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PRINCIPLES OF CHEMISTRY II CHEM 1212 CHAPTER 12 DR. AUGUSTINE OFORI AGYEMAN Assistant professor of chemistry Department of natural sciences Clayton state.

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Presentation on theme: "PRINCIPLES OF CHEMISTRY II CHEM 1212 CHAPTER 12 DR. AUGUSTINE OFORI AGYEMAN Assistant professor of chemistry Department of natural sciences Clayton state."— Presentation transcript:

1 PRINCIPLES OF CHEMISTRY II CHEM 1212 CHAPTER 12 DR. AUGUSTINE OFORI AGYEMAN Assistant professor of chemistry Department of natural sciences Clayton state university

2 CHAPTER 12 SOLUTIONS

3 - A homogeneous mixture of two or more substances Solvent - The substance present in the greatest quantity Solute - The other substance(s) dissolved in the solvent SOLUTION

4 - Solutions can exist in any of the physical states Solid Solution - dental fillings, metal alloys (steel), polymers Liquid Solution - sugar in water, salt in water, wine, vinegar Gas Solution - air (O 2, Ar, etc. in N 2 ), - NO x, SO 2, CO 2 in the atmosphere SOLUTION

5 - The amount of solute dissolved in a given quantity of solvent or solution Molarity (M: molar) - The number of moles of solute per liter of solution - A solution of 1.00 M (read as 1.00 molar) contains 1.00 mole of solute per liter of solution CONCENTRATION OF SOLUTIONS

6 Calculate the molarity of a solution made by dissolving 2.56 g of NaCl in enough water to make 2.00 L of solution - Calculate moles of NaCl using grams and molar mass - Convert volume of solution to liters - Calculate molarity using moles and liters CONCENTRATION OF SOLUTIONS

7 After dissolving 1.56 g of NaOH in a certain volume of water, the resulting solution had a concentration of 1.60 M. Calculate the volume of the resulting NaOH solution - Convert grams NaOH to moles using molar mass - Calculate volume (L) using moles and molarity CONCENTRATION OF SOLUTIONS

8 Mole Fraction (χ) - Fraction of moles of a component of solution CONCENTRATION OF SOLUTIONS The sum of mole fractions of all components = 1

9 Given that the total moles of an aqueous solution of NaCl and other solutes is 1.75 mol. Calculate the mole fraction of NaCl if the solution contains 4.56 g NaCl. CONCENTRATION OF SOLUTIONS

10 Percent Concentration - Percent by mass [mass-mass percent, %(m/m)] mass of solution = mass of solute + mass of solvent CONCENTRATION OF SOLUTIONS

11 A sugar solution is made by dissolving 5.8 g of sugar in 82.5 g of water. Calculate the percent by mass concentration of sugar. CONCENTRATION OF SOLUTIONS

12 volume of solution ≠ volume of solvent + volume of solute - Due differences in bond lengths and angles Percent Concentration - Percent by volume [volume-volume percent, %(v/v)] CONCENTRATION OF SOLUTIONS

13 Calculate the volume percent of solute if 345 mL of ethyl alcohol is dissolved in enough water to produce 1257 mL of solution CONCENTRATION OF SOLUTIONS

14 - Units are specified because they do not cancel Percent Concentration - Mass-volume percent [%(m/v)] CONCENTRATION OF SOLUTIONS

15 The concentration of a solution of NaCl is 0.92 %(m/v) used to dissolve drugs for intravenous use. What is the amount, in grams, of NaCl needed to prepare mL of the solution? g solute = [%(m/v)] x [volume of solution (mL)]/[100 %] = [(0.92 % g/mL) x (41.50 mL)]/(100 %) = 0.38 g CONCENTRATION OF SOLUTIONS

16 PARTS PER MILLION (PPM) Percent can be defined as parts per hundred 1 ppm ≈ 1 µg/mL or 1 mg/L

17 PARTS PER MILLION (PPM) If L of aqueous solution with a density of 1.00 g/mL contains 13.7 μg of pesticide, express the concentration of pesticide in ppm ppm = µg/mL L = 250 mL Density = 1.00 g/mL Implies mass solution = 250 g

18 PARTS PER BILLION (PPB) 1 ppb ≈ 1 ng/mL or 1 µg/L

19 PARTS PER MILLION (PPB) If L of aqueous solution with a density of 1.00 g/mL contains 13.7 μg of pesticide, express the concentration of pesticide in ppb ppm = µg/L Volume of solution = L Density = 1.00 g/mL Implies mass solution = 250 g

20 MOLALITY (m) Moles of solute per kg of solvent Unit: m or molal

21 MOLALITY (m) What is the molality of a solution that contains 2.50 g NaCl in g water? - Calculate moles NaCl - Convert g water to kg water - Divide to get molality

22 CONVERTING CONCENTRATION UNITS Calculate the molality of a 6.75 %(m/m) solution of ethanol (C 2 H 5 OH) in water Mass water = 100 g solution – 6.75 g ethanol = g water

23 CONVERTING CONCENTRATION UNITS Calculate the mole fraction of a 6.75 %(m/m) solution of ethanol (C 2 H 5 OH) in water Mass water = 100 g solution – 6.75 g ethanol = g water

24 CONVERTING CONCENTRATION UNITS Practice Question Given that the mole fraction of ammonia (NH 3 ) in water is Calculate the molality of the ammonia solution

25 CONVERTING CONCENTRATION UNITS - Molarity is temperature dependent (changes with change in temperature) - Volume increases with increase in temperature hence molarity decreases On the other hand - Molality - Mass percent - Mole fraction are temperature independent

26 - A measure of how much of a solute can be dissolved in a solvent - Grams of solute per 100 mL of solvent - Units: grams/100 mL Three factors that affect solubility - Temperature - Pressure - Polarity SOLUBILITY

27 Unsaturated Solution - More solute can still be dissolved at a given temperature - Concentration of the solute is less than the solubility Saturated Solution - No more solute can be dissolved at a given temperature - Concentration of the solute is equal to the solubility - Dynamic equilibrium is reached Supersaturated Solution - Too much solute has temporarily been dissolved - Concentration of solute is temporarily greater than the solubility - Unstable condition SOLUBILITY

28 The process of dissolving (known as dissolution) is contributed by factors such as - Enthalpy change due to solute-solvent interactions and - Change in disorder DISSOLUTION

29 - Change in enthalpy arises mainly from changes in intermolecular attractions Three types of intermolecular attractions are involved - Solute-solute - Solvent-solvent - Solute-solvent SOLUTE-SOLVENT INTERACTIONS

30 - The relative strengths of these interactions determine the formation of a solution by two substances - Substances with similar properties (strong solute-solvent interactions) tend to form solutions - Like dissolves like SOLUTE-SOLVENT INTERACTIONS

31 - Solvent molecules move apart to accommodate solute molecules - Energy is required to separate solvent molecules attracting each other (ΔH 1 ) - Energy is also required to separate solute molecules (ΔH 2 ) - Energy is released when solvent and solute molecules come together due to attractive forces between them (ΔH 3 ) SOLUTE-SOLVENT INTERACTIONS

32 Enthalpy of Solution - The overall enthalpy change that accompanies the dissolution of one mole of a solution ΔH soln = ΔH 1 + ΔH 2 + ΔH 3 SOLUTE-SOLVENT INTERACTIONS

33 Endothermic Heat of Solution - Energy released by solute-solvent interactions is less than the energy absorbed by separating the solvent and solute molecules ΔH soln is positive ΔH 3 < (ΔH 1 + ΔH 2 ) Example Ammonium nitrate in water SOLUTE-SOLVENT INTERACTIONS

34 ENTHALPY OF SOLUTION Enthalpy Solution Separated solute Solute + Separated solvent Endothermic Heat of Solution Separated solvent Solute + Solvent + ∆H 3 ∆H soln ∆H 2 ∆H 1

35 Exothermic Heat of Solution - Energy released by solute-solvent interactions is greater than the energy absorbed by separating the solvent and solute molecules ΔH soln is negative ΔH 3 > (ΔH 1 + ΔH 2 ) Example Sulfuric acid in water NaOH in water SOLUTE-SOLVENT INTERACTIONS

36 ENTHALPY OF SOLUTION Enthalpy Solution Separated solute Solute + Separated solvent Exothermic Heat of Solution Separated solvent Solute + Solvent + ∆H 3 ∆H soln ∆H 2 ∆H 1

37 Generally - Substances with similar intermolecular forces and hence similar properties have strong solute-solvent interactions - Such substances tend to form solutions - Like dissolves like Example CH 3 OH readily dissolves in H 2 O (hydrogen bonding in both) CCl 4 readily dissolves in C 7 H 16 (London forces in both) SOLUTE-SOLVENT INTERACTIONS

38 - Increase in disorder on mixing is another contributing factor in the dissolution process - Increase in disorder tends to occur spontaneously in processes - The main driving force in the formation of solutions Consider NH 4 NO 3 in H 2 O (used in cold packs) - Enthalpy change on mixing is positive (+28 kJ/mol) - NH 4 NO 3 dissolves to form solution due to increase in disorder SOLUTE-SOLVENT INTERACTIONS

39 Spontaneous Process - Takes place with no apparent cause Nonspontaneous Process - Requires something to be applied in order for it to occur (usually in the form of energy)

40 - Strong electrostatic attractions between oppositely charged ions hold ionic solids together - For soluble ionic compounds the enthalpy of attraction between solvent molecules and ions must be comparable to the crystal lattice enthalpy in the solid Example - NaCl solution contains Na + and Cl - ions - Each ion is surrounded by water molecules - Good conductor of electricity SOLUBILITY OF IONIC COMPOUNDS

41 Solvation Process (Hydration) - Ions in aqueous solution are surrounded by the H 2 O molecules -The O atom in each H 2 O molecule has partial negative charge and attract cations - The H atoms have partial positive charge and attract anions - Cations and anions are prevented from recombining - About 4 to 10 water molecules surround each cation SOLUBILITY OF IONIC COMPOUNDS

42 - There is an increase in disorder of the solute as it separates into ions - There is an increase or decrease in disorder of the solvent depending on the solute - Solubilities are difficult to predict due to these several contributing factors SOLUBILITY OF IONIC COMPOUNDS

43 - Most molecular compounds do not form ions in solution - The molecules disperse throughout the solution Example - Sucrose in water solution contains neutral sucrose molecules - Each molecule is surrounded by water molecules - Poor conductor of electricity - A few molecular compounds form ions in aqueous solution - HCl dissociates into H + (aq) and Cl - (aq) - HNO 3 dissociates into H + (aq) and NO 3 - (aq) SOLUBILITY OF MOLECULAR COMPOUNDS

44 Consider mixing hydrocarbons such as C 6 H 14 and C 7 H 16 - London dispersion forces dominate within the molecules - Attraction between C 6 H 14 and C 7 H 16 molecules are also due to London dispersion forces - These two substances mix because the attractions are close in energy - Increase in disorder is the controlling factor SOLUBILITY OF MOLECULAR COMPOUNDS

45 Consider mixing water and a hydrocarbon - Strong hydrogen bonding dominates intermolecular attractions between water molecules - London dispersion forces dominate intermolecular attractions between hydrocarbon molecules - Attraction between water and hydrocarbon molecules are due to weak London dispersion forces - Increase in disorder is not sufficient to overcome the unfavorable enthalpy change hence very low solubility results SOLUBILITY OF MOLECULAR COMPOUNDS

46 EFFECT OF PRESSURE ON SOLUBILITY - Solubilities of gases in liquids are sensitive to pressure changes - Increase in pressure increases solubility of gases - An increase in pressure of a saturated solution results in dissolving more gas molecules - Solubilities of liquids and solids change very little with pressure due to very little change in volume

47 Henry’s Law - The solubility of a gas is directly proportional to its partial pressure at any given temperature C = kP C = concentration of the gaseous solute k = proportionality constant (units depend on units of C) P = partial pressure of gaseous solute EFFECT OF PRESSURE ON SOLUBILITY

48 EFFECT OF TEMPERATURE ON SOLUBILITY - Effect of temperature depends on the sign of the enthalpy change - Solubility increases with increasing temperature when the enthalpy change is positive (+∆H, endothermic process) - Solubility decreases with increasing temperature when the enthalpy change is negative (-∆H, exothermic process) Generally - The more positive the ∆H the greater the change in solubility with temperature

49 - Solubility of most ionic solids increase with increase in temperature - Solubility of most gases decrease with increase in temperature - Enthalpy of solution of most gases in water is negative - There is little or no attraction between gas molecules but there are attractions between solvent and solute molecules - Hence the negative enthalpy change EFFECT OF TEMPERATURE ON SOLUBILITY

50 COLLIGATIVE PROPERTIES - The physical properties of a solution differ from those of the pure solvent - The physical properties of a solvent change when a solute is added to a solvent - Four physical properties change based on the amount of solute added but not the solute’s chemical identity

51 - These are known as the Colligative Properties - Vapor-pressure lowering - Boiling-point elevation - Freezing-point depression - Osmotic pressure COLLIGATIVE PROPERTIES

52 Vapor-Pressure Depression - When a nonvolatile solute (low tendency to vaporize) is added to a solvent, the vapor pressure of the resulting solution is lowered below that of the pure solvent at the same temperature COLLIGATIVE PROPERTIES

53 Vapor-Pressure Depression Raoult’s Law - The partial pressure of a substance in equilibrium with a solution is equal to the product of its mole fraction in the solution and the vapor pressure of the pure substance P solv = partial pressure exerted by solvent’s vapor above a solution X solv = mole fraction of the solvent in the solution P o solv = vapor pressure of the pure solvent COLLIGATIVE PROPERTIES

54 Vapor-Pressure Depression COLLIGATIVE PROPERTIES

55 Vapor-Pressure Depression - The vapor pressure depression (∆P) is proportional to the mole fraction (concentration) of solute - Raoult’s law only applies to dilute solutions COLLIGATIVE PROPERTIES

56 Vapor-Pressure Depression At 25 o C, the vapor pressure of pure benzene is 93.9 torr. What is the solute concentration in a benzene solution that has a vapor pressure of 92.1 torr? ∆P = 93.9 torr – 92.1 torr = 1.8 torr

57 Boiling-Point Elevation - When a nonvolatile solute (low tendency to vaporize) is added to a solvent, the boiling point of the resulting solution is raised above that of the pure solvent - Since vapor pressure is lowered, a higher temperature is needed to raise the depressed vapor pressure to atmospheric pressure ( a condition required for boiling) COLLIGATIVE PROPERTIES

58 Boiling-Point Elevation ΔT b = mk b ΔT b = increase in boiling point k b = molal boiling-point elevation constant (depends only on solvent) m = molality (molal concentration) - Applies to dilute solutions only COLLIGATIVE PROPERTIES

59 Boiling-Point Elevation What is the boiling point of a 0.21-molal aqueous solution of sodium chloride at 1 atm (k b of water = o C/m)? ΔT b = mk b = (0.21)(0.512 o C/m) = 0.11 o C Boiling point of pure water = 100 o C T b = 100 o C o C = o C COLLIGATIVE PROPERTIES

60 Freezing-Point Depression - When a nonvolatile solute (low tendency to vaporize) is added to a solvent, the freezing point of the resulting solution is lowered below that of the pure solvent - The triple-point temperature of a solution decreases with increasing concentration of solute (due to decrease in vapor pressure) - The solid-liquid equilibrium line moves to lower temperatures COLLIGATIVE PROPERTIES

61 Freezing-Point Depression ΔT f = mk f ΔT f = decrease in freezing point k f = molal freezing-point-depression constant (depends only on solvent) m = molality (molal concentration) - Applies to dilute solutions only COLLIGATIVE PROPERTIES

62 Freezing-Point Depression What is the freezing point of a 0.21-molal aqueous solution of sodium chloride at 1 atm (k f of water = 1.86 o C/m)? ΔT f = mk f = (0.21)(1.86 o C/m) = 0.39 o C Freezing point of pure water = 0 o C T f = 0 o C – 0.39 o C = – 0.39 o C

63 Osmosis - The passage of a solvent through a semipermeable membrane that separates a solution of lower solute concentration from a solution of higher solute concentration - Flow of solvent is in both directions so it is actually a net flow Semipermeable Membrane - Allows certain types of molecules to pass through but prohibits other types of molecules (usually based on size) COLLIGATIVE PROPERTIES

64 Osmotic Pressure - Pressure required to prevent osmosis by pure solvent - Pressure difference needed for no net transfer of solvent - Very sensitive and useful for measuring molar mass of large molecules - Aqueous solutions with higher osmotic pressure take up more water than aqueous solutions with lower osmotic pressure COLLIGATIVE PROPERTIES

65 Osmotic Pressure πV = nRT (similar to the ideal gas law) π = osmotic pressure V = volume of solution n = number of moles of solute R = ideal-gas constant T = absolute temperature M = molarity of solution COLLIGATIVE PROPERTIES

66 Electrolyte Solutions Electrolytes dissociate into ions in solution NaCl(aq) → Na + (aq) + Cl - (aq) 1 mole of NaCl in solution produces 2 moles of ions AlCl 3 (aq) → Al 3+ (aq) + 3Cl - (aq) 1 mole of AlCl 3 in solution produces 4 moles of ions For example - The osmotic pressure of NaCl is twice that of glucose - Glucose does not dissociate in solution COLLIGATIVE PROPERTIES

67 Electrolyte Solutions van‘t Hoff Factor (i) i = number of particles produced from the dissociation of one formula unit of solute (for dilute solutions) - The number of particles present determines the measured colligative property COLLIGATIVE PROPERTIES

68 Electrolyte Solutions Taking van‘t Hoff Factor (i) into account ΔT b = imk b ΔT f = imk f COLLIGATIVE PROPERTIES

69 Nonideal Solutions - The value of i tends to be smaller than expected at greater solution concentrations (> 0.01 m) - Some ions cluster in solution and behave as a single unit as a result of strong electrostatic attractions COLLIGATIVE PROPERTIES

70 Consider a solution of two volatile substances A and B - The vapor phase in equilibrium with the solution contains all the volatile components - According to Raoult’s law MIXTURES OF VOLATILE SUBSTANCES and

71 MIXTURES OF VOLATILE SUBSTANCES - Total vapor pressure is the sum of the partial pressures of the components P T = P A + P B - The vapor in equilibrium with the mixture is richer in the more volatile component - This is applied in fractional distillation to separate volatile substances

72 Ideal Solution - Obeys Raoult’s law throughout the entire range of composition Considering the mixture of substances A and B - Solution is ideal when A-B attractions are closer to A-A and B-B attractions - Nearly true for similar liquids Benzene and toleune Hexane and heptane - Strictly for very dilute solutions MIXTURES OF VOLATILE SUBSTANCES

73 Positive Deviation from Raoult’s Law - Occurs when the A-B attractions are weaker than the average A-A and B-B attractions - The observed vapor pressure is greater than expected MIXTURES OF VOLATILE SUBSTANCES Mole fraction Pressure - Straight lines (dotted) in ideal situation become curves

74 Negative Deviation from Raoult’s Law - Occurs when the A-B attractions are stronger than the average A-A and B-B attractions - The observed vapor pressure is less than expected - Straight lines (dotted) in ideal situation become curves MIXTURES OF VOLATILE SUBSTANCES Mole fraction Pressure


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