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Types of Chemical Reactions and Solution Stoichiometry. Classification of Matter. Solutions are homogeneous mixtures. Solute. A solute is the dissolved substance in a solution. Salt in salt water. Sugar in soda drinks. Carbon dioxide in soda drinks. Solvent.

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Types of Chemical Reactions and Solution Stoichiometry

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    1. Types of Chemical Reactionsand Solution Stoichiometry

    2. Classification of Matter Solutions are homogeneous mixtures

    3. Solute A solute is the dissolved substance in a solution. Salt in salt water Sugar in soda drinks Carbon dioxide in soda drinks Solvent A solvent is the dissolving medium in a solution. Water in salt water Water in soda

    4. Saturation of Solutions • A solution that contains the maximum amount of solute that may be dissolved under existing conditions is saturated. • A solution that contains less solute than a saturated solution under existing conditions is unsaturated. • A solution that contains more dissolved solute than a saturated solution under the same conditions is supersaturated.

    5. Electrolytes vs. Nonelectrolytes The ammeter measures the flow of electrons (current) through the circuit. • If the ammeter measures a current, and the bulb glows, then the solution conducts. • If the ammeter fails to measure a current, and the bulb does not glow, the solution is non-conducting.

    6. Definition of Electrolytes and Nonelectrolytes An electrolyte is: • A substance whose aqueous solution conducts • an electric current. A nonelectrolyte is: • A substance whose aqueous solution does not • conduct an electric current. Try to classify the following substances as electrolytes or nonelectrolytes…

    7. Electrolytes? • Pure water • Tap water • Sugar solution • Sodium chloride solution • Hydrochloric acid solution • Lactic acid solution • Ethyl alcohol solution • Pure, solid sodium chloride

    8. Answers… ELECTROLYTES: NONELECTROLYTES: • Tap water (weak) • NaCl solution • HCl solution • Lactate solution (weak) • Pure water • Sugar solution • Ethanol solution • Pure, solid NaCl But why do some compounds conduct electricity in solution while others do not…? We must understand the nature of water

    9. The Nature of Water • Water molecules are BENT (105⁰<) • H & O share electrons UNEVENLY • O pulls harder on e-s, so the O end of the molecule is slightly negative • H pulls less hard on e-s, so the H end of the molecule is slightly positive • This makes H2O a POLAR molecule (oppositely charged ends), like a little magnet • AKA “a dipole”

    10. The Nature of Water • The polarity of water gives it great ability to dissolve compounds

    11. 1. Ionic Compounds Ionize in Solution • + ions associate with the - (oxygen) end of the water dipole. • - ions associate with the + (hydrogen) end of the water dipole. • Ions tend to stay in solution where they can conduct a current rather than reforming a solid. • IONIC CPDS ARE ELECTROLYTES! • Dissociation of sodium chloride

    12. Many Ionic Compounds Dissociate (break apart) H2O NaCl(s)  Na+(aq) + Cl-(aq) H2O AgNO3(s)  Ag+(aq) + NO3-(aq) H2O MgCl2(s)  Mg2+(aq) + 2 Cl-(aq) H2O Na2SO4(s)  2 Na+(aq) + SO42-(aq) H2O AlCl3(s)  Al3+(aq) + 3 Cl-(aq) Note: write “H2O” above each arrow to show that the ionic compound is placed in water, NOT reacting with water.

    13. Practice Dissociation Equations • Study Guide p 106 #1, 2

    14. 2. Covalent (molecular) acids IONIZE in solution • Again, b/c water is polar…and also b/c • covalent acids are also polar! • For instance, hydrogen chloride molecules, which are polar, give up their hydrogens to water, • forming chloride ions (Cl-) • and hydronium ions (H3O+).

    15. Strong acids are completely ionized in solution (Strong Electrolytes) Examples of strong acids include: • Hydrochloric acid, HCl • Sulfuric acid, H2SO4 • Nitric acid, HNO3 • Hydroiodic acid, HI • Perchloric acid, HClO4 In general, we can assume nearly all other acids are weak

    16. Weak acids ionize only slightly (Weak Electrolytes) • HC2H3O2 (acetic acid or vinegar)is a weak acid Many of these weaker acids are “organic” acids that contain a “carboxyl” group. The carboxyl group does not easily give up its hydrogen.

    17. FYI: Because of the carboxyl group, organic acids are sometimes called “carboxylic acids”. Other organic acids and their sources include: • Citric acid – citrus fruit • Malic acid – apples • Butyric acid – rancid butter • Amino acids – protein • Nucleic acids – DNA and RNA • Ascorbic acid – Vitamin C This is an enormous group of compounds; these are only a few examples.

    18. 3.Some Other Polar Covalent Compounds are Weak Electrolytes • They ionize very slightly in water • Ex: Ammonia, NH3 • Only about 1% of the molecules dissociate!

    19. 4. MOST covalent compounds do not ionize at all in solution. (Nonelectrolyes) • Sugar (sucrose – C12H22O11), • and ethanol (ethyl alcohol – C2H5OH) NOTE: These molecular compounds DISSOLVE (the water pulls the molecules away from each other, but do not IONIZE (the molecules are not charged.)

    20. Practice Identifying Electrolytes • Study Guide, p 106 # 4 & 5

    21. Molarity The concentration of a solution measured in moles of solute per liter of solution. mol= M L

    22. Preparation of Molar Solutions Problem: How many grams of sodium chloride are needed to prepare 1.50 liters of 0.500 M NaCl solution? • Step #1: Ask “How Much?” (What volume to prepare?) • Step #2: Ask “How Strong?” (What molarity?) • Step #3: Ask “What does it weigh?” (Molar mass is?) 1.500 L 0.500 mol 58.44 g = 43.8 g 1 L 1 mol

    23. Practice calculating mass needed to make a solution • How many grams of NaOH are needed to make 3.5 L of a 2.5 molar solution? Unknown =?g NaOH Given 3.5 L x 2.5 mol NaoH x 1 L of soln 39.9g NaoH 1 mol NaOH

    24. Serial Dilution It’s not practical to keep solutions of many different concentrations on hand, so chemists prepare more dilute solutions from a more concentrated “stock” solution. Problem: What volume of stock (11.6 M) hydrochloric acid is needed to prepare 250. mL of 3.0 M HCl solution? (Note: must convert volumes to L) MstockVstock = MdiluteVdilute OR M1V1=M2V2 (11.6 M)(x Liters) = (3.0 M)(0.250 Liters) x Liters = (3.0 M)(0.250 Liters) 11.6 M = 0.065 L

    25. Molarity of Ions in Solution • When an ionic compound ionizes in solution, the number of ions formed may be different than the moles of compound. H2O(l) • Ex: CrCl3 (s)  Cr3+(aq)+ 3Cl-(aq) • Say “1 mole of CrCl3 ionizes to form 1 mole of Cr3+ ionsand 3 moles Cl- ions”

    26. Calculating Molarity of Ions in Solution 0.25 M CrCl3 What is the molarity of Cr3+ ions? Cl- ions? CrCl3 (s)  Cr3+(aq)+ 3Cl-(aq) • ANSWER Molarity of Cr3+ = molarity of CrCl3 = 0.25 M Molarity of Cl- = 3 x molarity of CrCl3 = 3 x 0.25M = 0.75M

    27. Practice Dilutions & Molarity Calcs • Study Guide, p 106 # 7, 9, 11, 13,

    28. Types of Reactions • 5 basic • Precipitation • Net ionic • Oxidation-reduction (redox) • Acid-Base

    29. 1. Single Replacement Reactions A + BX  AX + B BX + Y  BY + X Replacement of: • Metals by another metal • Hydrogen in water by a metal • Hydrogen in an acid by a metal • Halogens by more active halogens

    30. The Activity Series of the Metals • Lithium • Potassium • Calcium • Sodium* • Magnesium • Aluminum • Zinc • Chromium • Iron • Nickel • Lead • Hydrogen • Bismuth • Copper • Mercury • Silver • Platinum • Gold Metals can replace other metals provided that they are above the metal that they are trying to replace. Metals above hydrogen can replace hydrogen in acids. *Metals from sodium upward can replace hydrogen in water

    31. The Activity Series of the Halogens • Fluorine • Chlorine • Bromine • Iodine Halogens can replace other halogens in compounds, provided that they are above the halogen that they are trying to replace. 2NaCl(s) + F2(g)  ??? 2NaF(s) + Cl2(g) MgCl2(s) + Br2(g)  ??? No Reaction

    32. 2.Double Replacement Reactions The ions of two compounds exchange places in an aqueous solution to form two new compounds. AX + BY  AY + BX • One of the compounds formed is usually • a precipitate (an insoluble solid), • an insoluble gas that bubbles out of solution, • or a molecular compound, usually water.

    33. Highly Soluble Ionic Compounds (& their exceptions) MEMORIZE THESE RULES!

    34. Slightly Soluble (usually considered insoluble) Ionic Compounds MEMORIZE THESE RULES!

    35. Completing a Double replacement Rxn: Ex: Pb(NO3)2(aq) + 2KI(aq) PbI2 + 2KNO3 (aq) (s) 1. Determine the products formed when the ions are exchanged, & balance 2. Decide if the products are soluble (aq) or insoluble (s) PbI2 Insoluble, so it will be a solid KNO3 Soluble, so it will be in aqueous solution

    36. Complete Ionic Equation • Shows all soluble compounds as aqueous ions Pb(NO3)2(aq) 2KI  PbI2(s) + 2KNO3 (aq) + Pb2+(aq) + 2 NO3-(aq) + 2 K+(aq) +2 I-(aq)  2K+(aq) + 2 NO3-(aq) PbI2(s) + Shows all insoluble compounds as a unit, use the symbol (s) to show it is a solid.

    37. Net Ionic Equation • Eliminates all “spectator” ions (ions that appear identically on both sides of equation) Pb2+(aq) + 2NO3-(aq) + 2K+(aq) +2I-(aq)  2K+(aq) + 2NO3-(aq) PbI2(s) + Pb2+(aq) + 2 I-(aq)  PbI2(s)

    38. Practice Complete Ionic Equation & Net Ionic Equation CuSO4 + Na2S  CuS (insoluble) + Na2SO4 (soluble) Complete Ionic Net Ionic Cu2+(aq)+SO42-(aq) +2Na+(aq) +S2-(aq)  CuS(s) + 2Na+(aq)+SO42- (aq) Cu2+(aq) + S2-(aq)  CuS(s)

    39. Practice Problems • Study Guide: Please Note: show all work in your notebook, not your study guide! • Pp 108 #28

    40. Stoichiometry of Precipitation Rxns • It is helpful to be able to predict the amount of precipitant formed b/c it is often collected & used. • Ex: Calculate the mass of solid NaCl that must be added to 1.50L of a 0.100 M AgNO3 solution to precipitate all the Ag+ ions in the form of AgCl. • First, we need a balanced equation: NaCl(s) + AgNO3(aq)  AgCl(s) + NaNO3(aq) Given 1.5 L AgNO3 x 1 mol NaCl x 1 mol AgNO3 58.45 g NaCl = 1 mol NaCl 8.77 ___g NaCl 0.100 mol AgNO3 x 1 L AgNO3

    41. Stoichiometry of Precipitation Rxns, cont. • Please note: you will also encounter limiting reactant problems with precip. Rxns! • Remember, set up 2 separate equations to see which reactant forms the LEAST product. This is your limiting reactant. The equations will be identical to those on the prior page.

    42. Stoichiometry of Precipitation Rxns, cont. • HINT: label all equations with your givens & unknown. Example: When aqueous solutions of Na2SO4 and Pb(NO3)2 are mixed, PbSO4 precipitates. Calculate the mass of PbSO4 formed when 1.25 L of 0.0500 M Pb(NO3)2 and 2.00 L of 0.0250 M Na2SO4 are mixed. Na2SO4 (aq) + Pb(NO3)2 (aq)  PbSO4 (s) + 2NaNO3 (aq) V=2.00 L M=0.0250mol/L V=1.25 L M=0.0500mol/L m=? g

    43. Practice Problems: • Study Guide, pp __________ Q # _________ • Please show all your work in your notebook, not your study guide.

    44. Oxidation and Reduction Reactions (Redox) • Def: Rxns in which electrons are transferred • Ex: Na(s) + Cl2(g)  NaCl(s) • An electron transfers from the Na atom to the Cl atom. 

    45. What about molecular compounds? • Non-ionic compounds can also be formed from redox reactions. Even though e-s aren’t FULLY transferred, they can be assumed to involve a transfer…let’s see how!

    46. Oxidation and Reduction Reactions (Redox) • Ex:CH4(g) + 2O2(g)  CO2(g) + 2H2O + energy • Carbon is less EN than oxygen, so we assume there is a transfer of e-s from C to O. • NOTE: We will study electronegativity (EN) in Chapter 8. • EN is the strength with which an atom in a bond pulls on electrons. • (Check Figure 8.3 on p 353.) • This shows us the value for EN of O is 3.5 and the EN value for C is 2.5.

    47. Oxidation and Reduction Rxns, cont. • Oxidation States (Oxidation #s) provide a way to track e-s in redox reactions, especially in molecular substances.

    48. Rules for Assigning Oxidation NumbersRules 1 & 2 • The oxidation number of any uncombined element is zero 2. The oxidation number of a monatomic ion equals its charge

    49. Rules for Assigning Oxidation NumbersRules 3 & 4 3.The oxidation number of oxygen in compounds is -2 (except in peroxides, in which the oxidation number is -1) 4. The oxidation number of hydrogen in compounds is +1

    50. Rules for Assigning Oxidation Number Rule 5 5. The sum of the oxidation numbers in the formula of a compound is 0 2(+1) + (-2) = 0 H O (+2) + 2(-2) + 2(+1) = 0 Ca O H