Read Microsoft Word - Properties of Solutions notes text version

Properties of Solutions

Read chapter 12, p. 521-544

Driving Forces to the formation of solutions: 1. 2.

I.

Solution composition.

a) Homogeneous mixtures that can be made of gases, liquids and solids. We will focus on liquid solutions, made primarily of aqueous solutions. b)

molarity =

moles volume

(mass solute ) x100% (mass solution ) nA d) molefraction A = X A = n A + nB moles solute e) molality = ki log rams solvent

c)

mass % =

f)

Normality (N)= number of equivalents, 1 M hydrochloric acid has an equivalent of 1N, and 1 M sulphuric acid has an equivalent of 2N.

g) Solubility: amount of a substance that dissolves in a given volume of solvent at a given temperature. Example 1: A solution is made by mixing 1.00 g of ethanol with 100.0 g of water to give a final volume of 101mL. Calculate the molarity, mass percent, mole fraction, and molality of ethanol in this solution.

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Example 2: The electrolyte in automobile lead storage batteries is 3.75 M sulfuric acid solution that has a density of 1.230 g/mL. Calculate the mass percent, molality, and normality of the sulfuric acid.

II.

·

Energy considerations of forming solutions:

Enthalpy exchange between the system and its surroundings when 1 mol of solute dissolves in a solvent at constant pressure:

III.

Factors Affecting Solubility.

Like Dissolves Like!

a) Structure effects.

Solvation: surrounding a solute by a solvent. Hydration: surrounding a solute by water (special case of solvation).

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That is, polar solvents dissolve polar or ionic solutes, while non-polar solvents dissolve non-polar solutes. Example: Ethanol and water, ethanol and iodine, iodine and water:

b) Pressure Effects. · · · · Important for gas-liquid system. For a closed system, there is an equilibrium established. Gas + solvent solution Increase in gas solubility with an increase in pressure. Henry's Law: Amount of gas dissolved is directly proportional to the pressure of the gas above the solution.

C = kP T=constant

C= concentraion of dissolved gas. k= constant characteristic to a particular solution. P= partial pressure of the gas solute.

c) Temperature Effects. · The rate of dissolving of an ionic solute in water occurs faster at higher temperature, but the amount of solute that dissolves may not necessarily be greater. The solubility of sodium sulphate decreases as temperature increases. For gases, there is a decrease in solubility as temperature increases.

·

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IV.

Vapour Pressure of solutions. Colligative Properties.

a. Vapor pressure lowering · At any given temperature a liquid in a closed container will establish equilibrium with its vapor.

· ·

At this point the pressure created above the liquid is called its vapor pressure. Changes related to lowering of vapor pressure are governed by Raoults law, and fall into two categories. Those where the solute is non-volatile and those where the solution has two volatile components. A solution that contains a non-volatile solute

·

If the solute in a solution is non-volatile then the vapor pressure of the solution will be lower than that of the pure solvent. The vapor pressure falls since the solution has greater entropy (disorder) than the pure solvent and that is the preferred situation. Therefore in order to maintain that disordered state the solvent molecules tend not to leave the solution and the vapor pressure is lowered. Raoult's Law states

o Psolution = X solvent Psolvent

P = Vapor pressure X = Mole fraction of the solvent o P = Vapor pressure of the pure solvent

Vapour pressure of pure solvent Psoln

0

Mole fraction,Xsolvent

1

·

Solutions that obey Raoult's law, a plot between Psoln versus Xsolvent gives a straight line. These solutions are referred to as Ideal Solutions.

4

·

The phenomenon of lowering of the vapour pressure depends upon the number of solute particles in the solution. For example, 1 mole of NaCl dissolved in water lowers the vapour pressure approximately twice as much because the solid forms two ions per formula unit. The lowering of vapour pressure provides a means to experimentally determine molar masses.

·

Example: Calculate the expected vapour pressure at 25oC for a solution prepared by dissolving 158.0g of common table sugar (molar mass=342.3 g/mol) in 643.5 cm3 of water. At 25oC, the density of water is 0.9971 g/cm3 and the vapour pressure is 23.76 torr. (0.30 torr)

Example: Calculate the vapour pressure of a mixture of equal weights of water and ethyl alcohol at 350 C when the vapour pressure of the two substances are 42.0 and 100.0 mm Hg respectively.

Example: Predict the vapour pressure of a solution prepared by mixing 35.0 g of solid Na2SO4 with 175 g of water at 25oC. The vapour pressure of pure water at 25oC is 23.76 torr. (22.1 torr)

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b. A solution with two volatile components

As before, at a fixed temperature the vapour pressure of each component in the mixture is proportional to the mole fraction of that component. i.e.

o PA = X A PA

In addition, Dalton's Law of partial pressures states that the vapor pressure of the mixture is the sum of the individual vapor pressures. Ptotal= PA + PB + PC... For a mixture of two miscible liquids that obey Raoults Law ideally, we can plot two graphs. (i) Vapor pressure against mole fraction and,

Vapour pressure of ethanol and water at 350C.

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c. Deviations from Raoults law

· Liquid mixtures that obey Raoults Law are quite rare. When they do exist they are made up from two components that are chemically very similar such as heptane and hexane or benzene and methylbenzene. This is because in an ideal mixture the intermolecular attraction between different molecules is the same as the attraction between identical molecules. This means that when the two liquids are mixed together there is no change in the intermolecular bonding present in either of the two pure liquids. Sometimes when two liquids are mixed together, heat is released and the total volume is less than that of the individual components. This suggests that the two liquids attract one another quite strongly. E.g. propanone and trichloromethane.

·

The dipoles attraction is significant. Since the molecules attract one another strongly they cannot escape so readily, so the boiling point is increased and the vapour pressure is reduced. There is a slight reduction in volume when the liquids are mixed and since new bonds are formed an exothermic enthalpy change is observed. This is known as negative deviation and leads to a vapour pressure composition diagram shown below. Very large negative deviations lead to a minimum vapour pressure.

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·

Similarly sometimes when two liquids are mixed together, heat is absorbed and the total volume is greater than that of the individual components. This suggests that there has been a disruption of the intermolecular bonds present in one component when the liquids are mixed.

E.g. water and ethanol. Ethanol interferes with the strong hydrogen bonding in the water. · Since the intermolecular bonding has been decreased each of the components will vaporize more readily, (lower boiling point), and the vapor pressure will be increased. There is a slight increase in volume when the liquids are mixed and since bonds are broken an endothermic enthalpy change is observed. This is known as positive deviation and leads to a vapor pressure composition diagram shown below. Very large positive deviations lead to a maximum vapor pressure.

Task 17b

1. At 20oC the vapor pressures of methanol and ethanol are 95.0 and 45.0 mmHg respectively. An ideal solution contains 16.1g of methanol and 92.1g of ethanol. Calculate

i) the partial pressures of methanol and ethanol in the mixture ii) the total vapor pressure of the mixture iii) the composition of the vapor

2. In terms of intermolecular forces, explain the existence of positive and negative deviations from ideal behavior. Give one piece of experimental evidence in support of your explanation. 3. At 398K the vapor pressures of benzene and methylbenzene are 1520 and 571 mmHg respectively. A mixture of 0.20 mols of benzene and 0.80 moles of methylbenzene boils at 398K under atmospheric pressure.

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Calculate the partial pressures of the two components in equilibrium in the vapor phase with the liquid under these conditions. Hence calculate the mole fractions of each component in the vapor phase. Explain the difference in this composition and the original composition.

9

4. Boiling point elevation The boiling point of a solution is reached when the external atmospheric pressure is equal to its vapor pressure. Since a non-volatile solute lowers the vapor pressure of a solution it must, by definition, affect the boiling point. The change is summarized in the expression

Tb = K bm

Tb Kb m = Change in boiling point of the solvent = molal boiling point constant = molality of the solute = moles of solute / mass of solvent in kg

example: 1. 0.492 g of 1,3-dichlorobenzene is added to 10mL of cyclohexane (C6H12) which has a Kb = 2.79oC/m and a boiling point of 80.7oC. Given that the density of cyclohexane is 0.7785 g/mL and that the new boiling point is 81.9oC, what is the RMM of 1,3dichlorobenzene?

6. Freezing point Depression: The change caused by the addition of a non-volatile solute to the freezing point of a solvent is summarized by the expression

Tf = K f m

Tf Kf m = Change in freezing point of the solvent = molal freezing point constant of the solvent = molality of the solute = moles of solute / mass of solvent in kg

Examples: 1. Calculate the freezing point of a solution containing 650g of ethane-1,2-diol in 2500g of water. Kf for water = 1.86oC/m. 2. Calculate the RMM of compound X, given that a solution containing 5.02g of X in 89g of water has a freezing point 0.341oC below the normal freezing point of water. Kf for water = 1.86oC/m. 3. 7.82g of compound Y in 299g of benzene causes its freezing point to drop by 1.05oC. What is the RMM of Y given Kf for benzene = 5.12oC/m?

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7. Osmotic pressure Osmosis is the selective passage of solvent molecules through a porous semi-permeable membrane (s.p.m) from a dilute solution to one of a higher concentration. The s.p.m allows the passage of solvent but not solute. The osmotic pressure () is the pressure that must be applied in order to prevent osmosis occurring. It can be calculated using the expression

= MRT

M R T = Molarity = universal gas constant (0.0821 L atm mol K-1) = temperature in Kelvin

8. Colligative properties of Electrolyte Solutions:

·

Electrolyte solutions and the van't Hoff factor

Electrolytes require a slightly different approach than non-electrolyte solutions since they dissociate in solution and provide a greater number of solute particles than the undissolved solute. For example, one unit of MgCl2 splits into three particles when it is dissolved in solution (one Mg2+ ion and two Cl- ions) and should produce a freezing point depression three times greater than a non-electrolyte solute that does not produce a greater number of particles when dissolved like, say, molecular sucrose, C6H12O6. In order to account for this the van't Hoff factor (i) is added to the expressions already discussed.

Tb = i K bm Tf = i K f m

= i MRT

In each case the van't Hoff factor is given as

i=

actual number of particles in solution after dissociati on Number of formula units initially dissolved in solution

For non-electrolytes i = 1. Electrolytes usually have a van't Hoff factor less than the predicted one. For example, MgCl2 has i = 2.7 rather than the predicted 3, due to the pairing of some of the ions in solution. The pairing of ions is most prevalent when the ions involved have higher charges and the solution is relatively concentrated.

Task 17e

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1. At 25oC the osmotic pressure of a 0.01M solution of a compound is 0.466atm. Calculate the van't Hoff factor.

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Solutions and Colligative Properties Worksheet 1 1. As the temperature of water increases, the rate at which most solids dissolve in water: (A) Increases (B) Decreases (C) Remains the same (D) May either increase or decrease 2. When a nonvolatile solute is dissolved in water to form a solution, the vapor pressure of the solution (compared to pure water): (A) Increases (B) Decreases (C) Remains the same (D) May either increase or decrease 3. Which of the following is most soluble in water? (A) SiO2 (B) CH3-O-CH3 (C) CaCO3 (D) NaCH3CO2 (E) CO2 4. Which of the following will have the lowest freezing point? (A) 0.15 M glucose (B) 0.30 M sucrose (C) 0.15 M NaCl (D) 0.30 M CH3COOH 5. Which of the following will have the highest boiling point? (A) 0.25 M NaCl (B) 0.50 M glucose (C) 0.25 M NaCH3CO2 (D) 0.50 M MgBr2 (E) 0.10 M CaCl2 Short Answer 6. Explain why the concentration of solutions used for intravenous feeding must be controlled carefully. 7. Explain why, when making fudge (a supersaturated mixture), care must be taken to prevent it from getting "grainy." 8. Explain why fish in a lake seek deep, shaded places during summer afternoons.

9. Explain what causes the "bends" in divers. 10. Explain why champagne fizzes when poured into a glass.

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Solutions and Colligative Properties Worksheet 2 o 1. A solution of 5.0 g of X in 60.0 g of water freezes at -1.0 C. What is the molar mass of X? o (kf for water = 1.86 C/m)

2. A student dissolves 1.5 g of a compound in 75.0 g of cyclohexane. She then measures the freezing point of the solution to be 2.7° The freezing point of pure cyclohexane is C. o o 6.5 C (kf = 20.2 C/m). Find the molar mass of the compound.

3. A solution contains 102 g of sugar, C12H22O11 in 375 g of H2O. Calculate: (A) The mole fraction of sugar.

(B) The vapor pressure of the solution at 25 C (P of pure water = 23.75 mm Hg at 25 C)

o

o

4. A nonconducting solution contains 4.36 grams of an unknown compound dissolved in o 100 g of water. The freezing point of the solution is found to be -0.95 C. Calculate the molar mass of the compound. o (kf for water = 1.86 C/m)

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Solutions and Colligative Properties Answer Key WORKSHEET 1 1) A 2) B 3) D 4) D 5) D 6) If a concentration that is too high is used, water from body tissues would pass into the veins to dilute the feeding solution, dehydrating the body and causing the blood pressure to rise dramatically. 7) Supersaturated solutions are unstable; under proper conditions, the solute (sugar) will crystallize from the solution to give a saturated solution. Thus care must be taken to prevent the supersaturated fudge mixture from crystallizing out the sugar, making grainy fudge (a saturated mixture). 8) Solubility of oxygen in water decreases with increasing water temperature. Thus fish seek cool water so that they can breath more oxygenated water. 9) The solubility of N2 in blood (water) increases with increasing pressure. When a diver is in deep water, they are under tremendous hydrostatic pressure. In order to not be crushed, the insides of the diver must be at the same pressure as the outside. Thus divers breath pressurized gas. If a diver rises to the surface too quickly, the N2 that dissolved in the blood stream at high pressures will not have a chance to escape; bubbles of N2 form in the veins. 10) The fizzing is due to the escaping CO2. Inside the sealed bottle, the solubility of CO2 in champagne was increased by the very large over-pressure of CO2. At atmospheric pressure, CO2 has a much lower solubility in champagne. It begins to leave the champagne (as bubbles) as soon as the pressure over the liquid decreases. WORKSHEET 2 1) 155 gm/mol 2) 106 gm/mol 3a) .0141, b) 23.42 mmHg 4) 85.4 gm/mol

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AP Chemistry Old Exam Questions

1974 D Two beakers, one containing 100 milliliters of a 0.10 molal solution of sucrose (a nonvolatile nonelectrolyte) the other containing 100 milliliters of pure water, are placed side by side in a closed system, such as under a bell jar. Explain in terms of the principles involved what changes, if any, occur to bring the system to equilibrium. Answer: (1) Volume of sugar solution increases; (2) volume of pure water decreases; (3) water beaker finally empty. Raoult's Law, vapor pressure and volatility Description of process (rates of vaporization and condensation) 1975 D Alcohol dissolves in water to give a solution that boils at a lower temperature than pure water. Salt dissolves in water to give a solution that boils at a higher temperature than pure water. Explain these facts from the standpoint of vapor pressure. Answer: An alcohol-water solution has a higher than normal (pure water) vapor pressure because alcohol is a volatile solute and contributes substantially to the vapor of the solution. The higher the vapor pressure, the lower the boiling point. A salt-water solution has a lower than normal vapor because salt is a non-volatile solute and solute-solvent interaction decrease the vapor of the solution, the lower the vapor pressure, the higher the boiling point. 1976 B (a) Calculate the molality of a 20.0 percent by weight aqueous solution of NH4Cl. (Molecular weight: NH4Cl = 53.5) (b) (c) If this NH4Cl solution is assumed to be ideal and is completely dissociated into ions, calculate the pressure of this solution at 29.0°C.

Actually a solution of NH4Cl of this concentration is not ideal. Calculate the apparent degree of dissociation of the NH4Cl if the freezing point of this solution is -15.3°C? (Molal freezing point constant = 1.86°C) Answer:

20 .0

(a)

80 .0

1000 = 250 g NH 4 Cl in 1000g H 2 O

250 g 1 kg H 2O

(b) P1 = (P°)(X1) mol ions = (2)(4.67 mol) = 9.34 mol 1 kg water = 55.6 mol water

1 mol N H 4 Cl 53.5 g

= 4 .67 mol 1 kg = 4 .67 molal

X1 =

55 .6 55 .6 + 9 .34

= 0 .856

P1= (29.8 mm Hg)(0.856) = 25.5 mm Hg (c) Assume no dissociation.

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T = kfm = (1.86)(4.67) = 8.69°C i = 15.3 / 8.69 = 1.76 (1.76 - 1.00)(100) = 76% dissociated

1980 B (a) A solution containing 3.23 grams of an unknown compound dissolved in 100.0 grams of water freezes at -0.97°C. The solution does not conduct electricity. Calculate the molecular weight of the compound. (The molal freezing point depression constant for water is 1.86°C kg mole-1) (b) (c) Elemental analysis of this unknown compound yields the following percentages by weight H=9.74%; C=38.70%; O=51.56%. Determine the molecular formula for the compound.

Complete combustion of a 1.05 gram sample of the compound with the stoichiometric amount of oxygen gas produces a mixture of H2O(g) and CO2(g). What is the pressure of this gas mixture when it is contained in a 3.00 liter flask at 127°C? Answer: (a) Tf = kfm; 0.97°C = (1.86°Cm-1)(m) m = 0.52 mol solute/kg solvent In this solution, 3.23 g solute in 100.0 g water or 32.3 g solute in 1 kg of water

mol .wt. = 62g

(b)

32.3 g 1 kg s olvent

1 kg s olvent 0.52 mol 1 mol H

= 62 =

g mol

0.0974 g H 1g cmpd

6.0 mol H 1 mol cmpd 0.3870 g C

1 mol cmpd

1g H 62g

1 mol C

2.0 mol C

= =

1 mol cmpd 62g 1 mol cmpd

(c) = C2H6O2 2 C2H6O2 + 5 O2 6 H2O(g) + 4 CO2(g)

1g cmpd 0.5156 g O 1g cmpd

12 g C 1 mol O 16 g O

1 mol cmpd 2.0 mol O 1 mol cmpd

1 .0 5 g C 2 H 6 O 2

= 0.0847 mol gas P = (nRT) / V

1 mol C 2 H 6 O 2 10 mol g as = 62 g C 2 H 6 O 2 2 mol C 2 H 6 O 2

P=

L _atm ( 0 .0847 mol )( 0 .08205 mol K )( 400 .K )

_

3.00L

= 0.926 atm.

1984 C Give a scientific explanation for the following observations. Use equations or diagrams if they are relevant.

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(a) (b) (c) (d)

It takes longer to cook an egg until it is hard-boiled in Denver (altitude 1 mile above sea level) than it does in New York City (near sea level). Burn coal containing a significant amount of sulfur leads to ôacid rain.ö Perspiring is a mechanism for cooling the body.

The addition of antifreeze to water in a radiator decreases the likelihood that the liquid in the radiator will either freeze or boil. Answer: (a) Water boils at a lower temperature in Denver than in NYC because the atmospheric pressure is less at high altitudes. At a lower temperature, the cooking process is slower, so the time to prepare a hard-boiled egg is longer. (b) S + O2 SO2 (as coal is burned) SO2 + H2O H2SO3 (in the atmosphere) H2SO3 is sulfurous acid. (c) (d) Vaporization or evaporation of sweat from the skin is an endothermic process and takes heat from the body and so cool the skin. Colligative properties, which depend on the number of particles present, are involved. Solute (the antifreeze) causes the lowering of the vapor pressure of the solvent. When the vapor pressure of the solvent is lowered, the freezing point is lowered and the boiling point is raised.

1985 B The formula and the molecular weight of an unknown hydrocarbon compound are to be determined by elemental analysis and the freezing-point depression method. (a) The hydrocarbon is found to contain 93.46 percent carbon and 6.54 percent hydrogen. Calculate the empirical formula of the unknown hydrocarbon. (b) A solution is prepared by dissolving 2.53 grams of p-dichlorobenzene (molecular weight 147.0) in 25.86 grams of naphthalene (molecular weight 128.2). Calculate the molality of the p-dichlorobenzene solution. The freezing point of pure naphthalene is determined to be 80.2°C. The solution prepared in (b) is found to have an initial freezing point of 75.7°C. Calculate the molal freezing-point depression constant of naphthalene. A solution of 2.42 grams of the unknown hydrocarbon dissolved in 26.7 grams of naphthalene is found to freeze initially at 76.2°C. Calculate the apparent molecular weight of the unknown hydrocarbon on the basis of the freezing-point depression experiment above.

(c)

(d)

(e) What is the molecular formula of the unknown hydrocarbon? Answer: (a) Assume 100. g sample of the hydrocarbon

9 3.46 g C 6 .54 g H

1 mol C 12 .01 g C

= 7 . 782 mol C

1 mol H 1 .008 g H

= 6 . 49 mol H

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7.782 mol C 6.49 mol H

(b)

=

1 .20 1 .00

; C 1 .20 H1 .00 = C 6 H 5

m=

(c) (d) Tf = (80.2 - 75.7)°C = 4.5°C

mol s olute 1 .0 kg solvent

=

2 .53 g

1 mol 147.0 g

0 .0 2586 kg

= 0 . 666 mol al

kf = Tf /m = 4.5°C / 0.666 molal = 6.8°C/molal Tf = (80.2 - 76.2)°C = 4.0°C 1 6 .8 _ _k g solvent C

2 .43 g solute 0 .0267 kg solvent

=

4 .0_ C

1 mol s olute

=154 g/mol (e) C6H5 = 77 # empirical units/mol = 154/77 = 2 molecular formula = (C6H5)2 = C12H10

1995 D

The phase diagram for a pure substance is shown above. Use this diagram and your knowledge about changes of phase to answer the following questions. (a) What does point V represent? What characteristics are specific to the system only at point V?. (b) (c) What does each point on the curve between V and W represent? Describe the changes that the system undergoes as the temperature slowly increases from X to Y to Z at 1.0 atmosphere.

(d) In a solid-liquid mixture of this substance, will the solid float or sink? Explain. Answer: (a) Triple point. All three states of the substance coexist (equilibrium); the solid and the liquid have identical vapor pressures.

19

(b)

Curve VW represents the equilibrium between the liquid and its vapor. Along this line the liquid will be boiling. The points represent the vapor pressure of the liquid as a function of temperature. At point X the substance is a solid, as its temperature increases (at constant pressure), at point Y the solid is in equilibrium with its vapor and will sublime. From point Y to Z it exist only as a vapor. Sink. A positive slope of the solid-liquid line indicates that the solid is denser than its liquid and, therefore, will sink.

(c)

(d)

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