rules for Solubility • 1. Most nitrate (NO3) salts are soluble. • 2. Most alkali (group 1A) salts and NH4+are soluble. • 3. Most Cl, Br, and I salts are soluble (NOTAg+, Pb2+, Hg22+) • 4. Most sulfate salts are soluble (NOTBaSO4, PbSO4, HgSO4, CaSO4) • 5. Most OH salts are only slightly soluble (NaOH, KOH are soluble, Ba(OH)2, Ca(OH)2 are marginally soluble) • 6. Most S2, CO32, CrO42, PO43 salts are only slightly soluble.

  36. Solubility • Using the solubility rules, predict what will happen when the following pairs of solutions are mixed. • a) KNO3(aq) & BaCl2(aq) • b) Na2SO4(aq) & Pb(NO3)2(aq) • c) KOH(aq) & Fe(NO3)3(aq) No precipitate forms PbSO4(s) forms Fe(OH)3(s) forms

  37. Precipitation Calculations • When aqueous solutions of Na2SO4 & Pb(NO3)2 are mixed, PbSO4 precipitates. Calculate the mass of PbSO4 formed when 1.25 L of 0.0500 M Pb(NO3)2 & 2.00 L of 0.0250 M Na2SO4 are mixed. • 1. Species present are Na+, SO42-, Pb2+, & NO3-. • 2. Pb2+(aq) + SO42-(aq) ----> PbSO4(s)

  38. Precipitation CalculationsContinued • 3. (1.25L)(0.500mol Pb2+/L) = 0.0625 mol Pb2+ • (2.00L)(0.0250mol SO42-/L) = 0.0500 mol SO42- • 4. (0.0625mol Pb2+)(1mol SO42-/1mol Pb2+) = • 0.0625 mol SO42- • SO42- is the limiting reactant. • 5. (0.0500mol SO42-)(1mol PbSO4/1mol SO42-) • (303.3g/1mol) = 15.2 g PbSO4

  39. Performing Calculations for Acid-Base Reactions • 1. List initial species and predict reaction. • 2. Write balanced net ionic reaction. • 3. Calculate moles of reactants. • 4. Determine limiting reactant. • 5. Calculate moles of required reactant/product. • 6. Convert to grams or volume, as required.

  40. Acid-Base Calculations • What volume of a 0.100M HCl solution is needed to neutralize 25.0 mL of 0.350 M NaOH? • 1. H+, Cl-, Na+, & OH- • 2. H+(aq) + OH-(aq) ----> HOH(l) • (25.0mL)(0.350mol NaOH/1L)(1mol HCl/1mol NaOH)(1L/0.100mol) = 87.5 mL HCl solution

  41. Key Titration Terms • Titrant - solution of known concentration used in titration • Analyte - substance being analyzed • Equivalence point - enough titrant added to react exactly with the analyte • Endpoint - the indicator changes color so you can tell the equivalence point has been reached.

  42. Titration Calculations • If 41.20 mL of NaOH is required to neutralize 1.3009 g of KHP (KHC8H4O6), what is the concentration of the NaOH solution? Assume there is one acidic hydrogen in KHP. • HC8H4O6-(aq) + OH-(aq) ----> HOH(l) + C8H4O62-(aq) • (1.3009g KHP)(1mol/204.22g)(1mol NaOH/1mol KHP)(1/41.20mL)(1000mL/1L) = 0.1546 M NaOH

  43. Redox Reactions • Redox reactions are reactions in which electrons are transferred. • Decomposition and synthesis reactions may be redox. • Single replacement reactions are always redox. • Double replacement reactions are never redox. • Combustion reactions are always redox.

  44. OIL RIG • Oxidation Is Loss. • Reduction Is Gain.

  45. Redox • Oxidizing agent is the electron acceptor--usually a nonmetal. • Reducing agent is the electron donor--usually a metal. • CH4(g) + 2O2(g) ----> CO2(g) + 2HOH(g) • Carbon is oxidized. • Oxygen is reduced. • CH4 is the reducing agent. • O2 is the oxidizing agent.

  46. Rules for Assigning Oxidation States • 1. Oxidation state of an atom in an element = 0 • 2. Oxidation state of monatomic element = charge • 3. Oxygen = 2 in covalent compounds (except in peroxides where it = 1) • 4. H = +1 in covalent compounds • 5. Fluorine = 1 in compounds • 6. Sum of oxidation states = 0 in compounds Sum of oxidation states = charge of the ion

  47. Determining Oxidation States • SF6 [NO3]- • +6-6 = 0 +5 -6 = -1 (-1 for each F) (-2 for each O)

  48. Balancing by Half-Reaction Method • 1. Write separate reduction, oxidation reactions. • 2. For each half-reaction: • - Balance elements (except H, O) • - Balance O using H2O • - Balance H using H+ • - Balance charge using electrons

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