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CH 4: Chemical Reactions

CH 4: Chemical Reactions

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CH 4: Chemical Reactions

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  1. CH 4: Chemical Reactions Renee Y. Becker Valencia Community College CHM 1045

  2. Solutions • Solute – solid in liquid or lowest mass quantity of substance • Solvent- liquid solute is dissolved in or highest mass quantity of substance

  3. Solution Concentrations • Concentration: allows us to measure out a specific number of moles of a compound by measuring the mass or volume of a solution. • Molarity(M) = Moles of Solute • Liters of Solution • moles = M•L L = moles/M

  4. Example 1: Solution Concentrations • How many moles of solute are present in 125 mL of 0.20 M NaHCO3?

  5. Example 2: Solution Concentrations • How many grams of solute would you use to prepare 500.00 mL of 1.25 M NaOH?

  6. Solution Concentrations • Dilution: the process of reducing a solution’s concentration by adding more solvent. • Moles of solute(constant) = Molarity x Volume • Mi • Vi = Mf • Vf Vf = (Mi • Vi) / Mf • Mf = (Mi • Vi) / Vf

  7. Example 3: Solution Concentrations • What volume of 18.0 M H2SO4 is required to prepare 250.0 mL of 0.500 M H2SO4?

  8. Example 4: Solution Concentrations • What is the final concentration if 75.0 mL of • 3.50 M glucose is diluted to a volume of 400.0 mL?

  9. SolutionStoichiometry • Titration:a technique for determining the concentration of a solution • Standard solution: known concentration • If you have a known volume of standard solution and use it to titrate a known volume of an unknown concentrated solution you can calculate to find the number of moles in the unknown and therefore find it’s concentration

  10. Titration • When doing a titration you add titrant (standard solution) to the analyte (unknown concentration solution) until the endpoint or the equivalence point is reached. This point is when you have equal moles of titrant and analyte, from the volume of the titrant and analyte used and the molarity of the titrant, you can find the molarity of the analyte • Endpoint- based on an indicator • Indicator- a substance that changes color in a specific pH range • Equivalence point- not based on an indicator, usually a pH meter • Use Manalyte• Vanalyte = Mtitrant • Vtitrant

  11. Example 5: SolutionStoichiometry • A 25.0 mL sample of vinegar (dilute CH3CO2H) is titrated and found to react with 94.7 mL of a 0.200 M NaOH. What is the molarity of the acetic acid solution? • NaOH(aq) + CH3CO2H(aq) CH3CO2Na(aq) + H2O(l)

  12. Oxidation–Reduction Reactions • Assigning Oxidation Numbers:All atoms have an “oxidation number” regardless of whether it carries an ionic charge. 1. An atom in its elemental state has an oxidation number of zero. Elemental state as indicated by single elements with no charge. Exception: diatomics H2 N2 O2 F2 Cl2 Br2 and I2

  13. Oxidation–Reduction Reactions 2.An atom in a monatomic ion has an oxidation number identical to its charge.

  14. Oxidation–Reduction Reactions 3.An atom in a polyatomic ion or in a molecular compound usually has the same oxidation number it would have if it were a monatomic ion. A. Hydrogen can be either +1 or –1. B. Oxygen usually has an oxidation number of –2. In peroxides, oxygen is –1. C. Halogens usually have an oxidation number of –1. • When bonded to oxygen, chlorine, bromine, and iodine have positive oxidation numbers.

  15. Oxidation–Reduction Reactions 4. The sum of the oxidation numbers must be zero for a neutral compound and must be equal to the net charge for a polyatomic ion. A.H2SO4 neutral atom, no net charge SO42- sulfate polyatomic ion [SO4]2- [Sx O42-] = -2 X + -8 = -2 X = 6 so sulfur has an oxidation # of +6

  16. Oxidation–Reduction Reactions B.ClO4– , net charge of -1 [ClO4]-1 [Clx O42-] = -1 X + -8 = -1 X = 7 so the oxidation number of chloride is +7

  17. Example 6: Oxidation–Reduction Reactions Assign oxidation numbers to each atom in the following: A. CdS F. VOCl3 B. AlH3 G. HNO3 C. Na2Cr2O7 H. FeSO4 D. SnCl4 I. Fe2O3 E. MnO4– J. V2O3

  18. Electrolytes in Solution • Electrolytes:Dissolve in water to produce ionic solutions. • Nonelectrolytes:Do not form ions when they dissolve in water. a) NaCl sol’n conducts electricity, completes circuit (charged particles) b) C6H12O6 does not

  19. Electrolytes in Solution • Dissociation: • The process by which a compound splits up to form ions in the solution.

  20. Electrolytes in Solution • Strong Electrolyte:Total dissociation when dissolved in water. • Weak Electrolyte:Partial dissociation when dissolved in water.

  21. Types of Reactions • Precipitation • Acid-base neutralization • Oxidation-reduction (redox) • Double replacement • Single replacement • Combination • Decomposition

  22. Types of Chemical Reactions • Precipitation Reactions:A process in which an insoluble solid precipitate drops out of the solution. • Most precipitation reactions occur when the anions and cations of two ionic compounds change partners. (double replacement) Pb(NO3)2(aq) + 2 KI(aq) 2 KNO3(aq) + PbI2(s)

  23. Solubility Rules & Precipitation • Allow you to predict whether a reactant or a product is a precipitate. • Soluble compounds are those which dissolve to more than 0.01 M. • There are three basic classes of salts:

  24. Solubility Rules & Precipitation 1. Salts which are always soluble: • All alkali metal salts: Cs+, Rb+, K+, Na+, Li+ • All ammonium ion (NH4+) salts • All salts of the NO3–, ClO3–, ClO4–, C2H3O2–,and HCO3–ions

  25. Solubility Rules & Precipitation 2. Salts which are soluble with exceptions: • Cl–, Br–, I– ion salts except with Ag+, Pb2+, & Hg22+ • SO42– ion salts except with Ag+, Pb2+, Hg22+, Ca2+, Sr2+, & Ba2+

  26. Solubility Rules & Precipitation 3. Salts which are insoluble with exceptions: • O2– & OH– ion salts except with the alkali metal ions, and Ca2+, Sr2+, & Ba2+ ions • CO32–, PO43–, S2–, CrO42–, & SO32– ion salts except with the alkali metal ions and the ammonium ion • If not listed the compound is probably insoluble

  27. Example 7: Solubility Rules & Precipitation • Predict the solubility of the following in water: • CdCO3 • MgO • Na2S (d) PbSO4 (e) (NH4)3PO4

  28. Example 8: Solubility Rules & Precipitation • Write the balanced reaction and predict whether a precipitate will form for: • NiCl2 (aq) + (NH4)2S (aq) (b) Na2CrO4 (aq) + Pb(NO3)2 (aq) (c) AgClO4 (aq) + CaBr2 (aq)

  29. Equations • Molecular equation – Balanced reaction 2 FeBr3(aq) + 3 Pb(NO3)2(aq) 2 Fe(NO3)3(aq) + 3 PbBr2(s) • Complete ionic equation – All broken up into ions (only aqueous solutions) 2 Fe3+(aq) + 6 Br-(aq) + 3 Pb2+(aq) + 6 NO3-(aq) 2 Fe3+(aq) + 6 NO3-(aq) + 3 PbBr2(s) • Net ionic equation – Cancel out spectator ions 3 Pb2+(aq) +6 Br-(aq) 3 PbBr2(s)

  30. Net Ionic Equations for Precipitation Reactions • Write net ionic equation for the following reaction: 2 AgNO3(aq) + Na2CrO4(aq)  Ag2CrO4(s) + 2 NaNO3(aq) • Is it balanced? If not do it! (molecular equation) • Separate all aqueous sol’n into ions (complete ionic equation) • Cancel out spectator ions on both sides • Rewrite (net ionic equation)

  31. Example 9: • Write the ME, CIE, and NIE for the following reaction Na2CrO4 (aq) + Pb(NO3)2 (aq) NaNO3(aq) + PbCrO4(s)

  32. Types of Chemical Reactions • Acid–Base Neutralization:A process in which an acid reacts with a base to yield water plus an ionic compound called a salt. • The driving force of this reaction is the formation of the stable water molecule. HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)

  33. Acid–Base Concepts • Arrhenius Acid: A substance which dissociates in water to form hydrogen ions (H+). • Arrhenius Base: A substance that dissociates in, or reacts with, water to form hydroxide ions (OH–). Limitations: Has to be an aqueous solution and doesn’t account for the basicity of substances like NH3.

  34. Acid–Base Concepts • Brønsted Acid:Can donate protons (H+) to another substance. • Brønsted Base:Can accept protons (H+) from another substance. (NH3)

  35. Example 10: Conjugate acid-base pairs For the following reactions label the acid, base, conjugate acid, and conjugate base. CH3CO2H(aq) + H2O(l) H3O+(aq) + CH3CO2-(aq) NH3(aq) + H2O(l)  NH4+(aq) + OH-(aq)

  36. Acid–Base Concepts • Lewis Acid: Electron pair acceptor. Al3+, H+, BF3. • Lewis Base: Electron pair donor. H2O, NH3, O2–. • Bond formed is called a coordinate bond or dative bond.

  37. Example 11 • Which of the following is a Bronsted-Lowry base but not an Arrhenius base? • NaOH • NH3 • Mg(OH)2 • KOH

  38. Acids and Bases • Strong acid- st. electrolyte, almost completely dissociates in water • HCl, H2SO4, HNO3, HClO4, HI, HBr • Weak acid- wk. electrolyte, does not dissociate well in water • HF, HCN, CH3CO2H • Strong base - st. electrolyte, almost completely dissociates in water • Metal hydroxides • Weak base- does not dissociate well in water

  39. Acid–Base Concepts Other Weak bases – trimethyl ammonia N(CH3)3, C5H5N pyridine, ammonium hydroxide NH4OH, H2O water

  40. ME, CIE, NIE for Acids/Bases Strong Acid Strong Base ME: HCl(aq) + NaOH(aq) H2O(l) + NaCl(aq) Complete Ionic Equation: H+ + Cl- + Na+ + OH-H2O(l) + Na+ + Cl- Net Ionic Equation: H+ + OH- H2O(l) or H3O+ + OH- 2 H2O(l)

  41. ME, CIE, NIE for Acids/Bases Weak Acid Strong Base ME: HF(aq) + NaOH(aq) H2O(l) + NaF(aq) Complete Ionic Equation: HF + Na+ + OH-H2O(l) + Na+ + F- Net Ionic Equation: HF + OH- H2O(l) + F-

  42. Example 12: ME, CIE, NIE for Acids/Bases Write ME, CIE and NIE for the following: • NaOH(aq) + CH3CO2H(aq) (b) HCl(aq) + NH3(aq) • NaOH strong base will dissociate well • CH3CO2H weak acid doesn’t dissociate well • HCl is a strong acid and therefore a strong electrolyte • NH3 is a weak base and is a weak electrolyte

  43. Types of Chemical Reactions • Double Replacement: These are reactions where two reactants just exchange parts. (double displacement) AX + BYAY + BX BaCl2(aq) + K2SO4(aq) BaSO4(s) + 2 KCl(aq) This is also a ppt reaction, if I ask you what type of reaction is it, what is the best answer??

  44. Types of Chemical Reactions • Oxidation–Reduction (Redox) Reaction:A process in which one or more electrons are transferred between reaction partners. • The driving force of this reaction is the decrease in electrical potential. Mg(s) + I2(g) MgI2(s) Oxidation :Mg0 Mg2+ + 2 electrons Reduction: I20 + 2 electrons  I21-

  45. Example 12: • Which of the following is not an acid-base neutralization reaction? • HCl(aq) + NaOH(s) NaCl(aq) + H2O(l) • 2 HF(aq) + Mg(OH)2(aq) MgF2(aq) + 2 H2O(l) • Pb(NO3)2(aq) + 2 KI(aq)  PbI2 (s) + 2 KNO3(aq)

  46. Oxidation–Reduction Reactions • Redox reactions are those involving the oxidation and reduction of species. • Oxidation and reduction mustoccur together. They cannot exist alone. Fe2+ + Cu0 Fe0 + Cu2+ Reduced: Iron gained 2 electrons Fe2+ + 2 e  Fe0 Oxidized: Copper lost 2 electrons Cu0 Cu2+ + 2e • Remember that electrons are negative so if you gain electrons your oxidation # decreases and if you lose electrons your oxidation # increases

  47. Oxidation–Reduction Reactions Fe2+ + Cu0 Fe0 + Cu2+ • Fe2+ gains electrons, is reduced, and we call it an oxidizing agent • Oxidizing agent is a species that can gain electrons and this facilitates in the oxidation of another species. (electron deficient) • Cu0 loses electrons, is oxidized, and we call it a reducing agent • Reducing agent is a species that can lose electrons and this facilitates in the reduction of another species. (electron rich)

  48. Example 13: • Which is a reduction half reaction? • Fe  Fe2+ + 2e • Fe2+ Fe3+ + 1e • Fe  Fe3+ + 3e • Fe3+ + 1e  Fe2+

  49. Example 14: Oxidation–Reduction Reactions For each of the following, identify which species is the reducing agent and which is the oxidizing agent. • Ca(s) + 2 H+(aq) Ca2+(aq) + H2(g) • 2 Fe2+(aq) + Cl2(aq) 2 Fe3+(aq) + 2 Cl–(aq) C) SnO2(s) + 2 C(s) Sn(s) + 2 CO(g)

  50. Balancing Redox Reactions • Half-Reaction Method:Allows you to focus on the transfer of electrons. This is important when considering batteries and other aspects of electrochemistry. • The key to this method is to realize that the overall reaction can be broken into two parts, or half-reactions. (oxidation half and reduction half)