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Physical Properties of Solutions

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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, 4.12, 4.60, 4.70, 4.72, 4.74, 4.77, 4.88

2 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 12.1

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 Three types of interactions in the solution process:
solvent-solvent interaction solute-solute interaction solvent-solute interaction DHsoln = DH1 + DH2 + DH3 12.2

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 CCl4 in C6H6 polar molecules are soluble in polar solvents C2H5OH in H2O ionic compounds are more soluble in polar solvents NaCl in H2O or NH3 (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 x 100% mass of solute mass of solute + mass of solvent % by mass = = x 100% mass of solute mass of solution Mole Fraction (X) XA = moles of A sum of moles of all components 12.3

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

9 5.86 moles ethanol = 270 g ethanol
What is the molality of a 5.86 M ethanol (C2H5OH) solution whose density is g/mL? 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 = moles of solute mass of solvent (kg) = 5.86 moles C2H5OH 0.657 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, that’s 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 KNO3 contaminated with 10 g NaCl. Fractional crystallization: Dissolve sample in 100 mL of water at 600C Cool solution to 00C All NaCl will stay in solution (s = 34.2g/100g) 78 g of PURE KNO3 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
The solubility of a gas in a liquid is proportional to the pressure of the gas over the solution (Henry’s law). c is the concentration (M) of the dissolved gas c = kP P is the pressure of the gas over the solution k is a constant (mol/L•atm) that depends only on temperature low P high P low c high c 12.5

15 Chemistry In Action: The Killer Lake
8/21/86 CO2 Cloud Released 1700 Casualties Trigger? earthquake landslide strong Winds Lake Nyos, West Africa

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? M KI M KI volume KI moles KI grams KI 1 L 1000 mL x 2.80 mol KI 1 L soln x 166 g KI 1 mol KI x 500. mL = 232 g KI 4.5

17 4.5

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) after dilution (f) = MiVi MfVf = 4.5

19 How would you prepare 60.0 mL of 0.2 M
HNO3 from a stock solution of 4.00 M HNO3? MiVi = MfVf Mi = 4.00 Mf = 0.200 Vf = 0.06 L Vi = ? L Vi = MfVf Mi = 0.200 x 0.06 4.00 = L = 3 mL 3 mL of acid + 57 mL of water = 60 mL of solution 4.5

20 Gravimetric Analysis Dissolve unknown substance in water
React unknown with known substance to form a precipitate Filter and dry precipitate Weigh precipitate Use chemical formula and mass of precipitate to determine amount of unknown ion 4.6

21 Equivalence point – the point at which the reaction is complete
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 1.420 M NaOH solution is
Required to titrate mL of a 4.50 M H2SO4 solution? WRITE THE CHEMICAL EQUATION! H2SO4 + 2NaOH H2O + Na2SO4 M acid rx coef. M base volume acid moles acid moles base volume base 4.50 mol H2SO4 1000 mL soln x 2 mol NaOH 1 mol H2SO4 x 1000 ml soln 1.420 mol NaOH x 25.00 mL = 158 mL 4.7

23 Chemistry in Action: Metals from the Sea
CaCO3 (s) CaO (s) + CO2 (g) CaO (s) + H2O (l) Ca2+ (aq) + 2OH (aq) - Mg2+ (aq) + 2OH (aq) Mg(OH)2 (s) - Mg(OH)2 (s) + 2HCl (aq) MgCl2 (aq) + 2H2O (l) Mg2+ + 2e Mg 2Cl Cl2 + 2e- MgCl2 (l) Mg (l) + Cl2 (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 P1 = X1 P 1 P 1 = vapor pressure of pure solvent X1 = mole fraction of the solvent Raoult’s law If the solution contains only one solute: X1 = 1 – X2 P 1 - P1 = DP = X2 X2 = mole fraction of the solute 12.6

25 Fractional Distillation Apparatus
12.6

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

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

28 12.6

29 DTf = Kf m Kf water = 1.86 0C/m DTf = Kf m
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. DTf = Kf m Kf water = C/m = 3.202 kg solvent 478 g x 1 mol 62.01 g m = moles of solute mass of solvent (kg) = 2.41 m DTf = Kf m = C/m x 2.41 m = C DTf = T f – Tf Tf = T f – DTf = 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 P1 = X1 P 1 Boiling-Point Elevation DTb = Kb m Freezing-Point Depression DTf = Kf m Osmotic Pressure (p) p = MRT 12.6

31 Colligative Properties of Electrolyte Solutions
0.1 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 solution 0.2 m ions in solution van’t Hoff factor (i) = actual number of particles in soln after dissociation number of formula units initially dissolved in soln i should be nonelectrolytes 1 NaCl 2 CaCl2 3 12.7

32 Change in Freezing Point
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? sand, SiO2 Rock salt, NaCl Ice Melt, CaCl2

33 Colligative Properties of Electrolyte Solutions
Boiling-Point Elevation DTb = i Kb m Freezing-Point Depression DTf = i Kf m Osmotic Pressure (p) p = iMRT 12.7

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

35 Change in Boiling Point
Common Applications of Boiling Point Elevation

36 Freezing Point Depression
At what temperature will a 5.4 molal solution of NaCl freeze? Solution ∆TFP = Kf • m • i ∆TFP = (1.86 oC/molal) • 5.4 m • 2 ∆TFP = oC FP = 0 – 20.1 = oC

37 The Cleansing Action of Soap
12.8

38 Osmotic Pressure (p) 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 (p) is the pressure required to stop osmosis. more concentrated dilute 12.6

39 Osmotic Pressure (p) p = MRT High P Low P
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 hypotonic hypertonic
solution hypotonic solution hypertonic solution 12.6

41 Chemistry In Action: Desalination

42 Colloid versus solution
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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