Chemical Stoichiometry Aqueous Solutions and Solution Stoichiometry Chapters

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Chemical Stoichiometry, Aqueous Solutions and Solution Stoichiometry Chapters 3, 4 1

Chemical Stoichiometry, Aqueous Solutions and Solution Stoichiometry Chapters 3, 4 1

Previous knowledge you must know!!! 1. Mole concept. 2. Molar mass calculations. 3. Conversions

Previous knowledge you must know!!! 1. Mole concept. 2. Molar mass calculations. 3. Conversions between mass-moleparticles. 4. Limiting Reactant 5. Percent yield 6. Percent composition. 7. Empirical formula calculations. 8. Molecular formula calculations. 2

The Mole Avogadro’s number: Provides a connection between the number of moles in a

The Mole Avogadro’s number: Provides a connection between the number of moles in a pure sample and the number of particles or units in the sample. 1 mol = 6. 022 x 1023 particles atoms molecules formula units ions 3

The Molar mass: Mass in grams of 1 mol of a pure substance. n

The Molar mass: Mass in grams of 1 mol of a pure substance. n n 1 mol Ne atoms = 20. 18 g = 6. 022 x 1023 atoms 1 mol Cl- ions = 35. 45 g = 6. 022 x 1023 ions Difference between amu and mass n n n 1 amu = 1. 66 x 10 -24 g 1 carbon atom = 12. 01 amu 1 mol carbon atoms = 12. 01 g 4

Molar mass of a compound Problem 1: Calculate the molar mass for the following

Molar mass of a compound Problem 1: Calculate the molar mass for the following compounds: a) Fe(NO 3)2· 6 H 2 O b) (NH 4)3 PO 4 5

Converting moles to grams Problem 2: How many moles of iron (II) nitrate hexahydrate

Converting moles to grams Problem 2: How many moles of iron (II) nitrate hexahydrate are there in 48. 5 g of the salt? Problem 3: How many grams of ammonium phosphate are there in 1. 87 moles of the salt? 6

Converting grams to particles Problem 4: How many water molecules are contained within 56.

Converting grams to particles Problem 4: How many water molecules are contained within 56. 0 g of a pure sample of water? Problem 5: How many hydrogen atoms are contained within 23. 6 g of pure sucrose, C 12 H 22 O 11? 7

Conservation of Mass and Atoms in Chemical Reactions reactants yields products 1 formula unit

Conservation of Mass and Atoms in Chemical Reactions reactants yields products 1 formula unit + 3 molecules ≠ 1 mole + 3 moles ≠ 159. 7 g + = 5 atoms 84. 0 g 243. 7 + 6 atoms 11 atoms = 2 atoms 2 moles 111. 7 g + 3 molecules + 3 moles + 132 g 243. 7 2 atoms + 9 atoms 11 atoms 8

Predicting mass of products Problem 6: What mass of carbon dioxide can be produced

Predicting mass of products Problem 6: What mass of carbon dioxide can be produced by the reaction of 32. 0 g of iron (III) oxide with excess carbon monoxide? 9

Predicting mass of reactants Problem 7: What mass of sodium bicarbonate is needed to

Predicting mass of reactants Problem 7: What mass of sodium bicarbonate is needed to produce 10. 6 g of sodium carbonate? Na. HCO 3(aq) Na 2 CO 3(aq) + H 2 O(l) + CO 2(g) 10

Limiting Reactant Concept If you add more of one reactant you cannot make more

Limiting Reactant Concept If you add more of one reactant you cannot make more product. You must increase all reactants stoichiometrically. Look at a chemical limiting reactant situation. Zn + 2 HCl Zn. Cl 2 + H 2 11

Limiting Reactant Concept If quantities of two or more reactants is known, then one

Limiting Reactant Concept If quantities of two or more reactants is known, then one reactant is used up and some of the other(s) reactant(s) will be left over. Limiting Reactant: The reactant that is used up limits how far the reaction will proceed. Excess reactant: The reactant that is left over when the reaction is complete. 12

Limiting Reactant Concept Problem 8: Find the limiting reactant and the maximum mass of

Limiting Reactant Concept Problem 8: Find the limiting reactant and the maximum mass of sulfur dioxide that can be produced by the reaction of 38. 1 g of carbon disulfide with 52. 0 g of oxygen? CS 2 + O 2 CO 2 + SO 2 13

Limiting Reactant Concept Problem 9: Find mass of excess reactant that remains after the

Limiting Reactant Concept Problem 9: Find mass of excess reactant that remains after the reaction reached completion when 38. 1 g of carbon disulfide reacted with 52. 0 g of oxygen? CS 2 + O 2 CO 2 + SO 2 14

Percent Yields from Reactions Theoretical yield is calculated by assuming that the reaction goes

Percent Yields from Reactions Theoretical yield is calculated by assuming that the reaction goes to completion. n Determined from the limiting reactant calculation. Actual yield is the amount of a specified pure product made in a given reaction. n In the laboratory, this is the amount of product that is formed in your beaker, after it is purified and dried. Percent yield indicates how much of the product is obtained from a reaction. 15

Percent Yields from Reactions Problem 10: A 4. 60 g sample of ethanol, C

Percent Yields from Reactions Problem 10: A 4. 60 g sample of ethanol, C 2 H 5 OH, was boiled with excess acetic acid, CH 3 COOH, to produce 6. 40 g of ethyl acetate, CH 3 COOC 2 H 5. What is the percent yield? CH 3 COOH + C 2 H 5 OH CH 3 COOC 2 H 5 + H 2 O 16

Solving Stoichiometry problems without a calculator!!! a) Calculate the moles of precipitate produced from

Solving Stoichiometry problems without a calculator!!! a) Calculate the moles of precipitate produced from the reaction of 0. 0500 moles of Ba. Cl 2 with 0. 0300 moles of Na 2 CO 3. b) Calculate the moles of the excess reactant. Initial Ba. Cl 2(aq) + Na 2 CO 3(aq) Ba. CO 3(s) + 2 Na. Cl(aq) 0. 0500 0. 0300 0 0 Change -0. 0300 End 0. 0200 -0. 0300 0 +0. 0300 +0. 0600 0. 0300 0. 0600 c) Sketch the picture of how the reactant vessel will look like in terms of contents after the completion of the reaction. 17

Cl. Na+ Cl- Ba 2+ Na+ Cl- Cl. Cl- Na+ Ba. CO 3 18

Cl. Na+ Cl- Ba 2+ Na+ Cl- Cl. Cl- Na+ Ba. CO 3 18

Problem 11: Calculate the mass of carbon dioxide produced by the reaction of 2.

Problem 11: Calculate the mass of carbon dioxide produced by the reaction of 2. 5 moles of sodium carbonate with 5. 3 moles of hydrochloric acid. No calculator allowed 19

Derivation of Formulas from Elemental Composition Empirical Formula - smallest whole-number ratio of atoms

Derivation of Formulas from Elemental Composition Empirical Formula - smallest whole-number ratio of atoms present in a compound n CH 2 is the empirical formula for alkenes Molecular Formula - actual numbers of atoms of each element present in a molecule of the compound n n Ethene – C 2 H 4 Pentene – C 5 H 10 We determine the empirical and molecular formulas of a compound from the percent composition of the compound given by the mass spec. n percent composition is determined experimentally 20

Molecular and Empirical Formulas Common Name Molecular Formula Empirical Formula hydrogen peroxide H 2

Molecular and Empirical Formulas Common Name Molecular Formula Empirical Formula hydrogen peroxide H 2 O 2 HO ethylene C 2 H 4 CH 2 dextrose C 6 H 12 O 6 CH 2 O 21

Problem 12: A compound contains 24. 74% K, 34. 76% Mn, and 40. 50%

Problem 12: A compound contains 24. 74% K, 34. 76% Mn, and 40. 50% O by mass. What is its empirical formula? 22

Problem 13: A sample of a compound contains 6. 541 g of Co and

Problem 13: A sample of a compound contains 6. 541 g of Co and 2. 368 g of O. What is empirical formula for this compound? 23

Problem 14: A sample of caffeine was found to contain 49. 5% carbon, 28.

Problem 14: A sample of caffeine was found to contain 49. 5% carbon, 28. 9% nitrogen, 16. 5% oxygen, and 5. 1% hydrogen by mass. Find the empirical formula for caffeine. Problem 15. The molar mass for caffeine is 194. 2 g/mol. Find the molecular formula for caffeine. 24

Combustion Analysis § An organic compound is burnt in excess of oxygen. H 2

Combustion Analysis § An organic compound is burnt in excess of oxygen. H 2 O absorber O 2 Organic sample CO 2 absorbed § All of the hydrogen from the organic sample combines to form water and absorbed in the water absorber. § All of the carbon from the organic sample combines to form CO 2 and absorbed in the CO 2 absorber. § Oxygen is the only element that is added to the 25 sample and it must be calculated last.

Problem 16: A 2. 04 g sample containing C, H, and O underwent combustion

Problem 16: A 2. 04 g sample containing C, H, and O underwent combustion analysis. Find the empirical formula of the compound if 4. 49 g of CO 2 and 2. 45 g of H 2 O were produced. 26

Some Other Interpretations of Chemical Formulas What mass of ammonium phosphate, (NH 4)3 PO

Some Other Interpretations of Chemical Formulas What mass of ammonium phosphate, (NH 4)3 PO 4, would contain 15. 0 g of N? 27

Problem 17: How many grams of manganese are there in 57. 9 g of

Problem 17: How many grams of manganese are there in 57. 9 g of a pure sample of potassium permanganate? 28

Purity of Samples The percent purity of a sample of a substance is always

Purity of Samples The percent purity of a sample of a substance is always represented as 29

Purity of Samples A bottle of sodium phosphate, Na 3 PO 4, is 98.

Purity of Samples A bottle of sodium phosphate, Na 3 PO 4, is 98. 3% pure Na 3 PO 4. What are the masses of Na 3 PO 4 and impurities in 250. 0 g of this sample of Na 3 PO 4? 30

When gases involved…. PV = n. RT Problem 18: The number of moles and

When gases involved…. PV = n. RT Problem 18: The number of moles and mass of the gases can be calculated if the volume, temperature and pressure is given. Calcium carbonate decomposes to produce calcium oxide and carbon dioxide. The decomposition of Ca. CO 3 was accomplished by heating the sample at 300°C. 130. 1 m. L of the gas produced was collected at 25. 0°C and 751 mm Hg. Calculate the mass of calcium carbonate decomposed. 31

Chemical Equations Law of Conservation of Matter n n n There is no detectable

Chemical Equations Law of Conservation of Matter n n n There is no detectable change in quantity of matter in an ordinary chemical reaction. Balanced chemical equations must always include the same number of each kind of atom on both sides of the equation. This law was determined by Antoine Lavoisier. 32

Law of Conservation of Matter NH 3 burns in oxygen to form NO &

Law of Conservation of Matter NH 3 burns in oxygen to form NO & water 33

Reaction Types 34

Reaction Types 34

Combination Reaction between metal and a non-metal produces the binary ionic compound of most

Combination Reaction between metal and a non-metal produces the binary ionic compound of most stable oxidation state. Example: 2 Mg + O 2 2 Mg. O (Mg is 2+ and O is 2 -) Reaction between two non-metals produces a binary covalent compound. The compound depends on conditions. Example: 2 S + 3 O 2 (excess) 2 SO 3 S + O 2 (limited) SO 2 Reaction of metal oxides with water produces metal hydroxide. Example: Ca. O + H 2 O Ca(OH)2 35

Combination Reaction of non-metal oxides with water produces the oxyacid of the same oxidation

Combination Reaction of non-metal oxides with water produces the oxyacid of the same oxidation number. Example: SO 3 + H 2 O H 2 SO 4 (oxidation number of S in SO 3 and H 2 SO 4 is 6) SO 2 + H 2 O H 2 SO 3 (oxidation number of S in SO 2 and H 2 SO 3 is 4) Reaction of metal oxides and non-metal oxides produces the ternary salt. Example: Mg. O + CO 2 Mg. CO 3 Na 2 O + SO 2 Na 2 SO 3 36

Decomposition § Binary compounds generally decompose into the elements Example: n 2 KCl. O

Decomposition § Binary compounds generally decompose into the elements Example: n 2 KCl. O 3 2 KCl + 3 O 2 Metal hydroxides decompose into the metal oxide and water Example: n Na 2 CO 3 Na 2 O + CO 2 Metal chlorates decompose into metal chlorides and oxygen Example: n 2 Hg. O 2 Hg + O 2 SO 2 S + O 2 Metal carbonates or hydrogen carbonate decompose into metal oxide and CO 2 or metal oxide, CO 2 and H 2 O Example: n see decomposition of metal bicarbonate Ca(OH)2 Ca. O + H 2 O Oxyacids decomposed into non-metal oxides and water Example: H 2 CO 3 CO 2 + H 2 O H 2 SO 4 SO 3 + H 2 O 37

Aqueous solutions and chemical reactions

Aqueous solutions and chemical reactions

Electrolytes Substances that dissociate into ions when dissolved in water. A nonelectrolyte may dissolve

Electrolytes Substances that dissociate into ions when dissolved in water. A nonelectrolyte may dissolve in water, but it does not dissociate into ions when it does so.

Electrolytes Strong electrolytes. Weak electrolytes. Substances that • Substances that dissolve in water and

Electrolytes Strong electrolytes. Weak electrolytes. Substances that • Substances that dissolve in water and separate into moderately and ions. separate into ions Strong conductor of partially. electricity • Poor conductor of electricity. Soluble ionic compounds. • Ionic compounds with low solubility. n Solubility rules – Solubility rules Strong acids. • Weak acids. Strong bases. • Weak bases. Non electrolytes. • Substances that may dissolve in water but cannot separate into ions. • Non conductors of electricity. • Covalent compounds.

Strong Acids

Strong Acids

Strong Bases

Strong Bases

Strong Electrolytes Are… Strong acids Strong bases Soluble ionic salts. Memorize the simple solubility

Strong Electrolytes Are… Strong acids Strong bases Soluble ionic salts. Memorize the simple solubility rules: n All NH 4+ salts are soluble. n All NO 3 - , C 2 H 3 O 2 -, Cl. O 4 - salts are soluble. n All group 1 salts are soluble.

Dissociation When an ionic substance dissolves in water, the solvent pulls the individual ions

Dissociation When an ionic substance dissolves in water, the solvent pulls the individual ions from the crystal and solvates them. This process is called dissociation. Water is a polar molecule. Its positive ends (hydrogen sites) are attracted by the negative ions in the ionic compound, while the negative ends (oxygen sites) are attracted to the positive ions. Ionic compounds dissolve in water because the attraction of the ions to water is stronger than the attraction between the ions themselves.

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Dissociation Na. Cl(s) Na+(aq) + Cl-(aq) Na 2 CO 3(s) 2 Na+(aq) + CO

Dissociation Na. Cl(s) Na+(aq) + Cl-(aq) Na 2 CO 3(s) 2 Na+(aq) + CO 32 -(aq) Ba(NO 3)2 (s) Ba 2+(aq) + 2 NO 3 -(aq) (NH 4)3 PO 4(s) 3 NH 4+(aq) + PO 43 -(aq)

Problem 19: Complete the following dissociation equations a) Ca(CH 3 COO)2 b) (NH 4)2

Problem 19: Complete the following dissociation equations a) Ca(CH 3 COO)2 b) (NH 4)2 SO 4 c) Fe(NO 3)3 d) K 3 PO 3 47

Molecular compounds in water With the exception of strong acids, molecular compounds are non

Molecular compounds in water With the exception of strong acids, molecular compounds are non electrolytes or weak electrolytes. Non-electrolytes: C 12 H 22 O 11 (s) C 12 H 22 O 11 (aq) CH 3 OH (l) CH 3 OH (aq) Weak electrolytes are weak acids: CH 3 COOH(aq) HNO 2(aq) H+(aq) + CH 3 COO-(aq) H+(aq) + NO 2 -(aq) 48

Like dissolves like Polar molecules dissolve in polar solvents δ+ δ- δ+ δ+ δ+

Like dissolves like Polar molecules dissolve in polar solvents δ+ δ- δ+ δ+ δ+ δ- δ- δ+ δ+ 49

Double Replacement reactions Precipitation reactions (must know solubility rules). Neutralization reactions or acid-base reactions.

Double Replacement reactions Precipitation reactions (must know solubility rules). Neutralization reactions or acid-base reactions. Gas forming reactions. 50

Precipitation Reactions When one mixes ions that form compounds that are insoluble (as could

Precipitation Reactions When one mixes ions that form compounds that are insoluble (as could be predicted by the solubility guidelines), a precipitate is formed.

Molecular Equation Ag. NO 3 (aq) + KCl (aq) Ag. Cl (s) + KNO

Molecular Equation Ag. NO 3 (aq) + KCl (aq) Ag. Cl (s) + KNO 3 (aq)

Ionic Equation In the ionic equation all strong electrolytes (strong acids, strong bases, and

Ionic Equation In the ionic equation all strong electrolytes (strong acids, strong bases, and soluble ionic salts) are dissociated into their ions. This more accurately reflects the species that are found in the reaction mixture. Ag+ (aq) + NO 3 - (aq) + K+ (aq) + Cl- (aq) Ag. Cl (s) + K+ (aq) + NO 3 - (aq)

Net Ionic Equation To form the net ionic equation, cross out anything that does

Net Ionic Equation To form the net ionic equation, cross out anything that does not change from the left side of the equation to the right. Ag+(aq) + NO 3 -(aq) + K+(aq) + Cl-(aq) Ag. Cl (s) + K+(aq) + NO 3 -(aq)

Net Ionic Equation To form the net ionic equation, cross out anything that does

Net Ionic Equation To form the net ionic equation, cross out anything that does not change from the left side of the equation to the right. The only things left in the equation are those things that change (i. e. , react) during the course of the reaction. Ag+(aq) + Cl-(aq) Ag. Cl (s)

Net Ionic Equation Those things that didn’t change (and were deleted from the net

Net Ionic Equation Those things that didn’t change (and were deleted from the net ionic equation) are called spectator ions. K+ and NO 3 -

Problem 20: Write the balanced complete and net ionic equation for the reaction that

Problem 20: Write the balanced complete and net ionic equation for the reaction that takes place when an aqueous solution of barium nitrate is added to an aqueous solution of sodium sulfate. 57

Problem 21: Write the balanced complete and net ionic equation for the reaction that

Problem 21: Write the balanced complete and net ionic equation for the reaction that takes place when an aqueous solution of lead (II) nitrate and sodium iodide are mixed in a beaker. 58

Acids: Substances that increase the concentration of H+ when dissolved in water (Arrhenius). Proton

Acids: Substances that increase the concentration of H+ when dissolved in water (Arrhenius). Proton donors (Brønsted–Lowry).

Acids There are only seven strong acids: • • Hydrochloric (HCl) Hydrobromic (HBr) Hydroiodic

Acids There are only seven strong acids: • • Hydrochloric (HCl) Hydrobromic (HBr) Hydroiodic (HI) Nitric (HNO 3) Sulfuric (H 2 SO 4) Chloric (HCl. O 3) Perchloric (HCl. O 4)

Bases: Substances that increase the concentration of OH− when dissolved in water (Arrhenius). Proton

Bases: Substances that increase the concentration of OH− when dissolved in water (Arrhenius). Proton acceptors (Brønsted–Lowry).

Bases The strong bases are the soluble salts of hydroxide ion: • • Alkali

Bases The strong bases are the soluble salts of hydroxide ion: • • Alkali metals Calcium Strontium Barium

Neutralization Reactions Generally, when solutions of an acid and a base are combined, the

Neutralization Reactions Generally, when solutions of an acid and a base are combined, the products are a salt and water. HCl (aq) + Na. OH (aq) Na. Cl (aq) + H 2 O (l)

Neutralization Reactions When a strong acid reacts with a strong base, the net ionic

Neutralization Reactions When a strong acid reacts with a strong base, the net ionic equation is… HCl (aq) + Na. OH (aq) Na. Cl (aq) + H 2 O (l) H+ (aq) + Cl- (aq) + Na+ (aq) + OH- (aq) Na+ (aq) + Cl- (aq) + H 2 O (l)

Gas-Forming Reactions These metathesis reactions do not give the product expected. The expected product

Gas-Forming Reactions These metathesis reactions do not give the product expected. The expected product decomposes to give a gaseous product (CO 2 or SO 2). Ca. CO 3 (s) + HCl (aq) Ca. Cl 2 (aq) + CO 2 (g) + H 2 O (l) Na. HCO 3 (aq) + HBr (aq) Na. Br (aq) + CO 2 (g) + H 2 O (l) Sr. SO 3 (s) + 2 HI (aq) Sr. I 2 (aq) + SO 2 (g) + H 2 O (l)

Gas-Forming Reactions This reaction gives the predicted product, but you had better carry it

Gas-Forming Reactions This reaction gives the predicted product, but you had better carry it out in the hood, or you will be very unpopular! Just as in the previous examples, a gas is formed as a product of this reaction: Na 2 S (aq) + H 2 SO 4 (aq) Na 2 SO 4 (aq) + H 2 S (g)

Complexation Reactions n Transition metal ions (such as Fe, Co, Ni, Cu, Zn, Ag)

Complexation Reactions n Transition metal ions (such as Fe, Co, Ni, Cu, Zn, Ag) and Al will react with excess of molecules or ions with pairs of electrons in their Lewis structure (such as NH 3, H 2 O, OH-, CN-, Cl-, Br-, SCN-) to form coordination compounds. The amount of ligands is usually twice the charge of the metal ion. Example: Cu 2+ + 4 NH 3 [Cu(NH 3)4]2+ (the amount of NH 3 is twice the charge of Cu 2+) Fe 3+ + 6 CN- [Fe(CN)6]3 - (the amount of CN- is twice the charge of Fe 3+) 67

Oxidation-Reduction Reactions (REDOX) An oxidation occurs when an atom or ion loses electrons. LEO

Oxidation-Reduction Reactions (REDOX) An oxidation occurs when an atom or ion loses electrons. LEO A reduction occurs when an atom or ion gains electrons. GER Electrons are transferred. Single replacement and combustion reactions. See video

A piece or iron is immersed in a solution of copper(II) sulfate. Reduction Fe(s)

A piece or iron is immersed in a solution of copper(II) sulfate. Reduction Fe(s) + Cu 2+(aq) Fe 2+(aq) + Cu(s) Oxidation Electrons are transferred from the iron to the copper(II) ion. Fe gets oxidized. Cu 2+ gets reduced. 69

Oxidation Numbers To determine if an oxidation-reduction reaction has occurred, we assign an oxidation

Oxidation Numbers To determine if an oxidation-reduction reaction has occurred, we assign an oxidation number to each element in a neutral compound or charged entity.

Oxidation Numbers 1. Elements in their elemental form have an oxidation number of 0.

Oxidation Numbers 1. Elements in their elemental form have an oxidation number of 0. 2. In a binary compound, the oxidation number of the second element is the same as its ionic charge. 3. Oxygen has an oxidation number of − 2, except in the peroxide ion in which it has an oxidation number of − 1. 4. Hydrogen is − 1 when bonded to a metal, +1 when bonded to a nonmetal.

Oxidation Numbers 5. Fluorine always has an oxidation number of − 1. 6. The

Oxidation Numbers 5. Fluorine always has an oxidation number of − 1. 6. The other halogens have an oxidation number of − 1 when they are negative; they can have positive oxidation numbers, however, most notably in oxyanions. 7. The sum of the oxidation numbers in a neutral compound is 0. 8. The sum of the oxidation numbers in a polyatomic ion is the charge on the ion.

Problem 22: Identify the substance that gets oxidized and the one that gets reduced

Problem 22: Identify the substance that gets oxidized and the one that gets reduced in the following chemical reactions. a. Mn. O 4 -(aq) + H 2 O 2(aq) + H+ Mn 2+(aq) + O 2(g) + H 2 O(l) b. Ca(s) + Ni 2+(aq) Ca 2+(aq) + Ni(s) c. Cu(s) + HNO 3(aq) Cu 2+(aq) + NO 2(g) + H 2 O(l) d. KCl. O 3(l) + HNO 2(aq) KCl(aq) + HNO 3(aq)

Single Displacement Reactions In displacement reactions, ions oxidize an element. The ions, then, are

Single Displacement Reactions In displacement reactions, ions oxidize an element. The ions, then, are reduced.

Single Displacement n Reactive metals displace another metal ion from their compounds. Example: n

Single Displacement n Reactive metals displace another metal ion from their compounds. Example: n Reactive metals displace hydrogen from water forming the hydroxides. Example: n 2 Na + 2 H 2 O 2 Na. OH + H 2 Reactive metals react with acids to produce the salt of the metal and hydrogen. Example: n 2 Al + 3 Pb(NO 3)2 2 Al(NO 3)3 + 3 Pb (aluminum forms ions with a charge of 3+) Mg + 2 HCl Mg. Cl 2 + H 2 Halogens (Group 17) can displace another halogen from the compound only if it is higher in the periodic table. Example: Cl 2 + 2 KBr 2 KCl + Br 2 + 2 KCl No Reaction. Br 2 is not more reactive than Cl 2. 75

Single Displacement Reactions In this reaction, silver ions oxidize copper metal. Cu (s) +

Single Displacement Reactions In this reaction, silver ions oxidize copper metal. Cu (s) + 2 Ag+ (aq) Cu 2+ (aq) + 2 Ag (s)

Single Displacement Reactions The reverse reaction, however, does not occur. x Cu (s) +

Single Displacement Reactions The reverse reaction, however, does not occur. x Cu (s) + 2 Ag+ (aq) Cu 2+ (aq) + 2 Ag (s)

Activity Series Increasing ease of oxidation Au Pt Ag Cu H Fe Zn Mg

Activity Series Increasing ease of oxidation Au Pt Ag Cu H Fe Zn Mg Ca Ba K Li An element will oxidize anything that is below it in this series.

Problem 23: Write net ionic equations and identify the substance that gets oxidized and

Problem 23: Write net ionic equations and identify the substance that gets oxidized and the one reduced. a. A copper(II) nitrate solution is poured over solid zinc. b. Chlorine gas is bubbled through an aqueous solution containing potassium bromide. c. Sodium metal is placed in water. d. A solution of copper(I) chloride sits in a beaker for an extended period of time. e. Solid iron filings are added to a 0. 5 M solution of iron(III) chloride. 79

Combustion n Organic compounds (carbon containing compounds) react with oxygen to form carbon dioxide

Combustion n Organic compounds (carbon containing compounds) react with oxygen to form carbon dioxide and water Example: 2 C 3 H 6 O 2 + 7 O 2 6 CO 2 + 6 H 2 O 2 C 4 H 8 O + 11 O 2 8 CO 2 + 8 H 2 O 80

Balancing redox reactions Acidic solutions 1. 2. 3. 4. 5. 6. 7. Divide the

Balancing redox reactions Acidic solutions 1. 2. 3. 4. 5. 6. 7. Divide the equation into 2 halves: the oxidation half and the reduction half. For each half balance the atoms other than H and O. Add H+ whenever H is needed; add H 2 O whenever O is needed. Balance the charges. Add electrons if needed. Oxidation is the loss of electrons so e- should be in the products, while reduction id the gain of e-. Electrons must cancel. Add both halves to cancel the electrons. It may be necessary to multiply the halves to make the electrons the same. Cancel water molecules and H+ is they appear on both sides of the equation. Check final equation for balancing materially and electrically. 81

Common Reductions in Acidic solutions nitrate NO 3 - NO permanganate Mn. O 4

Common Reductions in Acidic solutions nitrate NO 3 - NO permanganate Mn. O 4 - Mn 2+ dichromate Cr 2 O 72 - Cr 3+ 82

Problem 24: Balance the following equation in acidic solution Mn. O 4 -(aq) +

Problem 24: Balance the following equation in acidic solution Mn. O 4 -(aq) + Sn 2+ (aq) Mn 2+ (aq) + Sn 4+ (aq) ½ oxidation: ½ reduction 83

Balancing redox reactions Basic solutions 1. 2. 3. 4. 5. 6. 7. 8. Divide

Balancing redox reactions Basic solutions 1. 2. 3. 4. 5. 6. 7. 8. Divide the equation into 2 halves: the oxidation half and the reduction half. For each half balance the atoms other than H and O. Add H+ whenever H is needed; add H 2 O whenever O is needed. Balance the charges. Add electrons if needed. Oxidation is the loss of electrons so e- should be in the products, while reduction id the gain of e-. Electrons must cancel. Add both halves to cancel the electrons. It may be necessary to multiply the halves to make the electrons the same. Cancel water molecules and H+ is they appear on both sides of the equation. Check final equation for balancing materially and electrically. Add (to both sides) as many OH- as H+ there are to cancel and form H 2 O. Cancel the waters. 84

Problem 25: Balance the following equation in basic solution. Mn. O 4 -(aq) +

Problem 25: Balance the following equation in basic solution. Mn. O 4 -(aq) + C 2 O 42 - (aq) Mn. O 2 (aq) + CO 2 (aq) ½ oxidation: ½ reduction 85

Molarity Two solutions can contain the same compounds but be quite different because the

Molarity Two solutions can contain the same compounds but be quite different because the proportions of those compounds are different. Molarity is one way to measure the concentration of a solution. moles of solute Molarity (M) = volume of solution in liters

Concentrations of Solutions Molarity changes with temperature. n If temperature increases, the volume increases

Concentrations of Solutions Molarity changes with temperature. n If temperature increases, the volume increases slightly so the molarity decreases.

Mixing a Solution See video

Mixing a Solution See video

Concentrations of Solutions Example: Calculate the molarity of a solution that contains 12. 5

Concentrations of Solutions Example: Calculate the molarity of a solution that contains 12. 5 g of sulfuric acid in 1. 75 L of solution.

Concentrations of Solutions Example: Determine the mass of calcium nitrate required to prepare 3.

Concentrations of Solutions Example: Determine the mass of calcium nitrate required to prepare 3. 50 L of 0. 800 M Ca(NO 3)2.

Problem 26: A 3. 75 g sample of Na. Cl is dissolved in water.

Problem 26: A 3. 75 g sample of Na. Cl is dissolved in water. The total volume of the solution is 768 m. L. What is the molarity of the solution? Problem 27: How many m. L of 0. 245 M Na. OH are needed to deliver 1. 75 moles of Na. OH?

Dilution

Dilution

Dilution of Solutions The relationship M 1 V 1 = M 2 V 2

Dilution of Solutions The relationship M 1 V 1 = M 2 V 2 is appropriate for dilutions, but not for chemical reactions. Example 3: Suppose you needed to prepare 100. 0 m. L of 1. 00 M NH 3 using 1. 25 M NH 3, distilled water, and a 100 m. L graduated cylinder. How would you do this? M 1 V 1 = M 2 V 2 1. 00 M * 100. 0 m. L = 1. 25 M * V 2 = 80. 0 m. L Pour 80. 0 m. L of 1. 25 M NH 3 into the graduated cylinder and add distilled water until you have 100. 00 of solution. This question can be adapted to whatever other container or accuracy on the concentration is required.

Problem 28: Describe how to prepare 100. 00 m. L of 3. 00 M

Problem 28: Describe how to prepare 100. 00 m. L of 3. 00 M HCl using 12. 6 M HCl, distilled water, and the glassware below. • 100 m. L graduated cylinder • 100 m. L volumetric flask • 25 m. L graduated cylinder • 25 m. L buret • 25 m. L volumetric pipet

Dilution of Solutions Common method to dilute a solution involves the use of volumetric

Dilution of Solutions Common method to dilute a solution involves the use of volumetric flask, pipet, and suction bulb.

Problem 29: If 10. 0 m. L of 5. 00 M NH 4 NO

Problem 29: If 10. 0 m. L of 5. 00 M NH 4 NO 3 is added to enough water to give 100. m. L of solution, what is the concentration of the solution? Problem 30: What volume of 18. 0 M sulfuric acid is required to make 2. 50 L of a 2. 40 M sulfuric acid solution?

Mole fraction = XA =. moles A total number of moles Mole fraction does

Mole fraction = XA =. moles A total number of moles Mole fraction does nor change with temperature.

Problem 31: Find the mole fraction of KOH and H 2 O in a

Problem 31: Find the mole fraction of KOH and H 2 O in a solution that is prepared by dissolving 1. 5 mol KOH in 1. 0 kg H 2 O.

Using Solutions in Chemical Reactions Combine the concepts of molarity and stoichiometry to determine

Using Solutions in Chemical Reactions Combine the concepts of molarity and stoichiometry to determine the amounts of reactants and products involved in reactions in solution.

Using Molarities in Stoichiometric Calculations

Using Molarities in Stoichiometric Calculations

Using Solutions in Chemical Reactions Example: What volume of 0. 500 M Ba. Cl

Using Solutions in Chemical Reactions Example: What volume of 0. 500 M Ba. Cl 2 is required to completely react with 4. 32 g of Na 2 SO 4?

Problem 32: a) What volume of 0. 200 M Na. OH will react with

Problem 32: a) What volume of 0. 200 M Na. OH will react with 50. 0 m. L 0 f 0. 200 M aluminum nitrate, Al(NO 3)3? b) What mass of Al(OH)3 precipitates in (a)?

Using Solutions in Chemical Reactions Titrations are a method of determining the concentration of

Using Solutions in Chemical Reactions Titrations are a method of determining the concentration of an unknown solutions from the known concentration of a solution and solution reaction stoichiometry. n Requires special lab glassware w Buret, pipet, and flasks n Must have an an indicator also

Titration

Titration

Titration

Titration

Titrations Equivalence point: Is the exact moment when: the moles of the titrant added

Titrations Equivalence point: Is the exact moment when: the moles of the titrant added = the moles of the titrated The use of an indicator which changes colors around the equivalence point. When the indicator changes color the volume of the titrant can be recorded. Knowing the molarity and the volume of the titrant, the moles of the titrant can be calculated.

Using Solutions in Chemical Reactions Example: What is the molarity of a KOH solution

Using Solutions in Chemical Reactions Example: What is the molarity of a KOH solution if 38. 7 m. L of the KOH solution is required to react with 43. 2 m. L of 0. 223 M HCl? Step 1: Convert to moles

Using Solutions in Chemical Reactions Step 2: Convert to moles of the unknown.

Using Solutions in Chemical Reactions Step 2: Convert to moles of the unknown.

Using Solutions in Chemical Reactions Step 3: Convert to molarity.

Using Solutions in Chemical Reactions Step 3: Convert to molarity.

Problem 33: What is the molarity of a barium hydroxide solution if 44. 1

Problem 33: What is the molarity of a barium hydroxide solution if 44. 1 m. L of 0. 103 M HCl is required to react with 38. 3 m. L of the Ba(OH)2 solution?

Problem 34: What volume of 0. 545 M of iron(II) nitrate is required to

Problem 34: What volume of 0. 545 M of iron(II) nitrate is required to react completely with 25. 00 m. L of 0. 200 M potassium permanganate?

Problem 35: What is the molarity of Na+ ions present in a solution prepared

Problem 35: What is the molarity of Na+ ions present in a solution prepared by mixing 10. 00 m. L of 0. 150 M Na. Cl with 25. 00 m. L of 0. 0500 M Na 2 SO 4?

Problem 36: What is the molarity of Cl- ions present in a solution prepared

Problem 36: What is the molarity of Cl- ions present in a solution prepared by mixing 100. 00 m. L of 0. 100 M Mg. Cl 2 with 125. 00 m. L of 0. 200 M Al. Cl 3?