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MEC Chapter 6

Transcript of Mec chapter 6

  • 1.Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.Chapter 6SolutionsDennistonToppingCaret7th Edition

2. 6.1 Properties of Solutions Solution - homogeneous mixture Solute - the substance in the mixture presentin lesser quantity Solvent - the substance present in the largestquantity Aqueous solution - solution where thesolvent is water Solutions can be liquids as well as solids andgases 3. 6.1 Properties of SolutionsExamples of Solutions Air - oxygen and several trace gases aredissolved in the gaseous solvent, nitrogen Alloys - brass and other homogeneousmetal mixtures in the solid state Focus on liquid solutions as many importantchemical reactions take place in liquidsolutions 4. 6.1 Properties of SolutionsGeneral Properties of LiquidSolutions Clear, transparent, no visible particles May have color Electrolytes are formed from solutes that aresoluble ionic compounds Nonelectrolytes do not dissociateNaCl(s ) H Na + (aq ) + Cl- (aq )2O Volumes of solute and solvent are not additive 1 L ethanol + 1 L water does not give 2 L of solution 5. 6.1 Properties of SolutionsSolutions and Colloids Colloidal suspension - contains soluteparticles which are not uniformlydistributed Due to larger size of particles (1nm - 200 nm) Appears identical to solution from thenaked eye Smaller than 1 nm, have solution Larger than 1 nm, have a precipitate 6. 6.1 Properties of SolutionsDegree of Solubility Solubility - how much of a particular solute candissolve in a certain solvent at a specifiedtemperature Factors which affect solubility: 1 Polarity of solute and solvent The more different they are, the lower the solubility 2 Temperature Increase in temperature usually increases solubility 3 Pressure Usually has no effect If solubility is of gas in liquid, directly proportionalto applied pressure 7. 6.1 Properties of SolutionsSaturation Saturated solution - a solution that contains all thesolute that can be dissolved at a particulartemperature Supersaturated solution - contains more solutethan can be dissolved at the current temperature How is this done? Heat solvent, saturate it with solute then cool slowly Sometimes the excess will precipitate out If it doesnt precipitate, the solution will be supersaturated 8. 6.1 Properties of SolutionsSolubility and Equilibrium If excess solute is added to a solvent, somedissolves At first, rate of dissolution is large Later, reverse reaction precipitation occursmore quickly When equilibrium is reached the rates ofdissolution and precipitation are equal, there issome dissolved and some undissolved solute A saturated solution is an example of a dynamicequilibrium 9. 6.1 Properties of Solutions Solubility of Gases: Henrys Law Henrys law the number of moles of a gasdissolved in a liquid at a given temperature isproportional to the partial pressure of the gasabove the liquid Gas solubility in a liquid is directly proportional tothe pressure of the gas in the atmosphere incontact with the liquid Gases are most soluble at low temperatures Solubility decreases significantly at highertemperatures Carbonated beverages CO2 solubility less when warm Respiration facilitates O2 and CO2 exchange in lungs 10. 6.2 Concentration Based on Mass 6 Concentration - amount of solute dissolvedin a given amount of solution Concentration of a solution has an effect on Physical properties Melting and boiling points Chemical properties Solution reactivity 11. 6.2 Concentration Based onWeight/Volume Percent Amount of solute = mass of solute in grams Amount of solution = volume in milliliters amount of soluteconcentration =Massamount of solution Express concentration as a percentage by multiplying ratio by 100% = weight/volume percent or % (W/V) W grams of solute % = 100% V milliliters of solution 12. 6.2 Concentration Based onCalculating Weight/VolumePercent Calculate the percent composition or % (W/V) of 2.00 x 102 mL containing 20.0 g sodium chloride20.0 g NaCl, mass of soluteMass2.00 x 102 mL, total volume of solution % (W/V) = 20.0g NaCl / 2.00 x 102 mL x 100% = 10.0% (W/V) sodium chloride 13. Calculate Weight of Solute from6.2 Concentration Based on Weight/Volume Percent Calculate the number of grams of glucose in 7.50 x 102 mL of a 15.0% solutionMassW grams of solute% = 100%V milliliters of solution 15.0% (W/V) = Xg glucose/7.50 x 102 mL x 100% Xg glucose x 100% = (15.0% W/V)(7.50 x 102 mL) Xg glucose = 113 g glucose 14. 6.2 Concentration Based on Weight/Weight PercentW grams solute% = 100%W grams solutions Weight/weight percent is most useful forMass solutions of 2 solids whose masses are easily obtained Calculate % (W/W) of platinum in gold ring with 14.00 g Au and 4.500 g Pt [4.500 g Pt / (4.500 g Pt + 14.00 g Au)] x 100% = 4.500 g / 18.50 g x 100% = 24.32% Pt 15. 6.3 Concentration of Solutions:Moles and Equivalents Chemical equations represent the relativenumber of moles of reactants producingproducts Many chemical reactions occur in solutionwhere it is most useful to representconcentrations on a molar basis 16. 6.3 Moles and EquivalentsMolarity The most common mole-basedconcentration unit is molarity Molarity Symbolized M Defined as the number of moles of solute perliter of solution moles soluteM=L solution 17. 6.3 Moles and Equivalents Calculating Molarity from Moles Calculate the molarity of 2.0 L of solution containing 5.0 mol NaOH Use the equationmoles soluteM=L solution Substitute into the equation:MNaOH = 5.0 mol solute2.0 L solution= 2.5 M 18. 6.3 Moles and Equivalents Calculating Molarity From Mass If 5.00 g glucose are dissolved in 1.00 x 102 mL ofsolution, calculate molarity, M, of the glucose solution Convert from g glucose to moles glucose Molar mass of glucose = 1.80 x 102 g/mol 5.00 g x 1 mol / 1.80 x 102 g = 2.78 x 10-2 mol glucose Convert volume from mL to L 1.00 x 102 mL x 1 L / 103 mL = 1.00 x 10-1 L Substitute into the equation: moles soluteM=L solution Mglucose = 2.78 x 10-2 mol glucose 1.00 x 10-1 L solution= 2.78 x 10-1 M 19. 6.3 Moles and Equivalents DilutionDilution is required to prepare a lessconcentrated solution from a moreconcentrated one M1 = molarity of solution before dilution M2 = molarity of solution after dilution V1 = volume of solution before dilution V2 = volume of solution after dilution moles soluteM=moles solute = (M)(L solution)L solution 20. 6.3 Moles and Equivalents Dilution In a dilution will thenumber of moles of solutechange? No, only fewer per unitvolume So, M1V1 = M2V2 Knowing any three termspermits calculation of thefourth 21. 6.3 Moles and EquivalentsCalculating MolarityAfter Dilution Calculate the molarity of a solution made bydiluting 0.050 L of 0.10 M HCl solution to avolume of 1.0 L M1 = 0.10 M molarity of solution before dilution M2 = X M molarity of solution after dilution V1 = 0.050 L volume of solution before dilution V2 = 1.0 L volume of solution after dilution Use dilution expressionM1V1 = M2V2 X M = (0.10 M) (0.050 L) / (1.0 L) 0.0050 M HCl OR 5.0 x 10-3 M HCl 22. 6.3 Moles and Equivalents Representation of Concentrationof Ions in SolutionTwo common ways of expressingconcentration of ions in solution:1. Moles per liter (molarity) Molarity emphasizes the number of individual ions2. Equivalents per liter (eq/L) Emphasis on charge 23. 6.3 Moles and EquivalentsComparison of Molarity and Equivalents1 M Na3PO4 What would the concentration of PO43- ions be? 1M Equivalent is defined by the charge One Equivalent of an ion is the number of gramsof the ion corresponding to Avogadros number ofelectrical charges molar mass of ion (g)One equivalent of an ion = number of charges on ion 24. 6.3 Moles and Equivalents Molarity vs. Equivalents 1 M Na3PO4 1 mol Na+ = 1 equivalent Na+ 1 mol PO43- = 3 equivalents PO43- Equivalents of Na+? 3 mol Na+ = 3 equivalents of Na+ Equivalents of PO43-? 1 mol PO43- = 3 equivalents of PO43- 25. 6.3 Moles and Equivalents Calculating Ion Concentration Calculate eq/L of phosphate ion, PO43- in asolution with 5.0 x 10-3 M phosphate Need to use two conversion factors: mol PO43-mol charge mol charge eq PO435.0 x 10-3 mol PO43- x 3 mol charge x 1 eq 1L 1 mol PO43- 1mol charge 1.5 x 10-2 eq PO43- /L 26. 6.4 Concentration-DependentSolution PropertiesColligative properties - properties of solutions that depend on the concentration of the solute particles, rather than the identity of the soluteFour colligative properties of solutions1. vapor pressure lowering2. boiling point elevation3. freezing point depression4. osmotic pressure 27. 6.4 Concentration-Dependent Vapor Pressure of a LiquidConsider Raoults law in molecular Solution Propertiesterms Vapor pressure of a solutionresults from escape of solventmolecules from liquid to gasphase Partial pressure of gas phasesolvent molecules increasesuntil equilibrium vaporpressure is reached Presence of solute moleculeshinders escape of solventmolecules, loweringequilibrium vapor pressure 28. 6.4 Concentration-Dependent Vapor Pressure Lowering Raoults law - when a nonvolatile solute is Solution Propertiesadded to a solvent, vapor pressure of the solventdecreases in proportion to the concentration ofthe solute Solute molecules (red below) serve as a barrier tothe escape of solvent molecules resulting in adecrease in the vapor pressure 29. 6.4 Concentration-Dependent Freezing Point Depression and Solution Properties Boiling Point Elevation Freezing point depression may be explainedconsidering the equilibrium between solid andliquid states Solute molecules interfere with the rate at which liquid water molecules associate to form the solid state Boiling point elevation can be explainedconsidering the definition as the temperature atwhich vapor pressure of the liquid equals theatmospheric pressure If a solute is present, then the increase in boiling temperature is necessary to raise the vapor pressure to atmospheric temperature 30. 6.4 Concentration-Dependent Freezing Point Depression Freezing point depression