AP CHEMISTRY CHAPTER 4 SOLUTION STOICHIOMETRY

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AP CHEMISTRY CHAPTER 4 SOLUTION STOICHIOMETRY Electrons aren t shared evenly in water (oxygen is more electronegative). Electrons spend more time close to O than to H.

The negative end of water (O) attracts the positive ions or the positive end of another polar molecule.

1 AP CHEMISTRY CHAPTER 4 SOLUTION STOICHIOMETRY Electrons aren t shared evenly in water (oxygen is more electronegative). Electrons spend more time close to O than to H. This uneven distribution of charge makes water polar. Because of this, water is a good solvent. The positive end (H) attracts negative ions or the negative end of another polar molecule. The negative end of water (O) attracts the positive ions or the positive end of another polar molecule. When water surrounds an ionic crystal, the H end attracts the anion and the O end attracts the cation. This process is called hydration. Hydration causes salts (ionic compounds) to dissolve. H 2 O also dissolves polar covalent substances such as C 2 H 5 OH. H 2 O doesn t dissolve nonpolar covalent substances because there is not enough attraction between the water and the nonpolar molecule. Show the association of the ions with some water molecules when 1 formula unit of KCl dissolves in excess water. K+ Cl 1

2 A solution is a homogeneous mixture. In a solution, a solute dissolves in the solvent. If the solute ionizes in the solution, electricity can be conducted and the solute is said to be an electrolyte. If the solute ionizes 100% or nearly 100%, it is called a strong electrolyte. Lesser ionization occurs with weak electrolytes. Svante Arrhenius determined that the extent to which a solution can conduct an electrical current depends directly on the number of ions present. Solubilityis usually shown as g/given volume solvent or moles/given volume solution Strong electrolytes 1.soluble salts 2.strong acids completely ionize HCl(aq), HNO 3 (aq), H 2 SO 4 (aq) Ex. Show how HCl dissociates when dissolved in water. HCl H + + Cl Acid (Arrhenius) a substance that produces H + ions in water solution 2

3.strong bases completely ionize contain OH bitter taste and slippery feel NaOH, KOH Weak electrolytes only ionize slightly (weak acids and bases) HC 2 H 3 O 2 H + + C 2 H 3 O 2 99% 1% Ammonia (NH 3

3 3.strong bases completely ionize contain OH bitter taste and slippery feel NaOH, KOH Weak electrolytes only ionize slightly (weak acids and bases) HC 2 H 3 O 2 H + + C 2 H 3 O 2 99% 1% Ammonia (NH 3 ) weak base NH 3 + H 2 O NH 4+ + OH Molarity (M) = moles of solute liters of solution Molarity is the most common unit of concentration used in Chemistry. We may also see mm (millimolar) = moles of solute ml of solution Ex. Calculate the molarity of a solution made by dissolving 23.4g of sodium sulfate in enough water to form 125 ml of solution g Na 2 SO 4 1 mol Na 2 SO 4 = mol Na 2 SO g Na 2 SO mol = 1.32 M L 3

Ex. How many grams of Na 2 SO 4 are required to make 350 ml of 0.50 M Na 2 SO 4? 0.350L 0.50 mol Na 2 SO 4 142.06g Na 2 SO 4 = 25g 1 L 1 mol Na 2 SO 4 Ex. What volume of 1.

4 Ex. How many grams of Na 2 SO 4 are required to make 350 ml of 0.50 M Na 2 SO 4? 0.350L 0.50 mol Na 2 SO g Na 2 SO 4 = 25g 1 L 1 mol Na 2 SO 4 Ex. What volume of M KNO 3 must be diluted with water to prepare ml of M KNO 3? Dilution problem (M 1 V 1 = M 2 V 2 ) (1.000M)(V 1 ) = (0.250M)(500.0mL) V 1 = 125 ml Remember, this formula can only be used for dilution! Never use it for a chemical reaction (stoichiometry)! Read procedure for using volumetric flasks and types of pipets. We will be using both in several labs this year. Let s Review Equation Writing from Chemistry I Some reactions fit neatly into a certain category of reaction type, some do not. DECOMPOSITION REACTION Reaction where a compound breaks down into two or more elements or compounds. Heat, electrolysis, or a catalyst is usually necessary. 4

5 A compound may break down to produce two elements. Ex. Molten sodium chloride is electrolyzed. 2NaCl(l) 2Na + Cl 2 A compound may break down to produce an element and a compound. Ex. A solution of hydrogen peroxide is decomposed catalytically. 2H 2 O 2 2H 2 O + O 2 A compound may break down to produce two compounds. Ex. Solid magnesium carbonate is heated. Metallic carbonates break down to yield metallic oxides and carbon dioxide. MgCO 3 MgO + CO 2 Metallic chlorates break down to yield metallic chlorides and oxygen. Metallic sulfites break down to yield metallic oxides and sulfur dioxide. 5

6 Hydrogen peroxide decomposes into water and oxygen. Sulfurous acid decomposes into water and sulfur dioxide. Carbonic acid decomposes into water and carbon dioxide. Hydrated salts decompose into the salt and water. Na 2 CO 3 H 2 O Na 2 CO 3 + H 2 O ADDITION REACTIONS (Also known as Synthesis, Combination, or Composition) A Group IA or IIA metal may combine with a nonmetal to make a salt. Two or more elements or compounds combine to form a single product. 6

7 A piece of lithium metal is dropped into a container of nitrogen gas. 6Li + N 2 2Li 3 N Two nonmetals may combine to form a molecular compound. C + O 2 CO 2 When an element combines with a compound, you can usually sum up all of the elements on the product side. Ex. PCl 3 + Cl 2 PCl 5 This is a trick that works because the common positive oxidation states of P are + 3 and +5. Two compounds combine to form a single product. Ex. Sulfur dioxide gas is passed over solid calcium oxide. SO 2 + CaO CaSO 3 A metallic oxide plus carbon dioxide yields a metallic carbonate. (Carbon keeps the same oxidation state) A metallic oxide plus sulfur dioxide yields a metallic sulfite. (Sulfur keeps the same oxidation state) 7

8 A metallic oxide plus water yields a metallic hydroxide. A nonmetallic oxide plus water yields an acid. Double Replacement (metathesis) Two compounds react to form two new compounds. No changes in oxidation numbers occur. All double replacement reactions must have a "driving force" that removes a pair of ions from solution. Formation of a precipitate: A precipitate is an insoluble substance formed by the reaction of two aqueous substances. Two ions bond together so strongly that water can not pull them apart. You must know your solubility rules to write these net ionic equations! Simple Rules for Solubility 1. Most nitrate (NO 3 ) salts are soluble. 2. Most alkali (group 1A) salts and NH 4+ are soluble. 3. Most Cl, Br, and I salts are soluble (NOT Ag +, Pb 2+, Hg 2 2+ ) 4. Most sulfate salts are soluble (NOT BaSO 4, PbSO 4, HgSO 4, CaSO 4 ) 5. Most OH salts are only slightly soluble (NaOH, KOH are soluble, Ba(OH) 2, Ca(OH) 2 are marginally soluble) 6. Most S 2, CO 3 2, CrO 4 2, PO 4 3 salts are only slightly soluble. SOLUBILITY SONG To the tune of My Favorite Things from The Sound of Music Nitrates and Group One and Ammonium, These are all soluble, a rule of thumb. Then you have chlorides, they re soluble fun, All except Silver, Lead, Mercury I. Then you have sulfates, except for these three: Barium, Calcium and Lead, you see. Worry not only few left to go still. We will do fine on this test. Yes, we will! Then you have the Insolubles Hydroxide, Sulfide and Carbonate and Phosphate, And all of these can be dried! Ex. Solutions of silver nitrate and lithium bromide are mixed. AgNO 3 (aq) + LiBr(aq) AgBr(s) + LiNO 3 (aq) 8

9 Formation of a gas: Gases may form directly in a double replacement reaction or can form from the decomposition of a product such as H 2 CO 3 or H 2 SO 3. H 2 CO 3 H 2 O and CO 2 H 2 SO 3 H 2 O and SO 2 NH 4 OH NH 3 and H 2 O Ex. Excess hydrochloric acid solution is added to a solution of potassium sulfite. 2HCl(aq) + K 2 SO 3 (aq) H 2 SO 3 H 2 O(l) + SO 2 (g) + 2KCl(aq) Remember that sulfurous acid decomposes into water and sulfur dioxide! Ex. A solution of sodium hydroxide is added to a solution of ammonium chloride. Remember that ammonium hydroxide does not exist. NaOH(aq) + NH 4 Cl(aq) NaCl(aq) + NH 4 OH NH 3 (g) + H 2 O(l) Formation of a molecular substance: When a molecular substance such as water or acetic acid is formed, ions are removed from solution and the reaction "works". Ex. Dilute solutions of lithium hydroxide and hydrobromic acid are mixed. LiOH(aq) + HBr(aq) LiBr(aq) +H 2 O(l) Ex. Gaseous hydrofluoric acid reacts with solid silicon dioxide. 4HF(g) + SiO 2 (s) SiF 4 (g) + 2H 2 O(l) This reaction occurs when glass is etched. 9

10 Single Replacement Reaction where one element displaces another in a compound. One element is oxidized and another is reduced. A + BC B + AC Active nonmetals replace less active nonmetals from their compounds in aqueous solution. Each halogen will displace less electronegative (heavier) halogens from their binary salts. Ex. Chlorine gas is bubbled into a solution of potassium iodide. Cl 2 (g) + 2KI(aq) 2KCl(aq) + I 2 (s) Activity Series of Nonmetals Most Active F 2 Cl 2 Br 2 Least Active I 2 Active metals replace less active metals or hydrogen from their compounds in aqueous solution. Use an activity series or a reduction potential table to determine activity. The more easily oxidized metal replaces the less easily oxidized metal. 10

11 Decreasing Activity ACTIVITY SERIES OF METALS Element Lithium Potassium Barium Calcium Sodium Magnesium Aluminum Zinc Iron Cadmium Nickel Tin Lead Hydrogen ( a nonmetal) Copper Mercury Silver Gold Platinum Group I,II,+III Transition Metals Hydrogen *Metals from Li to Na will replace H from water and acids; metals from Mg to Pb will replace H from acids only. Jewelry Metals Ex. Magnesium turnings are added to a solution of iron(iii) chloride. 3Mg(s) + 2FeCl 3 (aq) 2Fe(s)+3MgCl 2 (aq) Ex. Sodium is added to water. 2Na(s) + 2H 2 O(l) 2NaOH(aq) + H 2 (g) COMBUSTION REACTIONS Alkali metal demo Elements or compounds combine with oxygen to produce the oxides of each element. The oxide of H is H 2 O, oxide of S is usually SO 2, oxide of C is CO 2, etc. Hydrocarbons or alcohols combine with oxygen to form carbon dioxide and water. 11

12 Nonmetallic hydrides combine with oxygen to form oxides and water. Nonmetallic sulfides combine with oxygen to form oxides and sulfur dioxide. Ex. Carbon disulfide vapor is burned in excess oxygen. Ex. Ethanol (C 2 H 5 OH) is burned completely in air. CS 2 + 3O 2 CO 2 + 2SO 2 C 2 H 5 OH + 3O 2 2CO 2 + 3H 2 O Diethyl ether (C 2 H 5 OC 2 H 5 ) is burned in air. C 2 H 5 OC 2 H 5 + 6O 2 4CO 2 + 5H 2 O Writing netionic equations 1. molecular equation overall reaction stoichiometry 2. complete ionic equation all strong electrolytes are represented as ions 3. net ionic equation spectator ions are not included 12

13 1. NaCl(aq) + AgNO 3 (aq) NaNO 3 (aq) + AgCl(s) 2. Na + (aq) + Cl (aq) + Ag + (aq) + NO 3 (aq) Na + (aq) + NO 3 (aq) + AgCl(s) 3. Cl (aq) + Ag + (aq) AgCl(s) Selective precipitation process by which ions are caused to ppt one by one in sequence to separate mixtures of ions. Qualitative analysis process of separating and identifying ions Ex. Separate Ag +, Ba 2+, Fe 3+ (use a solubility rule chart) 1. Add Cl to remove Ag + as AgCl. 2. Add SO 4 2 to remove Ba 2+ as BaSO Add OH or S 2 to remove Fe 3+ as Fe(OH) 3 or Fe 2 S 3. Ex. Separate Pb 2+, Ba 2+, Ni Add Cl to remove Pb 2+ as PbCl Add SO 4 2 to remove Ba 2+ as BaSO Add OH or S 2 to remove Ni 2+ as Ni(OH) 2 or NiS. Quantitative analysisdetermines how much of a component is present. Gravimetric analysisquantitative procedure where a ppt containing the substance is formed, filtered, dried & weighed. Ex. The zinc in a g sample of foot powder was precipitated as ZnNH 4 PO 4. Strong heating of the ppt yielded g of Zn 2 P 2 O 7. Calculate the mass percent of zinc in the sample of the foot powder gZn 2 P 2 O 7 1 mol Zn 2 P 2 O 7 2 mol Zn 65.37g = g 1 mol Zn 2 P 2 O 7 1 mol Zn g Zn 100 = 14.62% Zn 1.200g sample 13

14 Ex. A mixture contains only NaCl and Fe(NO 3 ) 3. A 0.456g sample of the mixture is dissolved in water, and an excess of NaOH is added, producing a precipitate of Fe(OH) 3. The ppt is filtered, dried, & weighed. Its mass is 0.128g. Calculate: a. the mass of the iron b. the mass of Fe(NO 3 ) 3 c. the mass percent of Fe(NO 3 ) 3 in the sample AcidBase Reactions (type of double replacement) 0.128g Fe(OH) 3 1 mol Fe(OH) 3 1 mol Fe 55.85g Fe= g Fe 106.9g Fe(OH) 3 1 mol Fe(OH) 3 1 mol Fe g Fe 1 mol Fe 1 mol Fe(NO 3 ) g Fe(NO 3 ) 3 = 0.290g Fe(NO 3 ) g Fe 1 mol Fe 1 mol Fe(NO 3 ) g 100 = 63.6% Fe(NO 3 ) g BrønstedLowry acidbase definitions: acid proton donor base proton acceptor When a strong acid reacts with a strong base the net ionic rxn is: H + (aq) + OH (aq) H 2 O(l) When a strong acid reacts with a weak base or a weak acid reacts with a strong base, the reaction is complete (the weak substance ionizes completely.) HC 2 H 3 O 2 (aq) + OH (aq) H 2 O(l) + C 2 H 3 O 2 (aq) neutralization reaction acidbase rxn When just enough base is added to react exactly with the acid in a solution, the acid is said to be neutralized. (The ph is not necessarily 7) Volumetric Analysis titration process in which a solution of known concentration (standard solution) is added to analyze another solution (analyte). 14

15 titrant solution of known concentration (usually in buret) Titrations are most often used for acids and bases, but can be used for other types of reactions, also. equivalence point or stoichiometric pointpoint where just enough titrant has been added to react with the substance being analyzed Indicator chemical which changes color at or near the equivalence point End point point at which the indicator changes color Ex ml of M HClO 4 solution is required to neutralize 25.0 ml of an NaOH solution of unknown molarity. What is the concentration of the NaOH solution? HClO 4 + NaOH H 2 O + NaClO L HClO mol HClO 4 1 mol NaOH = 1 L HClO 4 1 mol HClO mol NaOH mol NaOH = M NaOH 0.025L OxidationReduction Reactions Electronegativity attraction for shared electrons Redox Rxns reactions in which one or more electrons are transferred. most electronegative elements F>O>N=Cl Phone Call These are most likely to have negative oxidation numbers. 15

16 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 Review oxidation state rules on page N 2 O PBr 3 HPO P 4 O 6 NH 2 Noninteger states are rare, but possible. 8/3 2 Fe 3 O 4 O = 4(2) = 8 Fe = 8/3 = 2 2/3 or Fe 2+, Fe 3+, Fe 3+ Oxidation loss of electrons increase in oxidation number Reduction gain of electrons decrease in oxidation number OIL RIG Oxidation Is Loss (of e ), Reduction Is Gain (of e ) LEO the lion goes GER Lose Electrons = Oxidation, Gain Electrons = Reduction 16

17 Oxidizing agent electron acceptor substance that is reduced Reducing agent electron donor substance that is oxidized The terms oxidizing agent and reducing agent are not tested on the AP test KI + F 2 2KF + I 2 oxidized I reduced F OA F 2 RA KI PbO 2 2PbO + O 2 oxidized: O reduced: Pb OA: PbO 2 RA: PbO 2 Balancing redox reactions by the halfreaction method 1.Write skeleton halfreactions. 2.Balance all elements other than O and H. 3.Balance O by adding H 2 O. 4.Balance H by adding H +. 5.Balance charge by adding e to the more positive side. 6.Make the # of e lost = # of e gained by multiplying each halfrxn by a factor. 7.Add halfreactions together. 8.Cancel out anything that is the same on both sides. 9.If the reaction occurs in basic solution, add an equal number of hydroxide ions to both sides to cancel out the hydrogen ions. Make water on the side with the hydrogen ions. Cancel water if necessary. 10.Check to see that charge and mass are both balanced. 17

18 Ex. Cr 2 O CH 3 OH HCO 2 H + Cr Write half reactions. Cr 2 O 7 2 Cr 3+ CH 3 OH HCO 2 H 2. Balance all elements other than oxygen and hydrogen. Cr 2 O Cr 3+ CH 3 OH HCO 2 H (acidic) You don t need to recopy each step! 3. Balance oxygen by adding water. Cr 2 O Cr H 2 O CH 3 OH + H 2 O HCO 2 H 4. Balance hydrogen by adding hydrogen ions. Cr 2 O H + 2 Cr H 2 O CH 3 OH + H 2 O HCO 2 H + 4 H + 5. Balance charge by adding electrons. Cr 2 O H e 2 Cr H 2 O CH 3 OH + H 2 O HCO 2 H + 4 H e 6. Multiply halfreactions by factors to make the number of electrons lost equal to the number of electrons gained. 2(Cr 2 O H e 2 Cr H 2 O) 3(CH 3 OH + H 2 O HCO 2 H + 4 H e ) You will start 7. Add halfreactions together. fresh here. 2(Cr 2 O H e 2 Cr H 2 O) 3(CH 3 OH + H 2 O HCO 2 H + 4 H e ) 2Cr 2 O H e + 3CH 3 OH + 3H 2 O 4Cr H 2 O + 3HCO 2 H + 12H e 8. Cancel out anything that is the same on both sides. 2Cr 2 O H e + 3CH 3 OH + 3H 2 O 4Cr H 2 O + 3HCO 2 H + 12H e 2Cr 2 O H + + 3CH 3 OH 4Cr H 2 O + 3HCO 2 H Copy it over so it is legible! 9. If the reaction occurs in basic solution, add hydroxide ions to both sides to cancel out the hydrogen ions. You will make water on the side with the hydrogen waters. Cancel water if necessary. NA 10. Check to see that both charge and mass (atoms) are balanced. Don t skip this 2Cr 2 O H + + 3CH 3 OH step!!!!! 4Cr H 2 O + 3HCO 2 H 4Cr, 17O, 28H, 3C, both sides +12 It is balanced. NO 3 + I 2 IO 3 + NO 2 1. NO 3 NO 2 I 2 IO 3 2. NO 3 NO 2 I 2 2IO 3 3. NO 3 NO 2 + H 2 O I 2 + 6H 2 O 2IO 3 4. NO 3 + 2H + NO 2 + H 2 O I 2 + 6H 2 O 2IO H + (acidic) 18

5. NO 3 + 2H + + e NO 2 + H 2 O I 2 + 6H 2 O 2IO 3 + 12H + +10e 6. 10(NO 3 + 2H + + e NO 2 + H 2 O) I 2 + 6H 2 O 2IO 3 + 12H + +10e 7.

19 5. NO 3 + 2H + + e NO 2 + H 2 O I 2 + 6H 2 O 2IO H + +10e 6. 10(NO 3 + 2H + + e NO 2 + H 2 O) I 2 + 6H 2 O 2IO H + +10e 7. 10NO H e + I 2 + 6H 2 O 10NO H 2 O + 2IO H + +10e 8. 10NO H e + I 2 + 6H 2 O 10NO H 2 O + 2IO H + +10e 10NO 3 + 8H + + I 2 10NO 2 +4H 2 O + 2IO 3 MnO 4 + I MnO 2 + I 2 1. MnO 4 MnO 2 I I 2 2. MnO 4 MnO 2 2 I I 2 3. MnO 4 MnO 2 + 2H 2 O 2 I I 2 4. MnO 4 + 4H + MnO 2 + 2H 2 O 2 I I 2 (basic) 5. MnO 4 + 4H e MnO 2 + 2H 2 O 2 I I 2 + 2e 6. 2(MnO 4 + 4H e MnO 2 + 2H 2 O) 3(2 I I 2 + 2e ) 7. 2MnO 4 + 8H e + 6 I 2 MnO 2 + 4H 2 O +3 I 2 + 6e 8. 2MnO 4 + 8H e + 6 I 2 MnO 2 + 4H 2 O +3 I 2 + 6e 9. 2MnO 4 + 8H I 2 MnO 2 + 4H 2 O +3 I MnO 4 + 8H I + 8OH 2 MnO 2 + 4H 2 O +3 I 2 + 8OH 2MnO 4 + 8H I + 8OH + 8H 2 O 2 MnO 2 + 4H 2 O +3 I 2 + 8OH 2MnO I H 2 O 2 MnO 2 + 4H 2 O +3 I 2 + 8OH 2MnO I + 4 H 2 O 2 MnO I 2 + 8OH OXIDATIONREDUCTION TITRATIONS Most common oxidizing agents: KMnO 4 & K 2 Cr 2 O 7 Potassium permanganate used to disinfect ponds and fish in Egypt. 19

MnO 4 in acidic solution: MnO 4 + 8H + + 5e Mn 2+ + 4H 2 O Purple colorless When you titrate with MnO 4, the solution is colorless until you use up all of the

20 MnO 4 in acidic solution: MnO 4 + 8H + + 5e Mn H 2 O Purple colorless When you titrate with MnO 4, the solution is colorless until you use up all of the reducing agent (substance being oxidized). In calculations, work redox titrations like acid base titrations. You must have a balanced reaction to know the mole ratio. 20

Chapter 4 Notes Types of Chemical Reactions and Solutions Stoichiometry A Summary

Chapter 4 Notes Types of Chemical Reactions and Solutions Stoichiometry A Summary 4.1 Water, the Common Solvent A. Structure of water 1. Oxygen s electronegativity is high (3.5) and hydrogen s is low (2.1)