Chapter Twelve General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 1 Chapter Twelve Physical Properties of.

Chapter TwelveGeneral Chemistry 4th edition, Hill, Petrucci, McCreary, PerryPrentice Hall © 2005 Hall © 2005

1

Chapter TwelveChapter Twelve

Physical Properties of Solutions


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Solution: Solute dispersed in a solvent.

Some Types of Solutions


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• Molarity: moles of solute/liter of solution

• Molality: moles of solute/kg of solvent

• Percent by mass: grams of solute/grams of solution (then multiplied by 100%)

• Percent by volume: milliliters of solute/milliliters of solution (then multiplied by 100%)

• Mass/volume percent: grams of solute/milliliters of solution (then multiplied by 100%)

Most concentration units are expressed as:

Solution Concentration

Amount of solvent or solution

Amount of solute


4

Example 12.1 How would you prepare 750 g of an aqueous solution that

is 2.5% NaOH by mass?

Example 12.2 At 20°C, pure ethanol has a density of 0.789 g/mL and

USP ethanol has a density of 0.813 g/mL. What is the mass percent ethanol in USP ethanol?


5Example 12.1

How would you prepare 750 g of an aqueous solution that is 2.5% NaOH by mass?

• ? g NaOH = 750 g solution x 2.5% = 19 g NaOH

• ? g H2O = 750 g solution – 19 g NaOH = 731 g H2O

Example 12.2

At 20 °C, pure ethanol has a density of 0.789 g/mL and USP ethanol (= 95% ethanol by volume) has a density of 0.813 g/mL. What is the mass percent ethanol in USP ethanol?

• Assume you a working with 100 ml sample of the USP ethanol, then:

• 95.0 ml ethanol x 78.9 g ethanol = 75.0 g ethanol

100 ml solution

• Mass percent ethanol = 75.0 g ethanol x 100 % = 92.3 %

81.3 g solution


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• Parts per million (ppm): grams of solute/grams of solution (then multiplied by 106 or 1 million)

• Parts per billion (ppb): grams of solute/grams of solution (then multiplied by 109 or 1 billion)

• Parts per trillion (ppt): grams of solute/grams of solution (then multiplied by 1012 or 1 trillion)

Amount of solution

Amount of solute

• ppm, ppb, ppt ordinarily are used when expressing extremely low concentrations (a liter of water that is 1 ppm fluoride contains only 0.001 g F–!)

Solution Concentration (cont’d)

Most concentration units are expressed as:


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PPM and PPB are Tiny Amounts

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Example 12.3 The maximum allowable level of nitrates in drinking water in

the United States is 45 mg NO3–/L. What is this level expressed

in parts per million (ppm)?


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Example 12.3The maximum allowable level of nitrates in drinking water in the United States is 45 mg NO3

–/L. What is this level expressed in parts per million (ppm)?

The density of water with only small amounts of dissolved substances is essentially 1.0 g/ml. So one liter of dilute aqueous nitrate solution has a mass of 1000 g.

45 mg NO3– 1 g water 45 mg NO3

–

1000 g water 1000 mg 1,000,000 mg waterX = 45 ppm NO3

–=


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Molality (m): moles of solute/kilograms of solvent.

• Molarity varies with temperature (expansion or contraction of solution).

• Molality is based on mass of solvent (not solution!) and is independent of temperature.

• We use molality in describing certain properties of solutions when temperature changes are occurring.


Most concentration units are expressed as: Amount of solvent or solution

Amount of solute


11

Example 12.4 What is the molality of a solution prepared by

dissolving 5.05 g naphthalene [C10H8(s)] in 75.0 mL of benzene, C6H6 (d = 0.879 g/mL)?

Example 12.5 How many grams of benzoic acid, C6H5COOH, must

be dissolved in 50.0 mL of benzene, C6H6 (d = 0.879 g/mL), to produce 0.150 m C6H5COOH?


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• Mole fraction (xi): moles of component i per moles of all components (the solution).

• The sum of the mole fractions of all components of a solution is ____.

• Mole percent: mole fraction times 100%.

Concentration expressed as:


Amount of solvent or solution

Amount of solute


13Example 12.6 An aqueous solution of ethylene glycol HOCH2CH2OH used as an

automobile engine coolant is 40.0% HOCH2CH2OH by mass and has a density of 1.05 g/mL. What are the (a) molarity, (b) molality, and (c) mole fraction of HOCH2CH2OH in this solution?

Example 12.7 An Estimation Example Without doing detailed calculations, determine which aqueous solution

has the greatest mole fraction of CH3OH: (a) 1.0 m CH3OH, (b)10.0% CH3OH by mass, or (c) xCH3OH = 0.10.


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Solution formation can be considered to take place in 3 steps:1. Move solvent molecules apart (breaking solvent-solvent

bonds) to make room for the solute molecules. H1 > 0 (endothermic)

2. Separate the molecules/ions of solute (breaking solute-solute bonds) . H2 > 0 (endothermic)

3. Allow the separated solute and solvent molecules to mix randomly (form solute-solvent bonds). H3 < 0 (exothermic)

Hsoln = H1 + H2 + H3

Enthalpy of Solution


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For dissolving to occur, the magnitudes of H1 + H2 and of H3 must be roughly

comparable.

Visualizing Enthalpy of Solution


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• An ideal solution exists when all intermolecular forces are of comparable strength, Hsoln = 0.

• When solute–solvent intermolecular forces are somewhat stronger than other intermolecular forces, Hsoln < 0.

• When solute–solvent intermolecular forces are somewhat weaker than other intermolecular forces, Hsoln > 0.

• When solute–solvent intermolecular forces are much weaker than other intermolecular forces, the solute does not dissolve in the solvent.

– Energy released by solute–solvent interactions is insufficient to separate solute particles or solvent particles.

Intermolecular Forcesin Solution Formation


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Intermolecular Forces in Solution

For a solute to dissolve, solvent–solvent forces …

… and solute–solute forces …

… must be comparable to solute–

solvent forces.


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Non-Ideal Solutions

50 mL of ethanol …

… and 50 mL of water …

… when mixed, give

less than 100 mL of solution.

In this solution, forces between ethanol and

water are > / < / = other intermolecular

forces.


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• The forces causing an ionic solid to dissolve in water are ion–dipole forces, the attraction of water dipoles for cations and anions.

• The attractions of water dipoles for ions pulls the ions out of the crystalline lattice and into aqueous solution.

• The extent to which an ionic solid dissolves in water is determined largely by the competition between:

– interionic attractions that hold ions in a crystal, and

– ion–dipole attractions that pull ions into solution.

Aqueous Solutions of Ionic Compounds


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Ion–Dipole Forces in Dissolution

Positive ends of dipoles attracted to anions.

Negative ends of dipoles attracted to cations.


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Example 12.8 Predict whether each combination is likely to be a

solution or a heterogeneous mixture:

(a) methanol, CH3OH, and water, HOH

(b) pentane, CH3(CH2)3CH3, and octane, CH3(CH2)6CH3

(c) sodium chloride, NaCl, and carbon tetrachloride, CCl4

(d) 1-decanol, CH3(CH2)8CH2OH, and water, HOH

Remember rule of thumb: “like dissolved like”


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• Liquids that mix in all proportions are called miscible.– Ones that don’t = immiscible

• When there is a dynamic equilibrium between an undissolved solute and a solution, the solution is saturated.

• The concentration of the solute in a saturated solution is the solubility of the solute.

• A solution which contains less solute than can be held at equilibrium is unsaturated.

Some Solubility Terms


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Formation of a Saturated Solution

Solid begins to dissolve.

As solid dissolves, some dissolved solute begins to crystallize.

Eventually, the rates of dissolving and of crystallization are equal; no

more solute appears to dissolve.

Longer standing does not change the amount

of dissolved solute.


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• Most ionic compounds have aqueous solubilities that increase significantly with increasing temperature

• A few have solubilities that change little with temperature

• A very few have solubilities that decrease with increasing temperature (see following slide)

• If solubility increases with temperature, a hot, saturated solution may be cooled carefully without precipitation of the excess solute. This creates a supersaturated solution.

• Supersaturated solutions ordinarily are unstable …

Solubility as Function of Temperature


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Some Solubility Curves

What is the (approx.) solubility of KNO3

per 100 g water at 90 °C? At 20 °C?


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A single “seed crystal” of

solute is added.

Solute immediately begins to crystallize …

… until all of the excess solute has

precipitated.

A Supersaturated Solution


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• Most gases become less soluble in liquids as the temperature increases. (Explanation**)

• At a constant temperature, the solubility (S) of a gas is directly proportional to the pressure of the gas (Pgas) in equilibrium with the solution.

S = k Pgas

The value of k depends on the particular gas and the solvent.

• The effect of pressure on the solubility of a gas is known as Henry’s law.

** http://antoine.frostburg.edu/chem/senese/101/solutions/faq/print-temperature-gas-solubility.shtml

The Solubilities of Gases

http://antoine.frostburg.edu/chem/senese/101/solutions/faq/print-temperature-gas-solubility.shtml


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Effect of Temperature on Solubility of Gases

Eco-ConnectionThermal pollution: as

river/lake water is warmed (when used by industry for

cooling), less oxygen dissolves, and fish no longer thrive.


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Pressure and Solubility of Gases(Henry’s Law)

… thus more frequent collisions of gas molecules with the surface …

… giving a higher concentration of dissolved gas.

Higher partial pressure means more molecules of gas per unit volume …

Example 12.9A 225-g sample of pure water is shaken with air under a pressure of 0.95 atm at 20°C. How many milligrams of Ar(g) will be present in the water when solubility equilibrium is reached? Use data from Figure 12.14 and the fact that the mole fraction of Ar in air is 0.00934.

According to the graph, argon has a solubility in water of 6 mg/atm•100 g H2O.

225 g H2O x 0089 atm x 6 mg/atm•100 g H2O

= 0.12 mg Ar


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• Colligative properties of a solution depend only on the concentration of solute particles, and not on the nature of the solute.

• Non-colligative properties include: color, odor, density, viscosity, toxicity, reactivity, etc.

• We will examine four colligative properties of solutions:

– Vapor pressure (of the solvent)

– Freezing point depression

– Boiling point elevation

– Osmotic pressure

Colligative Properties of Solutions


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Review: Vapor Pressure

• The pressure of the vapor phase of a substance above a liquid sample of the substance after a dynamic equilibrium has been established in a closed system.


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• The vapor pressure of solvent above a solution is less than the vapor pressure above the pure solvent.

• Raoult’s law: the vapor pressure of the solvent above a solution (Psolv) is the product of the vapor pressure of the pure solvent (P°solv) and the mole fraction of the solvent in the solution (xsolv):

Psolv = xsolv ·P°solv

• The vapor in equilibrium with an ideal solution of two volatile components has a higher mole fraction of the more volatile component than is found in the liquid.

Vapor Pressure of a Solution


34

Example 12.10

The vapor pressure of pure water at 20.0 °C is 17.5 mmHg. What is the vapor pressure at 20.0°C above a solution that has 0.250 mol sucrose (C12H22O11) and 75.0 g urea [CO(NH2)2] dissolved per kilogram of water?

Moles of urea = 75.0 g urea x 1 mol urea/60 g urea = 1.25 mol urea

Moles of solute = .250 mole sucrose + 1.25 mole urea = 1.50 mole solute

Moles solvent = 1000 g H20 x 1 mole H20 /18.0 g H20 = 55.6 moles

Mole fraction of solvent = 55.6 moles solvent/57.1 moles solution = 0.974

Psolv = xsolv ·P°solv

= 0.974 · 17.5 mmHg

Psolv = 17.0 mmHg


35Example 12.11 At 25°C, the vapor pressures of pure benzene (C6H6) and pure toluene (C7H8)

are 95.1 and 28.4 mmHg, respectively. A solution is prepared that has equal mole fractions of C7H8 and C6H6. Determine the vapor pressures of C7H8 and C6H6 and the total vapor pressure above this solution. Consider the solution to be ideal.

Since there are equal mole fractions, the vapor pressure of each substance will be ½ of the vapor pressure of the pure substance. The total vapor pressure will be the sum of the partial pressures of the two vapors (Dalton’s law of partial pressures).

Example 12.12 What is the composition, expressed as mole fractions, of the vapor in

equilibrium with the benzene–toluene solution of Example 12.11?

At constant temperature and number of moles, the pressures of gases are proportional to the molar amounts. So we can use the partial pressure values like moles to produce mole fractions of the two gases.

Fractional Distillation

The vapor here …

… is richer in the more volatile component than

the original liquid here …

… so the liquid that condenses here will also

be richer in the more volatile component.

FractionalDistillation of

Crude Oil

Fractional distillation on a very large scale

Example 12.13 A Conceptual Example Figure 12.16 (below) shows two different aqueous solutions

placed in the same enclosure. After a time, the solution level has risen in container A and dropped in container B. Explain how and why this happens.


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Vapor Pressure Lowering by a Nonvolatile Solute

Raoult’s Law: the vapor pressure from a solution (nonvolatile

solute) is lower than …

… the vapor pressure from the

pure solvent

Result: the boiling point of the solution

increases by Tb.


40

Tf = –Kf × m × i Tb = Kb × m × i

Freezing Point Depression and Boiling Point Elevation

Example 12.14

What is the freezing point of an aqueous sucrose solution that has 25.0 g C12H22O11 per 100.0 g H2O?

25.0 g x 1 mol sucrose/342 g = 0.0731 mol C12H22O11

` Tf = −1.86 ºC/m x m x 1 m = 0.0731 mol/0.100 kg = 0.731 mTf = − 1.86 ∙ 0.731 m ∙ 1Tf = −1.36ºC So, the freezing point is −1.36ºC

Example 12.15 Sorbitol is a sweet substance found in fruits and berries and sometimes used as

a sugar substitute. An aqueous solution containing 1.00 g sorbitol in 100.0 g water is found to have a freezing point of –0.102 °C. Elemental analysis indicates that sorbitol consists of 39.56% C, 7.75% H, and 52.70% O by mass. What are the (a) molar mass and (b) molecular formula of sorbitol?

• Use freezing point depression equation to calculate molality of solution• With m, moles of solute known. Molar mass = g/mol, and you’re given 1.00 g

of the solute, so 1.00 g sorbitol/moles from the calculation gives molar mass.

Tf = –Kf × m × i Tb = Kb × m × i


42

• A semipermeable membrane has microscopic pores, through which small solvent molecules can pass but larger solute molecules cannot.

• During osmosis, there is a net flow of solvent molecules through a semipermeable membrane, from a region of lower concentration to a region of higher concentration.

• The pressure required to stop osmosis is called the osmotic pressure (of the solution.

= (nRT/V) = (n/V)RT = M RT

Osmotic Pressure

This equation should look familiar …


43

Osmosis and Osmotic Pressure

Net flow of water from the outside (pure H2O) to the

solution.

The solution increases in

volume until … … the height of solution exerts the

osmotic pressure (π) of the solution.


44

Example 12.16

An aqueous solution is prepared by dissolving 1.50 g of hemocyanin, a protein obtained from crabs, in 0.250 L of water. The solution has an osmotic pressure of 0.00342 atm at 277 K. (a) What is the molar mass of hemocyanin? (b) What should the freezing point of the solution be?


45

Ordinarily a patient must be given intravenous fluids that are isotonic—have the same osmotic pressure as blood.

Practical Applications of Osmosis

External solution is hypertonic; produces

osmotic pressure > πint. Net flow of water out

of the cell.

External solution is hypotonic; produces osmotic pressure <

πint. Net flow of water into the cell.

Red blood cell in isotonic solution

remains the same size.


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• Reverse osmosis (RO): reversing the normal net flow of solvent molecules through a semipermeable membrane.

• Pressure that exceeds the osmotic pressure is applied to the solution.

• RO is used for water purification.

Pressure greater than π is applied here …

… water flows from the more concentrated solution, through

the membrane.

Practical Applications of Osmosis (cont’d)


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• Whereas electrolytes dissociate, the number of solute particles ordinarily is greater than the number of formula units dissolved. One mole of NaCl dissolved in water produces more than one mole of solute particles.

• The van’t Hoff factor (i) is used to modify the colligative-property equations for electrolytes:

Tf = i × (–Kf) × m

Tb = i × Kb × m

= i × M RT

• For nonelectrolyte solutes, i = 1.

• For electrolytes, we expect i to be equal to the number of ions into which a substance dissociates into in solution.

Solutions of Electrolytes


48At very low

concentrations, the “theoretical” values

of i are reached.

At higher concentrations, the values of i are significantly lower than the theoretical values; ion

pairs form in solution.


49

Example 12.17 A Conceptual Example

Without doing detailed calculations, place the following solutions in order of decreasing osmotic pressure:

(a) 0.01 M C12H22O11(aq) at 25 °C

(b) 0.01 M CH3CH2COOH(aq) at 37 °C

(c) 0.01 m KNO3(aq) at 25 °C

(d) a solution of 1.00 g polystyrene (molar mass: 3.5 × 105 g/mol) in 100 mL of benzene at 25 °C


50

• In a solution, dispersed particles are molecules, atoms, or ions (roughly 0.1 nm in size). Solute particles do not “settle out” of solution.

• In a suspension (e.g., sand in water) the dispersed particles are relatively large, and will settle from suspension.

• In a colloid, the dispersed particles are on the order of 1–1000 nm in size.

• Although they are larger than molecules/atoms/ions, colloidal particles are small enough to remain dispersed indefinitely.

Colloids


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Why are there no gas-in-gas colloids?


52

The Tyndall Effect

The dissolved Fe3+ ions are not large enough to scatter light; the beam is virtually invisible.

Light scattered by the (larger) colloidal particles of Fe2O3

makes the beam visible.


53

Formation and Coagulationof a Colloid

When a strong electrolyte is added to colloidal iron oxide, the charge on the

surface of each particle is partially neutralized …

… and the colloidal particles coalesce into a suspension

that quickly settles.


54

Cumulative Example

A 375-mL sample of hexane vapor in equilibrium with liquid hexane C6H14 (d = 0.6548 g/mL), at 25.0 °C is dissolved in 50.0 mL of liquid cyclohexane, C6H12, at 25.0 °C (d = 0.7739 g/mL, vp = 97.58 Torr). Use information found elsewhere in the text (such as Example 11.3) to calculate the total vapor pressure above the solution at 25.0 °C. How reliable is this calculation?

Chapter Twelve General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 1 Chapter Twelve Physical Properties of.

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