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Chapters 4 and 11: Chemical Rxns and Sol’n Chemistry

Chapters 4 and 11: Chemical Rxns and Sol’n Chemistry. Solutions. A solution is a homogeneous mixture of two or more substances in a single phase of matter. Examples of solutions include salt water, air and alloys. The dissolving medium (water) in a solution is called the solvent .

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Chapters 4 and 11: Chemical Rxns and Sol’n Chemistry

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  1. Chapters 4 and 11: Chemical Rxns and Sol’n Chemistry

  2. Solutions • A solution is a homogeneous mixture of two or more substances in a single phase of matter. • Examples of solutions include salt water, air and alloys. • The dissolving medium (water) in a solution is called the solvent. • The substance dissolved (salt) in a solution is called the solute. • The solute is generally designated as that component of a solution that is of lesser quantity. • PRACTICE: Identify the solute and solvent in a 1.00 M Sr(NO3)2 (aq) solution.

  3. A soluble solute will spread out in a solution until the concentration is the same everywhere. Insoluble solute will either float (lower density than solvent) or sink (greater density).

  4. Solutions are transparent Solute WILL NOT filter out if dissolved. The holes in the filter paper are way too big.

  5. Electrolytes • Solutions containing ions can be characterized by their electrical conductivity. • Solutions that contain strong electrolytes will conduct electricity, whereas solutions of weak or nonelectrolytes will conduct little or no electricity. • A strong electrolyte is a compound that dissociates completely in solution and produce mobile ions, such as NaCl. • A weak electrolyte is a compound that only partially dissociates (like vinegar, a weak acid), and a nonelectrolyte will not dissociate (an example is sugar). • http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/flashfiles/electroChem/conductivity.html

  6. STRONG ELECTROLYTES • Completely soluble salts are strong electrolytes, since all ions are dissociated. • Strong acids are strong electrolytes, since they are completely dissociated in water (ex: HCl, HBr, HI, HNO3, H2SO4, HClO4, HClO3) • Strong bases are strong electrolytes (ex: NaOH, and other Group 1 hydroxides). Some strong bases are not good electrolytes because of low solubility (ex: Ca(OH)2, and other lighter group 2 hydroxides).

  7. Solubility • Solubility is defined as the amount of a substance that can be dissolved in a given quantity of solvent. • Any substance whose solubility is less than 0.01 mol/L will be referred to as insoluble. • We can predict whether a precipitate will form when solutions are mixed if we know the solubilities of different substances.

  8. KISS (keep it simple solubility) Rules 1. All common compounds of Group I and ammonium ions are soluble. 2. All nitrates, acetates, and chlorates are soluble. • . All halogen compounds (other than F) with metals are soluble, except those of Ag+, Hg22+, and Pb2+. (Pb2+ halides are soluble in hot water.) 4. All sulfates are soluble, except those of barium, strontium, calcium, lead, silver, and mercury (I). The latter three are slightly soluble. * Except for rules above, other ions are generally insoluble.

  9. Solubility Table

  10. What determines solubility? Three factors will affect solubility: #1: Nature of solute and solvent. Remember the rule: Like Dissolves Like Polar solvents (partial + or – charges) will easily dissolve charged particles or polar molecules. Nonpolar solvents (no charges, equal sharing of e-) will dissolve nonpolar solutes.

  11. Factors Affecting Solubility #2: Temperature of solution. Generally warmer solutions will hold more solute (except for gases). #3: Pressure (gases only) on solution will increase solubility.

  12. Temp effects on solubility • This graph represents the solubility of NaCl, NaNO3, and KNO3 at different temps. • Notice that when temp increases, solubility increases.

  13. Sat. vs. Unsat. Solutions • A solution that cant dissolve any more solute is said to be saturated. • Additional solute will not dissolve if added to this solution. • An unsaturated solution can still hold more solute. It has not yet reached its capacity. • A supersaturated solution can be made by dissolving the solute under high temps and then carefully cooling them. These are unstable solutions and will suddenly precipitate if provoked.

  14. The Dissolving Process

  15. Ion hydration • Water is a polar solvent and is attracted to polar solutes. • Salt is polar. • Water molecules surround and isolate the surface ions. The ions become hydrated and move away from each other in a process called dissociation. • http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/flashfiles/thermochem/solutionSalt.html

  16. Solution Concentrations 1. Molarity: M = moles solute Liters solution (NOTE: solution is solute PLUS solvent) 2. molality: m = moles solute kg solvent (used when temperature may affect solution volume)

  17. Molarity Calculations Calculate the molarity of a solution prepared with 35.2 grams of CO2 in 500. mL solution. Step 1: convert 35.2 g of CO2 into moles: 35.2g CO2 x 1 mole CO2 = 0.800 mol CO2 44.01 g CO2 Step 2: divide moles by volume in liters 0.800 mol CO2 = 1.60 M 0.500 L

  18. Making Diluted Solutions • Often in chemistry, you will have to make a dilute solution using a stronger solution. • Use the following equation: M1V1 = M2V2 • M1 and M2 are the initial and final molar solutions. • V1 and V2 are the initial and final volumes of solutions. • How would you prepare a solution by dilution of a more concentrated solution?

  19. Example: How much 1.00 M HC2H3O2 is needed to create 200. mL of a 0.100 M solution? ANS: M1 = 1.00 M HC2H3O2 (concentrated) V1 = ? M2 = 0.100 M HC2H3O2 (diluted) V2 = 0.200 L M1V1 = M2V2 and V1= M2V2/M1 V1=(0.100M HC2H3O2 x 0.200L)/(1.00MHC2H3O2) =0.0200 L or 20.0 mL

  20. Overview: Aqueous Reaction Types • Reactions in aqueous solution will generally fall under one of three categories: • Precipitation reactions result in the formation of an insoluble product that separates from solution. • Acid-base reactions are characterized by the transfer of protons. • Redox reactions (or oxidation-reduction reactions) involve the transfer of electrons. *Remember, if you are not part of the solution, you’re part of the precipitate!

  21. Precipitation Reactions • A solid that forms from solution is called a precipitate. • Precipitates for when a combination of insoluble ions form: AgNO3(aq) + NaCl(aq)→ AgCl(s) + NaNO 3(aq) The insoluble solid AgCl will separate from the solution and precipitate out.

  22. Acid-base reactions • An acid can be defined as a proton (H+) donor, and a base as a proton acceptor. • When acids and bases combine, the reaction is called a neutralization reaction. The products of this reaction are water and a salt. example: HCl + NaOH → HOH + NaCl • The concentration of an unknown acid or base can be determined experimentally by a method called titration. In an acid-base titration, a neutralization reaction occurs.

  23. Titration In a titration, a buret is used to deliver a precise amount of an acid or base of KNOWN concentration (the titrant) to a KNOWN volume of a solution with an UNKNOWN concentration (the analyte). Using stoichiometry, the concentration of the unknown solution is determined at the equivalence point.

  24. How does an indicator work? Titration • The exact point at which the titrant reacts completely with the analyte is the equivalence point (or stoichiometric point). • This point is often marked by an indicator, which is a substance that changes color as close as possible to the equivalence point. The point at which the indicator changes is known as the endpoint of the titration.

  25. A-B Titration Problem #1 Calculate the concentration of 50.0 mL of a NaOH solution if the titration endpoint is reached at 34.6 mL with 0.250 M HCl: The stoichiometric ratio of H+:OH- is 1:1, so we can use the formula: MaVa=MbVb Therefore, Mb = (0.250 M)(0.0346L)/(0.050L) =0.173 M NaOH

  26. A-B Titration Problem #2 Calculate the concentration of 35.0 mL of a H2SO4 solution if the titration endpoint is reached at 22.4 mL with 0.150 M NaOH: The stoichiometric ratio of [H+]:[OH-] is 2:1 for H2SO4 and NaOH, so we can use the formula: (2Ma)Va=MbVb Therefore, Ma = (0.150 M)(0.0224L) 2(0.0350L) =0.0471 M NaOH * also referred to as NORMALITY (N)

  27. Redox Reactions • An atom’s oxidation number (or oxidation state) signifies the number of charges the atom would have in a molecule (or an ionic compound) if electrons were transferred completely. • Oxidation-reduction (redox) reaction is a chemical reaction that results in a change in oxidation number of one or more species due to electron transfer.

  28. Rules for determining oxidation #s • Free elements (uncombined) = 0 • Oxygen = usually -2 (except peroxide O2-2 = -1) • Hydrogen = +1 (or -1combined with metals) • Group 1 metals = +1 • Group 2 metals = +2 • Halogens = usually -1, but can be + when combined with oxygen • d-block metals = varies • Sum of all ox #’s in a compound must = 0 • Sum of all ox #’s in ion = ion charge

  29. Practice What are the oxidation numbers of the following: Oxygen in H2SO4 -2 Hydrogen in HCl +1 Iron metal 0 Iron in Fe2O3 +3 Lead in PbSO4 +2 Sulfur in H2SO4 +6 Hydrogen in H2 0 +1 Lithium in any cmpd

  30. More Practice Determine the oxidation number for atoms in each of the following: • Li2O (b) HNO3 (c) Cr2O72- ANSWER: • Li = +1, O= -2 • H = +1, (NO3) = -1, O = -2, N = +5 • O = -2, Cr = +6

  31. Examples of Redox Reactions • Oxidation = process that results in an increase in oxidation number (losing electrons) • Reduction = process that results in adecrease in oxidation number (gaining electrons) LEO says GER!

  32. Redox Terms • Higher oxidation number = species is oxidized, and is a good reducing agent • Lower oxidation number = gets reduced, goodoxidizing agent • Atoms that want to be reduced are strong oxidizers and atoms that want to be oxidized are good reducing agents. • Identify the redox pairs in these reactions: 0 0 +1 -1 Na=oxidized, reducing agent Cl=reduced, oxidizing agent 0 +2 0 +2 Zn=oxidized Cu=reduced 0 +1 0 +2 Cu=oxidized Ag=reduced

  33. Redox Half-Reactions • We can think of redox processes as two separate ½-reactions occurring at the same time. • Consider this single replacement reaction: Oxidation Reduction

  34. Steps to balancing Redox Reactions • First, chemically balance the oxidation and reduction ½- reactions. • Add H2O and H+ to balance (as needed)in acids. In bases, add OH-’s to neutralize H+ in final step. • Add e-’s to balance the charge on both sides. • Multiply both ½-reactions by integers so that the number of e-’s used are equal. • Add the newly balanced ½-reactions together and cancel species as necessary.

  35. Colligative Properties • Colligative comes from the Greek word kolligativ meaning glued together. • We use this term for the properties of substances (solutes and solvents) together. • Colligative properties of solutions depend only on the solvent and the concentration of the solute, not its identity.

  36. There are 4 colligative properties: • Vapor Pressure Lowering • Boiling Point elevation • Freezing Point Depression • Osmotic Pressure These properties will be investigated in the next few slides….

  37. Review: What is a heating curve? • A heating curve shows phase changes and specific heat capacities for a substance. ΔHvaporization ΔHfusion Cp heat capacities of solid, liquid and vapor

  38. What is a phase diagram?(phase diagram for water) Note the negative slope between solid and liquid indicates that the density of ice is LESS than the density of water Normal melting point: melting point at one atmosphere critical point: beyond this point the vapor cannot be liquified at any pressure Normal boiling Point: boiling point at one atmosphere triple point: T and P at which all three states coexist in equilibrium

  39. Phase diagram for CO2

  40. What is vapor pressure? Vapor pressure is the pressure of the vapor above a liquid present at equilibrium (saturated). Factors that affect vapor pressure: 1. Type of molecules: strong intermolecular forces will inhibit molecules from escaping and keep them stuck together in solution. 2. Temperature: warmer molecules will have more energy and can escape solution more easily. 3. The surface area DOES NOT affect vapor pressure:

  41. Vapor pressure and boiling point • The boiling point of a substance is defined as the temperature at which the vapor pressure of a solution is equal to the atmospheric pressure. • Equilibrium vapor pressure animation: • http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/vaporv3.swf vapor pressure: the pressure of the vapor present at equilibrium (saturated) vapor pressure equals atmospheric pressure, solvent will boil

  42. Recall the diagram for the vapor pressure of pure solvents:

  43. What happens to vapor pressure when solute is added? • The vapor pressure diagram for solutions shows that vapor pressure is lowered when a solute is added. • What will this do to the normal boiling point?

  44. The amount of lowering is found using Raoult’s Law: • Raoult’s law: Psoln = xsolventPosolvent where Psoln= solution vapor pressure xsolvent = mole fraction of solvent (=mol solvent/(total moles solute + solvent)) Posolvent = vapor pressure of pure solvent

  45. Molality revisited… • Recall the units for Molarity: M = moles solute L of solution • Molality is the measure of the number of moles of a solute per 1000g (1 Kg) of solvent. m = moles solute 1000 g (kg) solvent • Molality is best used to describe colligitive properties and is represented by m.

  46. Example Problem: • At 35oC, the vapor pressure of water is 43.4 mmHg. What is the vapor pressure of a 1.00 molal solution of NaCl? • ANSWER: 1.00 mol NaCl x 2 mol ions = 2.00 mol solute 1 mol NaCl 1000 g H2O x 1 mol H2O = 55.5 mol H2O 18.02 g H2O XH2O = 55.5 mol H2O /57.5 mol total = 0.965 Psoln = xsolventPosolvent = 0.965 x 43.4 mmHg Psoln = 41.9 mmHg

  47. Volatile solutes • So far we have only dealt with solutions in which the solute is nonvolatile and will not contribute to the vapor pressure. • For an ideal liquid-liquid solution in which both components are volatile, Raoult’s law is modified: • PTOTAL = PA + PB = xAPoA + xBPoB where PTOTAL = total mixture vapor pressure PA and PB = vapor pressure of solns A and B xA and xB = mol fractions of solvents A and B PoA and PoB =vapor pressures of pure A and B Ideal solutions obey Raoult’s Law

  48. Ideal solution vapor pressure Nonideal solutions Nonideal solution vapor pressure • Positive deviations occur when interactions between solute and solvent are WEAKER than in pure solvent and molecules escape more easily. Negative deviations will occur when solute-solvent interactions are stronger and hold molecules back. Vapor pressure of pure A Vapor pressure of pure B Vapor pressure of mixture at any ratio

  49. Example problem • A solution is prepared using 5.81 g acetone (C3H6O, molar mass= 58.1 g/mol) and 11.9 g chloroform (HCCl3, molar mass=119.4 g/mol). At 35ºC this solution has a total vapor pressure of 260. torr. Is this an ideal solution? The pure vapor pressures of acetone and chloroform are 345 and 293 torr at 35ºC, respectively. • ANSWER: PTOTAL = xAPoA + xCPoC 5.81 g acetone x 1 mol acetone = 0.100 mol acetone 58.1 g acetone 11.9 g chloroform x 1 mol chloroform = 0.100 mol chloroform 119.4 g chloroform xA= 0.500 and xC = 0.500 PTOTAL = (0.500)(345 torr) + (0.500)(293 torr) = 319 torr

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