Chapter 41 Aqueous Reactions and Solution Stoichiometry Chapter 4

2 Aqueous Reactions HCl(aq) + NaOH(aq)  H 2 O(l) + NaCl(aq) This is a reaction in which reactants are in solution Solution – homogeneous mixture composed of two parts: solute – the medium which is dissolved solvent – the medium which dissolves the solute.

Chapter 43 Molarity(M) Unit of concentration, moles of solute per liter of solution. Solutions

Chapter 44 Molarity(M) Unit of concentration, moles of solute per liter of solution. Solutions

Chapter 45 Example: What is the molarity(M) of a solution which contains 17.51g of NaCl in 751mL of solution? Solutions

Chapter 46 Example: What is the molarity(M) of a solution which contains 17.51g of NaCl in 751mL of solution? F.W. (NaCl): 58.45g/mol Solutions

Chapter 47 Example: What is the molarity(M) of a solution which contains 17.51g of NaCl in 751mL of solution? F.W. (NaCl): 58.45g/mol Solutions

Chapter 48 Example: What is the molarity(M) of a solution which contains 17.51g of NaCl in 751mL of solution? F.W. (NaCl): 58.45g/mol Solutions

Chapter 49 Example: What is the molarity(M) of a solution which contains 17.51g of NaCl in 751mL of solution? F.W. (NaCl): 58.45g/mol Solution volume Solutions

Chapter 410 Example: What is the molarity(M) of a solution which contains 17.51g of NaCl in 751mL of solution? F.W. (NaCl): 58.45g/mol Solution volume Solutions

Chapter 411 Example: What is the molarity(M) of a solution which contains 17.51g of NaCl in 751mL of solution? Solutions

Chapter 412 Example: What is the molarity(M) of a solution which contains 17.51g of NaCl in 751mL of solution? Solutions

Chapter 413 Example: What is the molarity(M) of a solution which contains 17.51g of NaCl in 751mL of solution? Solutions

Chapter 414 Solutions Molarity

Chapter 415 M dilute V dilute = M concentrated V concentrated Solutions Dilution

Chapter 416 Example: What volume of 6.00M NaOH is required to make 500mL of 0.100M NaOH? M concentrated = 6.00MM dilute = 0.100M V concentrated = ?V dilute = 500mL 0.100M(500mL) = 6.00M(V concentrated ) V concentrated = 8.33mL Solutions Dilution

Chapter 417 Some compounds conduct electricity when dissolved in water – electrolytes Those compounds which do not conduct electricity when dissolved in water are called – nonelectrolytes Properties of Solutes Compounds in Water

Chapter 418 The conductivity of the solution is due to the formation of ions when the compound dissolves in water Properties of Solutes Ionic Compounds in Water (Electrolytes)

Chapter 419 The conductivity of the solution is due to the formation of ions when the compound dissolves in water Properties of Solutes Ionic Compounds in Water (Electrolytes)

Chapter 420 The conductivity of the solution is due to the formation of ions when the compound dissolves in water These ions are not the result of a chemical reaction, they are the result of a dissociation of the molecule into ions that compose the solid. Properties of Solutes Ionic Compounds in Water (Electrolytes)

Chapter 421 Properties of Solutes Ionic Compounds in Water

Chapter 422 In this case no ions are formed, the molecules just disperse throughout the solvent. Properties of Solutes Molecular Compounds in Water(Nonelectrolytes)

Chapter 423 In this case no ions are formed, the molecules just disperse throughout the solvent. Properties of Solutes Molecular Compounds in Water(Nonelectrolytes)

Chapter 424 There are exceptions to this, some molecules are strongly attracted to water and will react with it. Properties of Solutes Molecular Compounds in Water(Nonelectrolytes)

Chapter 425 There are exceptions to this, some molecules are strongly attracted to water and will react with it. Properties of Solutes Molecular Compounds in Water(Nonelectrolytes)

Chapter 426 Strong electrolytes – A substance which completely ionizes in water. Properties of Solutes Strong and Weak Electrolytes

Chapter 427 Strong electrolytes – A substance which completely ionizes in water. For example: Properties of Solutes Strong and Weak Electrolytes

Chapter 428 Weak electrolyte: A substance which partially ionizes when dissolved in water. Properties of Solutes Strong and Weak Electrolytes

Chapter 429 Weak electrolyte: A substance which partially ionizes when dissolved in water. For example: Properties of Solutes Strong and Weak Electrolytes

Chapter 430 Weak electrolyte: A substance which partially ionizes when dissolved in water. For example: Notice that the arrow in this reaction has two heads, this indicates that two opposing reactions are occurring simultaneously. Properties of Solutes Strong and Weak Electrolytes

Chapter 431 Properties of Solutes Strong and Weak Electrolytes

Chapter 432 Since both reactions occur at the same time, this is called a chemical equilibrium Properties of Solutes Strong and Weak Electrolytes

Chapter 433 Acid - substance which ionizes to form hydrogen cations (H + ) in solution Examples: Hydrochloric AcidHCl Nitric AcidHNO 3 Acetic AcidCH 3 CO 2 H Sulfuric AcidH 2 SO 4 Notice that sulfuric acid can provide two H + ’s – Diprotic acid, the other acids can provide only one H + – Monoprotic acid. Acids, Bases, and Salts

Chapter 434 H 2 SO 4  H + + HSO 4 - HSO 4 -  H + + SO 4 2- Acids, Bases, and Salts Diprotic acid

Chapter 435 Bases - substances which reacts with H + ions formed by acids. Examples: ammoniaNH 3 sodium hydroxideNaOH Acids, Bases, and Salts Bases

Chapter 436 H + + OH -  H 2 O -Metal hydroxides (NaOH for example) provide OH - by disassociation. -Bases like ammonia make OH - by reacting with water (ionization) NH 3 + H 2 O  NH OH - Acids, Bases, and Salts Acid-Base Reactions

Chapter 437 Strong acids and bases are strong electrolytes. Weak acids and bases are weak electrolytes. Acids, Bases, and Salts Strong and Weak Acids and Bases

Chapter 438 -The strength of acids and bases are concerned with the ionization (or dissociation) of the substance, not its chemical reactivity. Example: Hydrofluoric acid (HF) is a weak acid, but it is very chemically reactive. - this substance can’t be stored in glass bottles because it reacts with glass (silicon dioxide). Acids, Bases, and Salts Strong Acids

Chapter 439 Common Strong Acids Hydrochloric AcidHCl Hydrobromic AcidHBr Hydroiodic acidHI Nitric AcidHNO 3 Perchloric AcidHClO 4 Chloric AcidHClO 3 Sulfuric AcidH 2 SO 4 Acids, Bases, and Salts Common Strong Acids and Bases

Chapter 440 Common Strong Bases Lithium HydroxideLiOH Sodium HydroxideNaOH Potassium HydroxideKOH Rubidium HydroxideRbOH Cesium HydroxideCsOH Calcium HydroxideCa(OH) 2 Strontium HydroxideSr(OH) 2 Barium HydroxideBa(OH) 2 Acids, Bases, and Salts Common Strong Acids and Bases

Chapter 441 -Reaction between an acid and a base. HCl(aq) + NaOH(aq)  H 2 O(l) + NaCl(aq) “The reaction between an acid and a metal hydroxide produces water and a salt” Salt – an ionic compound whose cation comes from a base and anion from an acid. Acids, Bases, and Salts Neutralization Reaction

Chapter 442 HCl(aq) + NaOH(aq)  H 2 O(l) + NaCl(aq) - despite the appearance of the equation, the reaction actually takes place between the ions. Acids, Bases, and Salts Neutralization Reaction

Chapter 443 HCl(aq) + NaOH(aq)  H 2 O(l) + NaCl(aq) Total Ionic Equation H + (aq) + Cl - (aq) + Na + (aq) + OH - (aq)  H 2 O(l) + Na + (aq) + Cl - (aq) Acids, Bases, and Salts Neutralization Reaction

Chapter 444 HCl(aq) + NaOH(aq)  H 2 O(l) + NaCl(aq) Total Ionic Equation H + (aq) + Cl - (aq) + Na + (aq) + OH - (aq)  H 2 O(l) + Na + (aq) + Cl - (aq) Some of the ions appear unchanged in both the reactants and products – spectator ion Acids, Bases, and Salts Neutralization Reaction

Chapter 445 HCl(aq) + NaOH(aq)  H 2 O(l) + NaCl(aq) Total Ionic Equation H + (aq) + Cl - (aq) + Na + (aq) + OH - (aq)  H 2 O(l) + Na + (aq) + Cl - (aq) Net Ionic Equation H + (aq) + OH - (aq)  H 2 O(l) Acids, Bases, and Salts Neutralization Reaction

Chapter 446 AX + BY  AY + BX Three driving forces for the reaction: -Formation of an insoluble solid (precipitation reaction) -Formation of a weak electrolyte or nonelectrolyte -Formation of a gas that escapes from solution Metathesis Reactions

Chapter 447 Precipitation Reaction Metathesis Reactions

Chapter 448 -A reaction which forms a solid (precipitate) AgNO 3 (aq) + NaCl(aq)  AgCl(s) + NaNO 3 (aq) -AgCl is classified as an insoluble substance Solubility – amount of substance that can be dissolved in a specific amount of water (g/L or mg/L) - for this textbook, any substance with a solubility less than 0.01mol/L is considered insoluble. Metathesis Reactions Precipitation Reaction

Chapter 449 Solubility Guidelines for Ionic Compounds Metathesis Reactions 1.Most nitrates (NO 3 - ) and acetates (CH 3 CO 2 - ) are soluble in water. 2.All chlorides are soluble except: Hg +, Ag +, Pb 2+, Cu + 3.All sulfates are soluble except: Sr 2+, Ba 2+, Pb 2+ 4.Carbonates (CO 3 2- ), Phosphates (PO 4 3- ), Borates (BO 3 3- ), Arsenates (AsO 4 3- ), and Arsenites (AsO 3 3- ) are insoluble. 5.Hydroxides (OH - ) of group Ia and Ba 2+ and Sr 2+ are soluble. 6.Most sulfides (S 2- ) are insoluble.

Chapter 450 Solubility Guidelines for Ionic Compounds Metathesis Reactions Predict the solubility of the following compounds: PbSO 4 AgCH 3 CO 2 KClO 4

Chapter 451 Solubility Guidelines for Ionic Compounds Metathesis Reactions Predict the solubility of the following compounds: PbSO 4 Insoluble AgCH 3 CO 2 KClO 4

Chapter 452 Solubility Guidelines for Ionic Compounds Metathesis Reactions Predict the solubility of the following compounds: PbSO 4 Insoluble AgCH 3 CO 2 Soluble KClO 4

Chapter 453 Solubility Guidelines for Ionic Compounds Metathesis Reactions Predict the solubility of the following compounds: PbSO 4 Insoluble AgCH 3 CO 2 Soluble KClO 4 Soluble

Chapter 454 Reactions Which Form Weak and Non- electrolytes Metathesis Reactions -One example of this type of reaction is a neutralization reaction. Mg(OH) HCl  MgCl H 2 O -Insoluble metal oxide (base) and a strong acid. MgO(s) + 2 HCl(aq)  MgCl 2 (aq) + H 2 O(l)

Chapter 455 Reactions Which Form Gases Metathesis Reactions 2 HCl(aq) + Na 2 S(aq)  2 NaCl(aq) + H 2 S(g)

Chapter 456 Reactions Which Form Gases Metathesis Reactions -Carbonates and Hydrogen Carbonates decompose to give CO 2 -Carbonate and Acid 2 HCl(aq) + Na 2 CO 3 (aq)  2 NaCl(aq) + CO 2 (g) + H 2 O(l) -Hydrogen Carbonate and Acid HCl(aq) + NaHCO 3 (aq)  NaCl(aq) + CO 2 (g) + H 2 O(l)

Chapter 457 Oxidation-Reduction Reactions -reaction where electrons are exchanged. 2 Na(s) + 2 H 2 O(l)  2 NaOH(aq) + H 2 (g) Na(s)  Na + (aq) + 1 e - oxidation – loss of electrons 2 H + (g) + 2 e -  H 2 (g) reduction – gain of electrons

Chapter 458 Oxidation-Reduction Reactions - Oxidation of metals by acids to form salts Mg(s) + 2HCl(aq)  MgCl 2 (aq) + H 2 (g) - Oxidation of metals salts Fe(s) + Ni 2+ (aq)  Fe 2+ (aq) + Ni(s) Notice that the Fe is oxidized to Fe 2+ and the Ni 2+ is reduced to Ni.

Chapter 459 Oxidation-Reduction Reactions - Oxidation of metals by acids to form salts Mg(s) + 2HCl(aq)  MgCl 2 (aq) + H 2 (g) - Oxidation of metals salts Fe(s) + Ni 2+ (aq)  Fe 2+ (aq) + Ni(s) but, Fe 2+ (aq) + Ni(s)  no reaction

Chapter 460 Oxidation-Reduction Reactions -Some metals are more easily oxidized than others. -The relative ability of an element to be oxidized is summarized in a table. Activity series - a list of metals arranged in decreasing ease of oxidation. -The higher the metal on the activity series, the more active that metal. -Any metal can be oxidized by the ions of elements below it. The Activity Series

Chapter 461 Oxidation-Reduction Reactions The Activity Series

Chapter 462 Solution Stoichiometry -We can now use molarity to determine stoichiometric quantities. Example How many grams of hydrogen gas are produced when 20.0 mL of 1.75M HCl is allowed to react with 15.0g of sodium metal? 2 HCl(aq) + 2 Na(s)  H 2 (g) + 2 NaCl(aq)

Chapter 463 Solution Stoichiometry 2 HCl(aq) + 2 Na(s)  H 2 (g) + 2 NaCl(aq) -convert quantities to moles

Chapter 464 Solution Stoichiometry 2 HCl(aq) + 2 Na(s)  H 2 (g) + 2 NaCl(aq) -convert quantities to moles

Chapter 465 Solution Stoichiometry 2 HCl(aq) + 2 Na(s)  H 2 (g) + 2 NaCl(aq) -convert quantities to moles -determine limiting reagent

Chapter 466 Solution Stoichiometry 2 HCl(aq) + 2 Na(s)  H 2 (g) + 2 NaCl(aq) -convert quantities to moles -determine limiting reagent

Chapter 467 Solution Stoichiometry 2 HCl(aq) + 2 Na(s)  H 2 (g) + 2 NaCl(aq) -calculate moles of H 2

Chapter 468 Solution Stoichiometry 2 HCl(aq) + 2 Na(s)  H 2 (g) + 2 NaCl(aq) -calculate moles of H 2

Chapter 469 Solution Stoichiometry 2 HCl(aq) + 2 Na(s)  H 2 (g) + 2 NaCl(aq) -calculate moles of H 2 -calculate grams of H 2

Chapter 470 Solution Stoichiometry 2 HCl(aq) + 2 Na(s)  H 2 (g) + 2 NaCl(aq) -calculate moles of H 2 -calculate grams of H 2

Chapter 471 Solution Stoichiometry Titrations

Chapter 472 Solution Stoichiometry Titrations How many milliliters of 0.155M HCl are needed to neutralize Completely 35.0 mL of M Ba(OH) 2 solution?

Chapter 473 Solution Stoichiometry Titrations How many milliliters of 0.155M HCl are needed to neutralize Completely 35.0 mL of M Ba(OH) 2 solution? 2 HCl(aq) + Ba(OH) 2 (aq) 

Chapter 474 Solution Stoichiometry Titrations How many milliliters of 0.155M HCl are needed to neutralize Completely 35.0 mL of M Ba(OH) 2 solution? 2 HCl(aq) + Ba(OH) 2 (aq)  BaCl 2 (aq) + 2 H 2 O(l)

Chapter 475 Solution Stoichiometry Titrations 2 HCl(aq) + Ba(OH) 2 (aq)  BaCl 2 (aq) + 2 H 2 O(l)

Chapter 476 Solution Stoichiometry Titrations 2 HCl(aq) + Ba(OH) 2 (aq)  BaCl 2 (aq) + 2 H 2 O(l)

Chapter 477 Solution Stoichiometry Titrations 2 HCl(aq) + Ba(OH) 2 (aq)  BaCl 2 (aq) + 2 H 2 O(l)

Chapter 478 Solution Stoichiometry Titrations 2 HCl(aq) + Ba(OH) 2 (aq)  BaCl 2 (aq) + 2 H 2 O(l)

Chapter 479 Solution Stoichiometry Titrations 2 HCl(aq) + Ba(OH) 2 (aq)  BaCl 2 (aq) + 2 H 2 O(l)

Chapter 480 Solution Stoichiometry Titrations 2 HCl(aq) + Ba(OH) 2 (aq)  BaCl 2 (aq) + 2 H 2 O(l)

Chapter 481 Solution Stoichiometry Titrations 2 HCl(aq) + Ba(OH) 2 (aq)  BaCl 2 (aq) + 2 H 2 O(l)

Chapter 482 Solution Stoichiometry Titrations 2 HCl(aq) + Ba(OH) 2 (aq)  BaCl 2 (aq) + 2 H 2 O(l)

Chapter , 4.12, 4.24, 4.32, 4.42, 4.46, 4.52, 4.64, 4.70, 4.72 End of Chapter Problems