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Chapter 4

Chapter 4. Types of Chemical Reactions and Solution Stoichiometry. Parts of solutions.  Solution – homogeneous mixture  Solute – part that dissolves  S olvent – causes the dissolving  S oluble – can be dissolved  M iscible – liquids dissolve in each other.

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Chapter 4

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  1. Chapter 4 Types of Chemical Reactions and Solution Stoichiometry

  2. Parts of solutions Solution – homogeneous mixture Solute – part that dissolves Solvent – causes the dissolving Soluble – can be dissolved Miscible – liquids dissolve in each other

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

  4. Aqueous Solutions • Dissolved in water • Water is a polar molecule • The oxygen atoms have a partial negative charge • The hydrogen atoms have a partial positive charge • The angle is 105o

  5. Hydration • The process of breaking apart ions of a salt. • The “+” end of water attracts the anion • The “-” end of water attracts the cation

  6. Solubility • The ability to dissolve in a given amount of water • Usually g/100mL • Varies greatly • Depends upon ion attraction • Will dissolve nonionic substances if they have polar bonds

  7. Electrolytes • Electrical current through a substance • Ions that are dissolved can move • Solutions are classified three ways

  8. Types of Solutions • Strong electrolytes –Completely ionized when dissolved in water many ions – conduct well Weak electrolytes – partially fall apart into ions few ions- conduct electricity slightly Non-electrolytes – don’t fall apart no ions – don’t conduct electricity

  9. Types of solutions continued • Acids – form H+ ion when dissolved • Strong acids fall apart completely H2SO4 HNO3HClHBr HI HClO4 Weak acids – do not dissociate completely Bases – form OH- when dissolved Strong bases – KOH NaOH

  10. Dissociation • Acids HCl H+ (aq) + Cl-(aq) HNO3H+(aq) + NO3- (aq) H2SO4 H+ (aq) + HSO4- (aq)

  11. Strong bases NaOH Na+ (aq) + OH- (aq) KOH  K+ (aq) + OH-(aq)

  12. Nonelectrolytes • Dissolve in water but do not produce any ions • Example is ethanol (C2H5OH) molecules disperse in the water but doesn’t conduct electricity

  13. Composition of Solutions • Concentration 1. Molarity (M) – moles of solute per volume of solution in liters 2. M = moles of solute liters of solution

  14. Preparation of Molar Solutions 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. 1.56 gHCl 1 1 mole HCl 36.5 gHCl = 0.0427 mole HCl 26.8 mL 1 1L 1000 mL = .0268 L M = 0.0427 mole HCl .0268 L = 1.60 M HCl

  15. Concentrations of Ions Give the concentration of each type of ion in 0.50 M Co(NO3)2 Co(NO3)2(s) Co+2(aq) + 2NO3-(aq) Co+2 1 x 0.50 M = 0.50 M Co+2 NO3- 2 x 0.50 M = 1.0 M NO3-

  16. Calculate the number of moles of Cl- ions in 1.75 L of 1.0 x 10-3 M ZnCl2. ZnCl2 Zn+2(aq) + 2Cl-(aq) 2 x 1.0 x 10-3 = 2.0 x 10-3 M Cl- 1.75 L 2.0 x 10-3 mole Cl- L = 3.5 x 10-3 mole Cl-

  17. Standard solution – a solution where the concentration is accurately known How much solid K2Cr2O7must by weighted out to make a solution? A chemist needs 1.00 L of an aqueous 0.200 M K2Cr2O7 solution.

  18. Dilution Water is added to achieve a particular M Moles of solute after = moles of solute before M1V1=M2V2 What volume of 16 M sulfuric acid must be used to prepare 1.5 L of a 0.10 M H2SO4 solution?

  19. Types of Solution Reactions Precipitation Reactions A solid forms from two solutions 2. Precipitate – the insoluble solid KNO3 (aq) + BaCl2 (aq) 

  20. Three Types of Equations Used to Describe Reactions in a Solution • The formula equation gives the overall reaction. • The complete ionic equation represents as ions all the reactants and products that are strong electrolytes. • The net ionic equation includes only those ions that undergo a change.

  21. Write the formula equation, complete ionic equation, and the net ionic equation for KCl (aq) + AgNO3 (aq) 

  22. Stoichiometry of Precipitation Reactions • Calculate the mass of solid NaCl that must be added to 1.50 L of a 0.100 M AgNO3 solution to precipitate all the Ag+ ions in the form of AgCl.

  23. When aqueous solutions of Na2SO4 and Pb(NO3)2 are mixed, PbSO4 formed when 1.25 L of 0.0500 M Pb(NO3)2 and 2.00 L of 0.0250 M Na2SO4are mixed.

  24. Acid-Base Reactions • Acid is a proton donor (H+) • Base is a proton acceptor usually OH— accepts a H+ • H+(aq)+ OH−(aq)

  25. Cations are surrounded and bound by water molecules protons are also solvated by water molecules Two ways to show this • H+ (aq) • H3O+ (aq) – hydronium ion

  26. Types of acid donors • Monoprotic HCl, HNO3 donates ____ H+ • Diprotic H2SO4 donates ____ H+ • Triprotic H3PO4 donates ____ H+

  27. Neutralization Reactions • What volume of a 0.100 M HCl solution is needed to neutralize 25.0 mL of 0.350 M NaOH? HCl(aq) + NaOH(aq)NaCl(aq) + H2O (l) H+(aq) + OH−(aq)  H2O (l) .0250 L 0.350 mole OH− = L NaOH

  28. Acid-Base Titrations • Volumetric analysis Process of determining the amount of a substance by titration • Titration Process of delivering solutions to one another • Equivalence point (stoichiometric point) Point at which the titration has occurred • Endpoint Point at which the indicator changes color

  29. What volume of 0.812 M HCl, in milliliters, is required to titrate 1.33 g of NaOH to the equivalence point?

  30. Oxidation-Reduction Reactions • Electrons are transferred • Spontaneous redoxrxns can transfer energy Electrons (electricity) Heat LEO the lion says GER

  31. Lose Electrons= Oxidation • Gain Electrons = Reduction • Redox Reactions Examples: 0 0 +1 --1 2Na + Cl2 2NaCl Each sodium atom loses one electron oxidation

  32. 0 -1 Cl + e__Cl • Each chlorine atom gains one electron: reduction

  33. Rules for Assigning Oxidation Numbers Rules 1 & 2 • 1. The oxidation number of any uncombined element is zero • 2. The oxidation number of a monatomic ion equals its charge • 3. The oxidation number of oxygen in compounds is -2 • 4. The oxidation number of hydrogen in compounds is +1

  34. 5. The sum of the oxidation numbers in the formula of a compound is 0. • 6. The sum of the oxidation numbers in the formula of a polyatomic ion is equal to its charge ex. NO3— SO42—

  35. Reducing Agents and Oxidizing Agents • The substance reduced is the oxidizing agent • The substance oxidized is the reducing agent 0 +1 Na  Na + e -- Sodium is oxidized – it is the reducing agent 0 -- 1 Cl + e--Cl Chlorine is reduced – it is the oxidizing agent

  36. Oxidation-Number Changes in reactions • Can you identify what is being oxidized and what is being reduced? 2AgNO3(aq) + Cu (s) Cu(NO3)2(aq) + 2Ag (s) The oxidation number of Ag decreases from +1 to 0 (reduction), copper’s oxidation number increase from 0 to +2 (oxidation)

  37. Balancing Redox Equations • Look at the reaction between solid copper and silver ions in aqueous solution: Cu (s) + Ag+ (aq)  Ag(s) + Cu+2 (aq) 1 e-- gained Cu + Ag+ Ag + Cu+2 0 +1 0 +2 2 e- lost

  38. Ultimately must have equal number of electrons gained and lost Cu (s) + 2Ag+ (aq) 2 Ag (s) + Cu+2 (aq) Try balancing: H+(aq) + Cl_(aq) + Sn(s) + NO3- SnCl6(aq) + NO2(g)+ H2O (l)

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