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Physical Properties of Solutions Chapter 12 Solution Stoichiometry end of Chapter 4 Problems: 12.12, 12.15, 12.16, 12.17, 12.18, 12.21, 12.22, 12.28, 12.36,

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Presentation on theme: "Physical Properties of Solutions Chapter 12 Solution Stoichiometry end of Chapter 4 Problems: 12.12, 12.15, 12.16, 12.17, 12.18, 12.21, 12.22, 12.28, 12.36,"— Presentation transcript:

1 Physical Properties of Solutions Chapter 12 Solution Stoichiometry end of Chapter 4 Problems: 12.12, 12.15, 12.16, 12.17, 12.18, 12.21, 12.22, 12.28, 12.36, 12.38, 12.51, 12.54, 12.55, 12.57, 12.60, 12.65, 12.76, 12.77, , 4.60, 4.70, 4.72, 4.74, 4.77, 4.88

2 12.1 A solution is a homogenous mixture of 2 or more substances The solute is(are) the substance(s) present in the smaller amount(s) The solvent is the substance present in the larger amount

3 An electrolyte is a substance that, when dissolved in water, results in a solution that can conduct electricity. A nonelectrolyte is a substance that, when dissolved, results in a solution that does not conduct electricity. nonelectrolyte weak electrolyte strong electrolyte 4.1

4 A saturated solution contains the maximum amount of a solute that will dissolve in a given solvent at a specific temperature. An unsaturated solution contains less solute than the solvent has the capacity to dissolve at a specific temperature. A supersaturated solution contains more solute than is present in a saturated solution at a specific temperature. Sodium acetate crystals rapidly form when a seed crystal is added to a supersaturated solution of sodium acetate. 12.1

5 12.2 Three types of interactions in the solution process: solvent-solvent interaction solute-solute interaction solvent-solute interaction H soln = H 1 + H 2 + H 3

6 like dissolves like Two substances with similar intermolecular forces are likely to be soluble in each other. non-polar molecules are soluble in non-polar solvents CCl 4 in C 6 H 6 polar molecules are soluble in polar solvents C 2 H 5 OH in H 2 O ionic compounds are more soluble in polar solvents NaCl in H 2 O or NH 3 (l) 12.2

7 Concentration Units The concentration of a solution is the amount of solute present in a given quantity of solvent or solution. Percent by Mass % by mass = x 100% mass of solute mass of solute + mass of solvent = x 100% mass of solute mass of solution 12.3 Mole Fraction (X) X A = moles of A sum of moles of all components

8 Concentration Units Continued M = moles of solute liters of solution Molarity (M) Molality (m) m = moles of solute mass of solvent (kg) 12.3

9 What is the molality of a 5.86 M ethanol (C 2 H 5 OH) solution whose density is g/mL? m =m = moles of solute mass of solvent (kg) M = moles of solute liters of solution Assume 1 L of solution: 5.86 moles ethanol = 270 g ethanol 927 g of solution (1000 mL x g/mL) mass of solvent = mass of solution – mass of solute = 927 g – 270 g = 657 g = kg m =m = moles of solute mass of solvent (kg) = 5.86 moles C 2 H 5 OH kg solvent = 8.92 m 12.3

10 Convert % mass to Molarity What is the Molarity of a 95% acetic acid solution? (density = g/mL) If you assume 1 L, that amount of solution = 1049 g 95% of the solution is acetic acid 1049 g solution x 0.95 = 997 g solute 997 g X 1 mol/60.05 g = 16.6 mol solute Since we assumed 1 L, thats 16.6 mol / 1 L or 16.6 M

11 Temperature and Solubility Solid solubility and temperature solubility increases with increasing temperature solubility decreases with increasing temperature 12.4

12 Fractional crystallization is the separation of a mixture of substances into pure components on the basis of their differing solubilities. Suppose you have 90 g KNO 3 contaminated with 10 g NaCl. Fractional crystallization: 1.Dissolve sample in 100 mL of water at 60 0 C 2.Cool solution to 0 0 C 3.All NaCl will stay in solution (s = 34.2g/100g) 4.78 g of PURE KNO 3 will precipitate (s = 12 g/100g). 90 g – 12 g = 78 g 12.4

13 Temperature and Solubility Gas solubility and temperature solubility usually decreases with increasing temperature 12.4

14 Pressure and Solubility of Gases 12.5 The solubility of a gas in a liquid is proportional to the pressure of the gas over the solution (Henrys law). c = kP c is the concentration (M) of the dissolved gas P is the pressure of the gas over the solution k is a constant (mol/Latm) that depends only on temperature low P low c high P high c

15 Chemistry In Action: The Killer Lake Lake Nyos, West Africa 8/21/86 CO 2 Cloud Released 1700 Casualties Trigger? earthquake landslide strong Winds

16 Solution Stoichiometry (Chapter 4) The concentration of a solution is the amount of solute present in a given quantity of solvent or solution. M = molarity = moles of solute liters of solution What mass of KI is required to make 500. mL of a 2.80 M KI solution? volume KImoles KIgrams KI M KI 500. mL= 232 g KI 166 g KI 1 mol KI x 2.80 mol KI 1 L soln x 1 L 1000 mL x 4.5

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18 Dilution is the procedure for preparing a less concentrated solution from a more concentrated solution. Dilution Add Solvent Moles of solute before dilution (i) Moles of solute after dilution (f) = MiViMiVi MfVfMfVf = 4.5

19 How would you prepare 60.0 mL of 0.2 M HNO 3 from a stock solution of 4.00 M HNO 3 ? M i V i = M f V f M i = 4.00 M f = 0.200V f = 0.06 L V i = ? L 4.5 V i = MfVfMfVf MiMi = x = L = 3 mL 3 mL of acid + 57 mL of water= 60 mL of solution

20 Gravimetric Analysis Dissolve unknown substance in water 2.React unknown with known substance to form a precipitate 3.Filter and dry precipitate 4.Weigh precipitate 5.Use chemical formula and mass of precipitate to determine amount of unknown ion

21 Titrations In a titration a solution of accurately known concentration is added gradually added to another solution of unknown concentration until the chemical reaction between the two solutions is complete. Equivalence point – the point at which the reaction is complete Indicator – substance that changes color at (or near) the equivalence point Slowly add base to unknown acid UNTIL the indicator changes color 4.7

22 What volume of a M NaOH solution is Required to titrate mL of a 4.50 M H 2 SO 4 solution? 4.7 WRITE THE CHEMICAL EQUATION! volume acidmoles acidmoles basevolume base H 2 SO 4 + 2NaOH 2H 2 O + Na 2 SO mol H 2 SO mL soln x 2 mol NaOH 1 mol H 2 SO 4 x 1000 ml soln mol NaOH x mL = 158 mL M acid rx coef. M base

23 Chemistry in Action: Metals from the Sea CaCO 3 (s) CaO (s) + CO 2 (g) Mg(OH) 2 (s) + 2HCl (aq) MgCl 2 (aq) + 2H 2 O (l) CaO (s) + H 2 O (l) Ca 2+ (aq) + 2OH (aq) - Mg 2+ (aq) + 2OH (aq) Mg(OH) 2 (s) - Mg e - Mg 2Cl - Cl 2 + 2e - MgCl 2 (l) Mg (l) + Cl 2 (g) Now back to Chapter 12…

24 Colligative Properties of Nonelectrolyte Solutions Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles. Vapor-Pressure Lowering Raoults law If the solution contains only one solute: X 1 = 1 – X 2 P P 1 = P = X 2 P = vapor pressure of pure solvent X 1 = mole fraction of the solvent X 2 = mole fraction of the solute 12.6 P 1 = X 1 P 1 0

25 Fractional Distillation Apparatus 12.6

26 Boiling-Point Elevation T b = T b – T b 0 T b > T b 0 T b > 0 T b is the boiling point of the pure solvent 0 T b is the boiling point of the solution T b = K b m m is the molality of the solution K b is the molal boiling-point elevation constant ( 0 C/m) 12.6

27 Freezing-Point Depression T f = T f – T f 0 T f > T f 0 T f > 0 T f is the freezing point of the pure solvent 0 T f is the freezing point of the solution T f = K f m m is the molality of the solution K f is the molal freezing-point depression constant ( 0 C/m) 12.6

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29 What is the freezing point of a solution containing 478 g of ethylene glycol (antifreeze) in 3202 g of water? The molar mass of ethylene glycol is g. T f = K f m m =m = moles of solute mass of solvent (kg) = 2.41 m = kg solvent 478 g x 1 mol g K f water = C/m T f = K f m = C/m x 2.41 m = C T f = T f – T f 0 0 = C – C = C 12.6

30 Colligative Properties of Nonelectrolyte Solutions Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles Vapor-Pressure Lowering P 1 = X 1 P 1 0 Boiling-Point Elevation T b = K b m Freezing-Point Depression T f = K f m Osmotic Pressure ( ) = MRT

31 Colligative Properties of Electrolyte Solutions m NaCl solution 0.1 m Na + ions & 0.1 m Cl - ions Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles. 0.1 m NaCl solution0.2 m ions in solution vant Hoff factor (i) = actual number of particles in soln after dissociation number of formula units initially dissolved in soln nonelectrolytes NaCl CaCl 2 i should be 1 2 3

32 Which would you use for the streets of Bloomington to lower the freezing point of ice and why? Would the temperature make any difference in your decision? a)sand, SiO 2 b)Rock salt, NaCl c)Ice Melt, CaCl 2 Change in Freezing Point

33 Boiling-Point Elevation T b = i K b m Freezing-Point Depression T f = i K f m Osmotic Pressure ( ) = iMRT Colligative Properties of Electrolyte Solutions 12.7

34 Change in Freezing Point Common Applications of Freezing Point Depression Propylene glycol Ethylene glycol – deadly to small animals

35 Change in Boiling Point Common Applications of Boiling Point Elevation

36 At what temperature will a 5.4 molal solution of NaCl freeze? Solution T FP = K f m i T FP = (1.86 o C/molal) 5.4 m 2 T FP = (1.86 o C/molal) 5.4 m 2 T FP = 20.1 o C T FP = 20.1 o C FP = 0 – 20.1 = o C FP = 0 – 20.1 = o C Freezing Point Depression

37 The Cleansing Action of Soap 12.8

38 Osmotic Pressure ( ) 12.6 Osmosis is the selective passage of solvent molecules through a porous membrane from a dilute solution to a more concentrated one. A semipermeable membrane allows the passage of solvent molecules but blocks the passage of solute molecules. Osmotic pressure ( ) is the pressure required to stop osmosis. dilute more concentrated

39 High P Low P Osmotic Pressure ( ) = MRT M is the molarity of the solution R is the gas constant T is the temperature (in K) 12.6

40 A cell in an: isotonic solution hypotonic solution hypertonic solution 12.6

41 Chemistry In Action: Desalination

42 A colloid is a dispersion of particles of one substance throughout a dispersing medium of another substance. Colloid versus solution collodial particles are much larger than solute molecules collodial suspension is not as homogeneous as a solution 12.8

43 Colloids Brownian motion Tyndall Effect

44 Suspensions These are mixed, but not dissolved in each other Will settle over time Particles are bigger than 1 micrometer (larger than colloid) Examples: dust in air, muddy water


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