1 / 40

Chapter 4

Chapter 4. Chemical Reactions and Solution Stoichiometry. Water: The Common Solvent. Essential for most biological reactions Polar covalent molecule Partial charges create poles (dipole moments) Polarity gives water the ability to dissolve compounds. Water Hydration of a Solute.

april
Download Presentation

Chapter 4

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Chapter 4 Chemical Reactions and Solution Stoichiometry

  2. Water: The Common Solvent • Essential for most biological reactions • Polar covalent molecule • Partial charges create poles (dipole moments) • Polarity gives water the ability to dissolve compounds

  3. WaterHydration of a Solute Hydration-the interaction between solute particles and water molecules

  4. WaterHydration of a Solute • When ions dissolve, the break up into individual ions • NH4NO3(s) NH4+(aq) + NO3-(aq) • HOWEVER, solubility varies greatly among ionic compounds • Solubility depends on relative attractions of ions for each other (to be covered in CHpt 11) • Take home message: • An ionic solid dissolves into dispersed ions in water upon hydration

  5. Non-ionic solubility • Non ionic compounds can be soluble if they have a polar bond. • Examples: • ethanol C2H5OH • Methanol CH3OH • Acetone (CH3)2CO • Rule of Thumb • Polar molecules are soluble, “like dissolves like”

  6. Non soluble Compounds/ions • Carbonate (CO32- ) • Phosphate (PO43- ) • ETC……………………………………………………………………………………………………………………………………………… • Non-polar molecules (animal fat)

  7. Characterization of Solutions • Electrical Conductivity • Strong electrolytes: dissolve and completely ionize in water • Strong conductors of electricity • Soluble Salts, Strong Acids, Strong Bases • Weak electrolytes: small degree of ionization • Weak conductors of electricity (but STILL CONDUCT!!) • Weak acids (Acetic Acid), weak bases (NH3) • Non-electrolytes: dissolve but no ionization • Ie; Sucrose (C12H22O11), Ethanol (C2H5OH)

  8. Composition of Solutions • Molarity= • What is the molarity of a 0.30 liter solution containing 0.50 moles of NaCl? • How many liters of solution are needed to make a 1.66 M solution containing 2.11 moles of KMnO4? Moles of solute liters of solution

  9. Composition of Solutions • What is the molarity in 650. ml of solution containing 63 grams of NaCl? • How many grams of Ca(OH)2 are needed to produce 500. ml of 1.66 M Ca(OH)2 solution? • What volume of a 0.88 M solution can be made using 130. grams of FeCl2?

  10. Composition of Solutions • Dilutions • Moles solute before = moles after M1V1=M2V2 • How do you prepare a 250.-ml of a 2.35 M HF dilution from a 15.0 M stock solution? • If 65.5 ml of HCl stock solution is used to make 450.-ml of a 0.675 M HCl dilution, what is the molarity of the stock solution?

  11. Tonight’s Homework • Worksheet I just handed you  • Pg 180-#’s 15,16,21-25 • Read Chapter 4, Sections 1-5 (20 pgs) • DO IT OR YOU WILL BE BEHIND TOMORROW!!!!

  12. Types of Solutions • Precipitation Reactions- aqueous solutions form an insoluble precipitate (duh…) • Involves using solubility rules (pg 150 Table 4.1) to determine the solid precipitate. FOR EXAMPLE Pb(NO3)2(aq)+ KCl(aq)

  13. Pb(NO3)2(aq)+ KCl(aq) • We know that: • Most nitrates salts are soluble. • Most Group I salts are soluble. • Most Chloride salts are soluble with the exception of Pb+2 and others. • Therefore, we determine the products to be: Pb(NO3)2(aq)+ KCl(aq) PbCl2(s) + KNO3(aq)

  14. Describing reactions/Writing equations • Molecular Equation: • Pb(NO3)2(aq)+ 2KCl(aq) PbCl2(s) + 2KNO3(aq) • Complete ionic Equation • Pb+(aq) + 2NO3-(aq) + 2K+(aq) + 2Cl-(aq)  Pb+2(s) + 2Cl-(s) + 2K+(aq) + 2NO3-(aq) • Net Ionic Equation (get rid of the spectator ions) • Spectator ions: present in the aqueous solution before and after reaction. In this case, K+andNO3- • Indicate ions that form precipitate. Pb+(aq) + 2Cl-(aq)  PbCl2(s)

  15. Solution Stoichiometry • Steps to solving Solution Stoichiometry • 1. Identify Species, determine reaction • 2. Write balanced net ionic equation • 3. Calculate moles of reactants • 4. Determine limiting reactant • 5. Calculate moles of product • 6. Convert to proper unit, sig figs, etc…

  16. Calculate the mass of solid NaCl that must be added to 1.5L of a .100M AgNO3 solution to precipitate all the Ag+ ions in the form of AgCl • Step 1: NaCl(s) + AgNO3(aq) AgCl(s) + NaNO3(aq) • Step 2: Ag+(aq) + Cl-(aq) AgCl(s) • Step 3: 1.5L x .1M Ag+ = ? • Step 4? • Step 5? • Step 6?

  17. Calculate the mass of solid NaCl that must be added to 1.5L of a .100M AgNO3 solution to precipitate all the Ag+ ions in the form of AgCl 1Ag+(aq) + 1Cl-(aq) 1AgCl(s) • Step 3: Using a 1:1 mole ratio from the balanced equation • 1.5L x .1M Ag+ = .150 mol Ag+ • .150 mol NaCl x (58.45g NaCl/1mol)= 8.77g NaCl • Step 4? • Step 5? • Step 6?

  18. More homework  • Pg 181 #’s 29-36, 39-44

  19. Acid-Base Reactions • Acids-produce H+ ions when dissolved in water. A proton donor. • Bases-produce OH- ions when dissolved in water. A proton acceptor H+(aq) + OH-(aq) H20(l)

  20. Solution Stoichiometry for Acid-Base Reactions • Steps to solving Solution Stoichiometry • 1. Identify Species, determine reaction • 2. Write balanced net ionic equation • 3. Calculate moles of reactants • 4. Determine limiting reactant • 5. Calculate moles of product • 6. Convert to proper unit, sig figs, etc…

  21. Solution Stoichiometry for Acid-Base Reactions FOR EXAMPLE What Volume of a .100M HCl solution is needed to neutralize 25.0 mL of .350M NaOH? In reality, all you are trying to do is match the moles of hydroxide ions with moles of hydrogen ions. • 1. Identify Species, determine reaction • 2. Write balanced net ionic equation • Involves using solubility rules (pg 150 Table 4.1) to determine a solid precipitate and spectator ions

  22. What Volume of a .100M HCl solution is needed to neutralize 25.0 mL of .350M NaOH? • 1. Identify Species, determine reaction, mole ratio HCl(aq) + NaOH(aq) NaCl(s)+ H20(l) • Mole ratio of 1:1:1:1 • Remember your solubility rules, • Remember that you cannot dissolve water in water, so water is always in the “liquid” state of matter. (Sounds stupid but you would be surprised).

  23. What Volume of a .100M HCl solution is needed to neutralize 25.0 mL of .350M NaOH? • 2. Write balanced net ionic equation • Involves using solubility rules (pg 150 Table 4.1) to determine spectator ions. • Complete ionic equation: H+(aq) + Cl-(aq) + Na+ (aq) + OH- (aq)  H2O(l) + Na+(aq) + Cl-(aq) • Na+ and Cl- are soluble, so they exist as ions in solution. Therefore, the actual reaction does not take place. They are spectator ions. • H+(aq) + OH-(aq)  H2O(l)

  24. What Volume of a .100M HCl solution is needed to neutralize 25.0 mL of .350M NaOH? • 3. Calculate moles of reactants (don’t forget your mole ratios in this step!) • M = n / V • .350M HCL = n / 0.025L • n= .00875 moles  account for sig figs 8.75 x 10-3 moles (preferred by AP graders)

  25. What Volume of a .100M HCl solution is needed to neutralize 25.0 mL of .350M NaOH? • 4. Determine limiting reactant • We are adding a reactant, therefore we do not care about a limiting reactant. But in reality it is a 1:1 mole ratio so there isn’t one.

  26. What Volume of a .100M HCl solution is needed to neutralize 25.0 mL of .350M NaOH? • 5. Calculate moles of reactant needed (HCl solution) • We have a 1:1 mole ratio of hydrogen ions to hydroxide ions. • Moles H+ = Moles OH- in order to reach neutralization point • Therefore, we need 8.75 x 10-3 mol H+ (present in HCl)

  27. What Volume of a .100M HCl solution is needed to neutralize 25.0 mL of .350M NaOH? • 6. Convert to proper unit, sig figs, etc… • The question asks us for the volume of .1M HCl • So: M = n / V  V = n / M VHCl= 8.75 x 10-3 mol H+ .100 mol H+ / L VHCl= 8.75 x 10-2

  28. More homework… • Pg 182 #’s 45-50

  29. Acid-Base titrations • Key vocabulary • Volumetric Analysis-technique for determining the amount of substance through titration • Titration-delivery of a measured volume of a known solution into a substance of unknown concentration • Titrant • Analyte • Equivalence point-the point at which enough titrant has been added to react exactly with the analyte • Indicator-a substance that changes color at or near the equivalence point • Endpoint-point actual point where the indicator changes color

  30. Acid-Base titrations • Requirements for sucessful titrations • 1. exact reaction between titrant and analyte must be known • 2. Equivalence point must be marked accurately (.15 mL of titrant added, etc…) • 3. Volume of titrant required to reach equivalence point must be known accurately (to calculate percent error)

  31. Acid-Base Titrations • Basically, a titration is a neutralization reaction, so we follow the same rules as before. • The goal is to react an equal hydrogen ions to react with hydroxide ions.

  32. Acid-Base Titrations For Example: A student dissolves 1.3009g KHC8H4O4 (KHP) in distilled H20, adds Phenolphthaleine as an indicator, and titrates with NaOH. It was found that 41.20mL of NaOH is required to react with the 1.3009g KHP. What is the concentration of the sodium hydroxide?

  33. A student dissolves 1.3009g KHC8H4O4 (KHP) in distilled H20, adds Phenolphthaleine as an indicator, and titrates with NaOH. It was found that 41.20mL of NaOH is required to react with the 1.3009g KHP. What is the concentration of the sodium hydroxide? • 1. Identify Species, determine reaction, mole ratios • KHC8H4O4(aq) + NaOH(aq)  KOH + NaHC8H4O4(aq) • 2. Write balanced net ionic equation • K+ + HC8H4O4-(aq) + Na+ + OH-(aq)  K+ + OH- + NaHC8H4O4(aq)

  34. Oxidation-Reduction Reactions • Redox reaction-one or more electrons are transferred • Oxidation state (number)-the imaginary charges the atoms would have if the shared electrons were divided equally between identical atoms bonded to each other (huh?)

  35. Rules for assigning Oxidation States • Pg 167, Table 4.2 of your textbook • Memorize it… • More importantly, memorize exceptions to the norm

  36. Oxidation number Practice • Assign Oxidation states to all atoms in the following • CO2 • SF6 • NO3- • CH3- • H2O2 • H2O • START WITH FIRST DRAWING THE MOLECULE, oxidation states do not have to be whole numbers (for some atoms) • KMnO4 • NiO2 • K4Fe(CN)6 • N2H4 • FeO4 • Na2C2O4

  37. Oxidation-Reduction Reactions • Oxidation-increase in oxidation state (loss of electron). Ie, going from -1 to +1 • Reduction- decrease in oxidation state (gain of an electron). Ie, going from +1 to -1 • Oxidizing agent-the electron acceptor • Reducing agent-electron donor

  38. Identification of Oxidizers, reducers • 2Al (s) + 3I2 2AlI3 • Step 1: balance equation, assign oxidation states to reactants • Step 2: assign oxidation states to products • Step 3: Identify who gained electrons (reducing agent • Step 4: Identify who lost electrons (oxidizing agent) This was an easy problem

  39. Identification of Oxidizers, reducers • Identify the oxidizing and reducing agents • PbO(s) + CO2(g) Pb(s) + CO2(g) • CH4(g) + O2(g)  CO2(g) + H2O(l) • Cr2O72-(aq) + HCl  ZnCl2(aq) + H2(g)

  40. Homework  • Pg 182-183, #s 57-62

More Related