Solutions
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Solutions

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  • Figure 14.3 Dissolution of sodium chloride in water. Polar water molecules are attracted to Na and Cl- ions in the salt crystal, weakening the attraction between the ions. As the attraction between the ions weakens, the ions move apart and become surrounded by water dipoles. The hydrated ions slowly diffuse away from the crystal to become dissolved in solution.
  • Figure 14.4 Solubility of various compounds in water. Solids are shown in red and gases are shown in blue.
  • Pouring root beer into a glass illustrates the effect of pressure on solubility. The escaping carbon dioxide produces the foam.
  • Figure 14.8 Vapor pressure curves of pure water and water solutions, showing (a) boiling point elevation and (b) freezing point depression effects (concentration: 1mol solute/1 kg water).

Solutions Solutions Presentation Transcript

  • Chapter 14Foundations of College Chemistry, 13eJohn Wiley & Sons, IncMorris Hein and Susan ArenaSolutionsBrass, a solidsolution of zinc andcopper, is used tomake musicalinstruments andmany other objects.
  • Chapter OutlineCopyright 2011 John Wiley & Sons, Inc14-214.1 General Properties ofSolutions14.2 Solubility14.3 Factors Related toSolubility14.4 Rate of DissolvingSolids14.5 Solutions: A ReactionMedium14.6 Concentration ofSolutions14.7 Colligative Propertiesof Solutions14.8 Osmosis and OsmoticPressure
  • General Properties of Solutions A solution is a homogeneous mixture of one ormore solutes and the solvent. The solute is the substance being dissolved. The solvent is the dissolving agent and isusually the most abundant substance in themixture. Air is a solution of N2(g), O2(g), Ar(g), CO2(g)... What substance is the solvent in air? N2(g), since 78% of air is N2.Copyright 2011 John Wiley & Sons, Inc14-3
  • Common Types of SolutionsCopyright 2011 John Wiley & Sons, Inc14-4What gas is the solute in soft drinks?carbon dioxideWhat is another solute in soft drinks?sugar andflavorings
  • Properties of a True Solution1. A homogeneous mixture of 2 or morecomponents whose ratio can be varied.2. The dissolved solute is molecular or ionic insize (less than 1 nm).3. Liquid or gaseous solutions can be coloredor colorless and are usually transparent.4. The solute will not settle out of the solution.5. The solute can be separated from thesolvent by physical means.Copyright 2011 John Wiley & Sons, Inc14-5
  • Your Turn!Sweet tea is prepared by dissolving aninstant tea packet in water. Whichsubstance is the solvent?a. sugarb. teac. waterCopyright 2011 John Wiley & Sons, Inc14-6
  • Your Turn!A solution of alcohol and water is preparedby adding 25 mL of water to 75 mL methylalcohol. Which substance is the solute?a. methyl alcoholb. waterCopyright 2011 John Wiley & Sons, Inc14-7
  • Solubility Solubility describes the amount of asubstance that will dissolve in a specifiedamount of solvent at a particulartemperature. For example: 36 g NaCl/100 g H2O at20°C Miscible is the term used if 2 liquids willdissolve in each other. Immiscible is used if the liquids will notdissolve in each other.Copyright 2011 John Wiley & Sons, Inc14-8
  • Ionic Compound SolubilityRules NaCl soluble AgNO3 soluble AgCl insoluble AgOH insolubleCopyright 2011 John Wiley & Sons, Inc14-9
  • Your Turn!Use the ionic compound solubility rules topredict the solubility of barium sulfate.a. solubleb. insolubleCopyright 2011 John Wiley & Sons, Inc14-10
  • Your Turn!Use the ionic compound solubility rules topredict the solubility of ammoniumcarbonate.a. solubleb. insolubleCopyright 2011 John Wiley & Sons, Inc14-11
  • Factors Related to Solubility“Like dissolves like” Polar compounds dissolve in polarsolvents, like water and alcohol(CH3CH2OH) Acetone [(CH3)2CO] dissolves in waterbecause it has a net dipole on the O to Cbond, making it polar. Nonpolar compounds dissolve in nonpolarsolvents, like petroleum ether and CCl4 Hexane [CH3(CH2)4CH3] dissolves inpetroleum ether because they are bothnonpolar.Copyright 2011 John Wiley & Sons, Inc14-12
  • Ionic Solubility Many ionic compounds dissolvein water because they form ionto dipole forces with water (astrong intermolecular force). The ions become surroundedby water (become hydrated). The cation is attracted to thepartially negative O in water The anion is attracted to thepartially positive H in water.Copyright 2011 John Wiley & Sons, Inc14-13
  • Temperature and SolubilityMost solids’solubility increaseswith increasingtemperature. (Seered lines.)All gases solubilitydecreases withincreasingtemperature. (Seeblue lines.)Copyright 2011 John Wiley & Sons, Inc14-14
  • Pressure and Solubility Pressure does not affect thesolubility of solids or liquids, butthere is a large effect with gases. The solubility of gas in a liquid isproportional to the pressure of thegas over the liquid. Sodas are canned under highpressure. When you open a can,the pressure decreases andbubbles form, releasing the excessgases.Copyright 2011 John Wiley & Sons, Inc14-15
  • Your Turn!Under what conditions are gases mostsoluble in liquids?a. high temperature, high pressureb. high temperature, low pressurec. low temperature, high pressured. low temperature, low pressureCopyright 2011 John Wiley & Sons, Inc14-16
  • Saturated and UnsaturatedSolutions Saturated solutions contain as muchdissolved solute as the solvent will hold ata given temperature. Saturated solutions are always inequilibrium with undissolved solute. undissolved solute dissolved solute Any point on the solubility curverepresents a saturated solution of thatsolute. Unsaturated solutions contain less solutethan the amount needed to saturate theCopyright 2011 John Wiley & Sons, Inc14-17→→
  • Supersaturated Solutions Supersaturated solutionscontain more solute than theamount needed to saturatethe solution at a particulartemperature. Supersaturated solutions areunstable – stirring, adding acrystal of solute – will causethe excess solute to come outof solution.Copyright 2011 John Wiley & Sons, Inc14-18
  • Your Turn!Copyright 2011 John Wiley & Sons, Inc14-19What mass of thiscompound willdissolve at 30°C?a. 5.0 gb. 5.4 gc. 5.8 gd. 6.0 g
  • Your Turn!Copyright 2011 John Wiley & Sons, Inc14-206.0 g of solute isdissolved in 100 g ofwater at 60°C. Thesolution is allowed tocool to 25°C. Nocrystals form. Thesolution is:a. saturatedb. unsaturatedc. supersaturated
  • Your Turn!The addition of a crystal sodium acetate toa sodium acetate solution causesadditional crystals of sodium acetate toprecipitate. The original solution wasa. Saturatedb. Supersaturatedc. UnsaturatedCopyright 2011 John Wiley & Sons, Inc14-21
  • Rate of Dissolving SolidsParticle SizeA solid can dissolve only at the surface that isin contact with the solvent.Smaller crystals have a larger surface tovolume ratio than large crystals.Smaller crystals dissolve faster than largercrystals.Copyright 2011 John Wiley & Sons, Inc14-22
  • Surface AreaCopyright 2011 John Wiley & Sons, Inc14-23
  • Rate of Dissolving Solids Temperature Increasing the temperature increases therate at which most compounds dissolve.This occurs because solvent molecules strike thesurface of the solid more frequently, causingthe solid to dissolve more rapidly.The dissolved solute particles are also carriedaway from the solid by the higher kineticenergy solvent molecules, allowing moresolvent to hit the surface. Copyright 2011 John Wiley & Sons, Inc14-24
  • Rate of Dissolving Solids Concentration of solutionCopyright 2011 John Wiley & Sons, Inc14-25
  • Rate of Dissolving Solids Agitation or Stirring Stirring rapidly distributes the dissolvedsolute throughout the solution, eliminatingthe saturated solution that forms at thesurface of the solid. Moving dissolved solute away from thesurface increases the contact betweenwater molecules and the solid andincreases the rate of dissolving.Copyright 2011 John Wiley & Sons, Inc14-26
  • Your Turn!Which would most likely increase thesolubility of a solid in water?a. Stirringb. Grind the solid to increase its surfaceareac. Increase the pressured. Increase the temperaturee. All of the aboveCopyright 2011 John Wiley & Sons, Inc14-27
  • Solutions: A ReactionMedium Dissolving reactants allows them to comeinto solution. Combining two solids usually will not resultin any significant reaction: KCl(s) + AgNO3 (s)  no reaction But if you dissolve those same reactants inwater, the silver ion can collide with thechloride ion, resulting in solid AgCl. KCl(aq) + AgNO3(aq)  AgCl(s) +KNO3(aq)Copyright 2011 John Wiley & Sons, Inc14-28
  • Concentration of Solutions Qualitative expressions of concentration:• A dilute solution contains a relatively smallamount of dissolved solute.• A concentrated solution contains arelatively large amount of solute. Hydrochloric acid is sold as aconcentrated 12 M (moles/ L) solution. Adilute 0.1 M solution is commonly found inlabs.Copyright 2011 John Wiley & Sons, Inc14-29
  • Concentration of SolutionsCopyright 2011 John Wiley & Sons, Inc14-30Quantitative expressions of concentration:
  • Mass Percent Calculate the mass % NaCl in a solutionprepared by dissolving 50. g NaCl in 150.g H2O.Copyright 2011 John Wiley & Sons, Inc14-31mass of solutemass % = 100%mass of solution50 g NaClmass % 100%200 g soln= 25% NaClKnownsCalculateSolving for50. g NaCl (solute)150. g H2O (solvent)50. g NaCl + 150. g H2O =200. g mass of solution
  • Mass PercentCalculate the mass of Na2CO3 and waterneeded to make 350. g of a 12.3%solution.Copyright 2011 John Wiley & Sons, Inc14-32mass solute100%mass solution2 312.3 g Na CO350. g soln100 g soln= 43.1 gNa2CO3KnownsCalculateSolving for350. g solution12.3% solutionmass of solute (Na2CO3) and massof H2Omass of H2O = 350. g – 43.1 g = 307 gH2O
  • Mass-Volume PercentNormal saline is a 0.90 m/v % NaCl solution.What mass of sodium chloride is neededto make 50. mL of normal saline?Copyright 2011 John Wiley & Sons, Inc14-33g solute100%ml solution0.90 g NaCl50. mL soln100 mL soln= 0.45 g NaClKnownsCalculateSolving for50. mL solution0.90 m/v% solutionmass of solute (NaCl)
  • Volume Percent What volume of beer that is 6.0 % byvolume alcohol contains 200. mlCH3CH2OH (ethyl alcohol)?Copyright 2011 John Wiley & Sons, Inc14-34volume solute100%volume solution100 mL beer 1 L6.0 ml EtOH 1000 mL200 mLEtOHKnownsCalculateSolving for200. mL EtOH (solute)6.0 volume % solutionvolume of solutionvolume of solution = 3.3 Lbeer
  • Your Turn! A 20.0 % solution of KCl has a mass of 400.g. What mass of KCl is contained in thissolution?a. 20.0 gb. 80.0 gc. 320. gd. 400. gCopyright 2011 John Wiley & Sons, Inc14-35
  • Your Turn! A solution is prepared by mixing 20.0 mL ofpropanol with enough water to produce400.0 mL of solution. What is the volumepercent of propanol in this solution? A. 0.500 % B. 4.76 % C. 5.00 % D. 5.26 %Copyright 2011 John Wiley & Sons, Inc14-36
  • Molarity A 1.0 M KClsolution isprepared bydissolving 1.0moles KCl inenough waterto make 1.0 Lof solution.Copyright 2011 John Wiley & Sons, Inc14-37moles of soluteL of solution
  • Molarity Calculate the molarity of a solutionprepared by dissolving 9.35 g KCl inenough H2O to make 250. mL solution.Copyright 2011 John Wiley & Sons, Inc14-38moles of solute=L of solutionM = 0.502 M KClKnownsCalculateSolving for9.35 g KCl (solute)250. mL solutionmoles of soluteL of solution9.35 g KClM =250 mL soln1 mol KCl74.551 g KCl1000 mL1 L
  • Solution StoichiometryHow many milliliters of 0.175 M Hg(NO3)2 isneeded to completely precipitate 2.50 gKI?Copyright 2011 John Wiley & Sons, Inc14-39Plan1 mol KI2.50gKI166.00g KI3 2Hg(NO )1 mol2 mol KI 3 2Hg(NO )1000 mL soln0.175 molg KI  mol KI  mol Hg(NO3)2 mL solnHg(NO3)2 (aq) + 2 KI(aq)  2KNO3(aq) + HgI2(s)= 43.0 mL of 0.175 MHg(NO3)2
  • Dilution Dilution: Adding solvent to aconcentrated solution to make a moredilute solution. When you dilute a concentratedsolution, only the volume of solutionchanges. The quantity of solute remainsthe same. Volume (V) × Molarity (M) = moles ofsoluteCopyright 2011 John Wiley & Sons, Inc14-40V1 × M1 = V2 × M2
  • Dilution How many milliliters of 12 M HCl areneeded to make 500. mL of 0.10 M HCl?Copyright 2011 John Wiley & Sons, Inc14-41V1 × M1 = V2 × M2KnownsSolving for12 M HCl (concentrated solution) M10.10 M HCl (dilute solution) C2500. mL (dilute solution) V2volume of 12 M HCl V12 211V MVM500 mL×0.10=12MM4.2 mL of 12 MHClCalculate
  • Your Turn! What is the molarity of a solution in which5.85 g of NaCl is dissolved in 200. mL ofsolution?a. 0.500 Mb. 1.00 Mc. 2.00 Md. 4.00 MCopyright 2011 John Wiley & Sons, Inc14-42
  • Your Turn! What is the molarity of the resultingsolution when 300. mL of a 0.400 Msolution is diluted to 800. mL?a. 0.109 Mb. 0.150 Mc. 1.07 Md. 1.47 MCopyright 2011 John Wiley & Sons, Inc14-43
  • Colligative Properties ofSolutions A colligative property is any property of asolution that depends on the number ofsolute particles, and not on the nature ofthe particles.Solutions will have Lower vapor pressures than the puresolvent Higher boiling points than the pure solvent Lower freezing points than the puresolventOsmosis and osmotic pressure are alsoCopyright 2011 John Wiley & Sons, Inc14-44
  • Vapor Pressure Lowering Dissolving solute in a solvent lowers thevapor pressure of the solvent, decreasingthe boiling point (graph a) and thefreezing point (graph b) of the solvent.Copyright 2011 John Wiley & Sons, Inc14-45
  • Molality Since we are looking at properties thatdepend on the number of particles in thesolvent, we use molality, which is thenumber of moles of solute per kg ofsolvent. What is the molality of a solution preparedby dissolving 0.10 mol sugar in 0.50 kgwater?Copyright 2011 John Wiley & Sons, Inc14-46moles of solutekg of solvent20.10 mol= 0.20 m0.50 kg H O
  • Colligative Properties To calculate the effect of dissolving un-ionized andnonvolatile solutes on the boiling point or freezingpoint of the solvent, we use the equation:Copyright 2011 John Wiley & Sons, Inc14-47change in temp = molality constantt m K
  • Boiling Point Elevation What is the boiling point of a solutionprepared by dissolving 0.10 mol sugar in0.50 kg water? The normal boiling point ofwater is 100.0°C and the boiling pointconstant for water is 0.512 °C/m. The boiling point goes up, so we need toadd 0.10°C to the boiling point of purewater. Copyright 2011 John Wiley & Sons, Inc14-48b bt mKb0.512°CΔt = 0.20 × = 0.10°Cmm100.0°C + 0.10°C = 100.1°C
  • Freezing Point Elevation What is the freezing point of a solutionprepared by dissolving 0.10 mol sugar in0.50 kg water? The normal freezing pointof water is 0.0°C and the freezing pointconstant for water is 1.86 °C/m. The freezing point goes down, so we needto subtract 0.37°C from the freezing pointof pure water. Copyright 2011 John Wiley & Sons, Inc14-49b bt mKb1.86°CΔt = 0.20 × = 0.37°Cmm0°C - 0.37°C = -0.37°C
  • Your Turn! What is the boiling point of a 4.00 maqueous solution of a nonvolatilenonelectrolyte? (The boiling pointelevation constant for water is 0.512°C/m.)a. 100.00° Cb. 102.05° Cc. 97.95° Cd. 2.05° CCopyright 2011 John Wiley & Sons, Inc14-50
  • Your Turn! When compared to pure water, aqueoussolutions always havea. Higher boiling point and higher freezingpointb. Lower boiling point and lower freezingpointc. Higher boiling point and lower freezingpointd. Lower boiling point and higher freezingpointCopyright 2011 John Wiley & Sons, Inc14-51
  • Osmosis Osmosis is thediffusion ofwater, either from adilute solution or frompure water, througha semipermeablemembrane into asolution of higherconcentration.Copyright 2011 John Wiley & Sons, Inc14-52
  • Osmotic Pressure The osmotic pressure of a solutioncan be measured by applyingenough pressure to stop the flow ofwater due to osmosis. The difference between theapplied pressure and theatmospheric pressure is the osmoticpressure.Copyright 2011 John Wiley & Sons, Inc14-53
  • Blood and Osmosis Isotonic – same concentration of dissolved particles(0.9% saline) Hypertonic – higher concentration (for example, 1.6%saline) Hypotonic – lower concentration (for example, 0.2%saline) The effect of different concentrations on red blood cells.Copyright 2011 John Wiley & Sons, Inc14-54isotonic hypertonic hypotonic