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Study Guide Chapter 13...

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Chapter 13 Physical Properties of Solutions 13.1 Types of Solutions 13.2 A Molecular View of the Solution Process 

The Importance of Intermolecular Forces

 Energy and Entropy in Solution Formation 13.3 Concentration Units 

Molality



Percent by Mass

 Comparison of Concentration Units 13.4 Factors that Affect Solubility 

Temperature

 Pressure 13.5 Colligative Properties 

Vapor-Pressure Lowering



Boiling-Point Elevation



Freezing-Point Depression



Osmotic Pressure

 Electrolyte Solutions 13.6 Calculations Using Colligative Properties 13.7 Colloids

13.1 Types of Solutions The main subject of this chapter concerns the formation and properties of liquid solutions. Recall from earlier work that a solution is a homogeneous mixture of two or more substances. The component in greater quantity is called the solvent and the component in lesser amount is called the solute. In aqueous solutions, water is the solvent. Solutes can be liquids, solids, or gases. Several terms are used to describe the degree to which a solute will dissolve in a solvent: 1.

2.

When two liquids are completely soluble in each other in all proportions, they are said to be miscible. For example, ethanol and water are miscible. If the liquids do not mix, they are said to be immiscible. Oil and water, for example, are immiscible. A solution that contains the maximum concentration of a solute at a given temperature is a saturated solution. An unsaturated solution has a concentration of solute that is less than the maximum concentration. Supersaturated solutions have a concentration of solute that is greater than that of a saturated solution.

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Chapter 13: Physical Properties of Solutions

13.2

The Solution Process

Dissolving is a process that takes place at the molecular level and can be discussed in molecular terms. When one substance dissolves in another, the particles of the solute disperse uniformly throughout the solvent. The solute particles occupy positions that are normally taken by solvent molecules. The ease with which a solute particle may replace a solvent molecule depends on the relative strengths of three types of interactions:  Solvent-solvent interaction 

Solute-solute interaction



Solvent-solute interaction

Imagine the solution process as taking place in three steps as shown in the figure below. Step 1 is the separation of solvent molecules. Step 2 is the separation of solute molecules. These steps require inputs of energy to overcome attractive intermolecular forces. Step 3 is the mixing of solvent and solute molecules; it may be exothermic or endothermic. According to Hess's law, the heat of solution is given by the sum of the enthalpies of the three steps: Step 2

Step 1

H1

H2 Solute

Solvent Step 3

H3

Solution

∆Hsoln = ∆H1 + ∆H2 + ∆H3 The solute will be soluble in the solvent if the solute-solvent attraction is stronger than the solvent-solvent attraction and solute-solute attraction. Such a solution process is exothermic. Only a relatively small amount of the solute will be dissolved if the solute-solvent interaction is weaker than the solvent-solvent and solute-solute interaction; then the solution process will be endothermic. It turns out that all chemical processes are governed by two factors. The first of these is the energy factor. In other words, does the solution process absorb energy or release energy? Disorder or randomness is the other factor that must be considered. Processes that increase randomness or disorder are also favored. In the pure state, the solvent and solute possess a fair degree of order. Here, we mean the ordered arrangement of atoms, molecules, or ions in a three-dimensional crystal. Where order is high, randomness is low. The order in a crystal is lost when the solute dissolves and its molecules are dispersed in the solvent. The solution process is accompanied by an increase in disorder or randomness. It is the increase in disorder of the system that favors the solubility of any substance. The most general solubility rule is that "like dissolves like." In this rule, the term "like" refers to molecular polarity. "Like dissolves like" means that substances of like or similar polarity will mix to form solutions. Substances of different polarity will be immiscible or will only form very dilute solutions. The rule

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Chapter 13: Physical Properties of Solutions

predicts that two polar substances will form a solution and two nonpolar substances will form a solution, but a polar substance and a nonpolar substance will tend not to mix. EXAMPLE 13.2A: SOLUBILITY Which would be a better solvent for molecular I2(s): CCl4 or H2O? Solution Using the like-dissolves-like rule, first we identify I2 as a nonpolar molecule. Therefore, it will be more soluble in the nonpolar solvent CCl4 than in the polar solvent H2O. EXAMPLE 13.2B: SOLUBILITY In which solvent will NaBr be more soluble : benzene or water? Solution Benzene is a nonpolar solvent and water is a polar solvent. Since NaBr is an ionic compound, it cannot dissolve in nonpolar benzene, but is very soluble in water. ________________________________________________________________________________________ PRACTICE EXERCISES 1. Isopropyl alcohol and water dissolve in each other regardless of the proportions of each. What term describes the solubilities of these liquids in each other? 2.

Explain why hexane (C 6H14), even though a liquid, is not miscible with water.

3.

Indicate whether each compound listed is soluble or insoluble in water. a. CH3OH b. LiBr c. d.

C8H18 CCl4

e. f.

BaCl2 HOCH2CH2OH

4.

Explain why ammonia gas, NH3, is very soluble in water, but not in hexane, C 6H14.

5.

What is the difference between solvation and hydration?

6.

Predict which substance of the following pairs will be more soluble in water. a. b. c.

7.

NaCl(s) or I2(s) CH4(g) or NH3(g) CH3OH(ℓ) or C6H6(benzene)

For each of the following pairs, predict which substance will be more soluble in CCl 4(ℓ). a. I2(s)

b. KBr(s) ________________________________________________________________________________________

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Chapter 13: Physical Properties of Solutions

13.3

Concentration Units

The term concentration refers to how much of one component of a solution is present in a given amount of solution. You have already been introduced to molarity as a concentration unit.

molarity =

moles of solute liters of solution

Three new concentration units are introduced in this chapter. The percent by mass of solute:

Percent by mass of solute =

=

mass of solute  100% mass solute + mass solvent

mass of solute  100% mass of solution

The mole fraction of a component of the solution:

mole fraction of component A = XA =

moles of A sum of moles of all compounds

The molality of a solute in a solution:

molality =

moles of solute mass of solvent(kg)

Notice that molality is the only one of the mentioned concentration units which has the amount of solvent (and not solution) in the denominator. This is essential to keep in mind. It will be very important to be able to convert between the different concentration units. Let’s look at an example of how to deal with these conversions. EXAMPLE 13.3A: CONCENTRATION UNITS The dehydrated form of Epsom salts is magnesium sulfate. a.

What is the percent MgSO4 by mass in a solution made from 16.0 g MgSO4 and 100 mL of H2O at 25 °C? The density of water at 25 °C is 0.997 g/mL.

b. c.

What is the mole fraction of each component? Calculate the molality of the solution.

a.

Write the equation for percent by mass.

Solution

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Chapter 13: Physical Properties of Solutions

percent MgSO4 =

mass MgSO 4  100% mass MgSO 4  mass water

mass H2O = 100 mL 

percent MgSO4 =

0.997 g = 99.7 g H2O 1 mL

16.0 g 16.0 g  100% =  100% 115.7 g 16.0 g  99.7 g percent MgSO4 = 13.8%

b.

Write the formula for the mole fraction of MgSO4.

XMgSO4 =

mol MgSO4 mol MgSO4  mol H 2 O

Convert the masses in part (a) into moles to substitute into the equation.

16.0 g MgSO4 

1 mol MgSO4 = 0.133 mol MgSO4 120.4 g MgSO 4

99.7 g H2O 

1 mol H2 O = 5.54 mol H2O 18.0 g H 2O

The mole fractions of MgSO4 and H2O are

XMgSO4 =

0.133 mol = 0.0235 0.133 mol  5.54 mol

XH2O =

5.54 mol = 0.977 5.67 mol

Notice that the sum of the two mole fractions is 1.000. The sum of the mole fractions of all solution components is always 1.00. c.

Write the formula for molality:

molality =

moles of MgSO4 kilograms of H2O

Substitute into this equation the quantities previously calculated in parts (a) and (b).

molality =

103 g 0.133 mol MgSO4 x = 1.33 m 1 kg 99.7 g

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Chapter 13: Physical Properties of Solutions

EXAMPLE 13.3B: CONVERSIONS BETWEEN UNITS Concentrated hydrochloric acid is 36.5 percent HCl by mass. Its density is 1.18 g/mL. Calculate: a. the molality of HCl b.

the molarity of HCl.

a.

Write the formula for molality.

Solution

molality =

mol HCl kg H2 O

Next, find the number of moles of HCl per kilogram of H2O. We take 100 g of solution, and determine how many moles of HCl and how many kilograms of the solvent it contains. A solution that is 36.5 percent HCl by mass corresponds to 36.5 g HCl/100 g solution. Since 100 g of solution contains 36.5 g HCl, the difference 100 g – 36.5 g must equal the mass of water which is 63.5 g. We have two ratios:

36.5 g HCl 100 g soln

and

36.5 g HCl 63.5 g H2 O

The moles of HCl is given by:

moles HCl = 36.5 g HCl 

1 mol HCl = 1.00 mol HCl 36.5 g HCl

The kilograms of water is given by:

kg H2O = 63.5 g H2O 

1 kg = 0.0635 kg H2O 1×10 3 g

Then we calculate molality:

molality =

b.

1.00 mol HCl = 15.7 m 0.0635 kg H2 O

Write the formula for molarity:

molarity =

moles HCl liters soln

Find the number of moles of HCl and liters of solution present in 100 g of solution

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Chapter 13: Physical Properties of Solutions

36.5 g HCl 100 g soln Convert 100 g of solution to volume of solution using the density. (Note: 36.5 g HCl = 1.00 mol.) Volume of soln = (100 g soln) 

molarity =

10 3 L 1 mL = 0.0847 L  1 mL 1.18 g soln

1.00 mol HCl = 11.8 M 0.0847 L soln

________________________________________________________________________________________ PRACTICE EXERCISES 8. Calculate the percent by mass of the solute in the following aqueous solutions. a. 6.50 g NaCl in 75.2 g of water. b. 27.2 g ethanol in 250 g of solution. c. 9.

2.0 g I2 in 125 g methanol.

Calculate the molality of each of the following solutions. a. b.

6.50 g NaCl in 75.2 g of water. 27.5 g glucose (C6H12 O6) in 425 g of water.

10. Calculate the molarity of a 2.44 m NaCl solution given that its density is 1.089 g/mL. 11. Calculate the percent AgNO3 by mass in a 0.650 m AgNO3 solution. 12. Calculate the molality of a 5.5% AgNO3 solution. ________________________________________________________________________________________ 13.4

Factors that Affect Solubility

Temperature Temperature changes strongly affect the solubility of most solid solutes. The effect of temperature on the solubilities of some common salts is shown in Figure 13.4 (text). In most cases, the solubility of a solid in water increases with increasing temperature. However, this is not always true. Temperature effects must be determined experimentally. For gases, the dependence of solubility on temperature is different. The solubility of gases in water decreases with increasing temperature. The solubility of an unreactive gaseous solute is due to intermolecular attractive forces between solute molecules and solvent molecules. As the temperature of such a solution is increased, more solute molecules attain sufficient kinetic energy to break away from these attractive forces and enter the gas phase and so the solubility decreases. Pressure The solubility of gases in liquids is directly proportional to the gas pressure. Henry's law relates gas concentration c,

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Chapter 13: Physical Properties of Solutions

in moles per liter, to the gas pressure P in atmospheres. c = kP where k is a constant for each gas and has units of mol/L.atm. The greater the value of k, the greater the solubility of the gas. It is common in a “Henry’s Law” proble m to be given a concentration for a corresponding pressure in order to find the Henry’s law constant. Once k is determined, it can be used to convert between pressure of a gas and its concentration in solution. EXAMPLE 13.4: HENRY'S LAW What is the concentration of O2 in air-saturated water at 25 °C and atmospheric pressure of 645 mmHg? Assume the mole fraction of oxygen in air is 0.209 and that the Henry’s law constant for O 2 at 25°C is

1.28 10 3 mol

L atm

.

Solution Henry's law states that the concentration of dissolved O2 (c O 2 ) is proportional to its partial pressure (P O2 ) in atm.

c O 2  k  PO 2 The partial pressure of O2 in air is found by using Dalton's law of partial pressures.

PO2   O2 P T   0.209 645 mmHg 

1 atm  0.177 atm 760 mmHg

Finally, substitute the values into the formula:

c O 2  k P O 2  1.28 10 3 mol/L atm  0.177 atm   2.27 10 4 mol/L ____________________________________________________________________________________________ PRACTICE EXERCISES 13. The solubility of KNO3 at 70 °C is 135 g per 100 g of water. At 10 °C the solubility is 21 g per 100 g of water. What mass of KNO3 will crystallize out of solution if exactly 100 g of its saturated solution at 70 °C is cooled to 10 °C? 14. As temperature increases, the solubility of all gases in water _____________ (increases/decreases). 15. The Henry's law constant for argon is 1.5 x 10–3 mol/L·atm at 20 °C. Calculate the solubility or argon in water at 20 °C and 7.6 mmHg. ____________________________________________________________________________________________

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Chapter 13: Physical Properties of Solutions

13.5

Colligative Properties

The properties of solutions that depend on the number of solute particles in solution are called colligative properties. The four colligative properties of interest here are: Vapor-Pressure Lowering Vapor-pressure lowering, ∆P, is given by: ∆P =  2 P1  where  2 is the mole fraction of the solute, and P1  is the vapor pressure of the pure solvent. Recall that the vapor pressure of a liquid is the pressure exerted by a vapor in equilibrium, with its liquid phase. For a solution containing a nonvolatile solute, the vapor pressure due to the solvent is less than it is for the pure solvent. The amount of lowering of the vapor pressure can be seen to depend on  2 , the mole fraction of the solute. EXAMPLE 13.5A: VAPOR PRESSURE LOWERING Calculate the vapor pressure of an aqueous solution at 30 °C made from 100 g of sucrose (C12H22O11), and 100 g of water. The vapor pressure of water at 30 °C is 31.8 mmHg. Solution Sucrose is a nonvolatile solute, so the vapor pressure over the solution will be due to H 2O molecules. The problem can be worked in two ways. a.

The vapor-pressure lowering is proportional to the mole fraction of sucrose  2 , and P1  , the vapor pressure of pure water at 30 °C. ∆P =  2 P1  First, we calculate the mole fraction of sucrose,  2 .

2 =

n2 n1  n 2

n2 = 100 g 

1 mol = 0.292 mol 342 g

n1 = 100 g 

1 mol = 5.55 mol 18.0 g

2 =

0.292 0.292  0.292  5.55 5.84

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Chapter 13: Physical Properties of Solutions

 2 = 0.0500 The vapor pressure lowering is: ∆P = (0.0500) (31.8 mmHg) = 1.59 mmHg The vapor pressure, P1, is P1  – ∆P. P1 = 31.8 – 1.59 = 30.2 mmHg b.

Alternatively Raoult's law can be used to calculate the vapor pressure of the solvent: P1 =  1 P1  where P1  and P1 are the vapor pressures of pure solvent and of the solvent in solution, respectively. From part a,  2 = 0.0500, therefore  1 = 1.00 –  2 = 0.95, and P1 = (0.95)(31.8 mmHg) = 30 mmHg

Boiling-Point Elevation The boiling point of a solution is higher than that of pure solvent because the vapor pressure of a solution is always less than the vapor pressure of pure solvent. Thus, a solution must be hotter than a pure solvent, if both vapor pressures are to be 1 atm. The boiling-point elevation ∆Tb of a solution of molality m is given by: ∆Tb = Kbm where Kb is a constant called the molal boiling-point elevation constant and has the units °C/m. There is a Kb for each solvent, and Table 13.2 in the textbook lists Kb values for five solvents. The magnitude of ∆Tb, the increase in boiling point of the solution over the boiling point of the pure solvent, is proportional to the solute concentration m. The preceding equation is independent of the solute, but requires that the solute be nonvolatile and a nonelectrolyte. EXAMPLE 13.5B: BOILING POINT ELEVATION What is the boiling point of an "antifreeze/coolant" solution made from a 50-50 mixture (by volume) of ethylene glycol, C2H6O2 (density 1.11 g/mL), and water? Solution The boiling point depends on the molality of the 50-50 mixture. ∆T = Kbm. For simplicity, assume 100 mL of the solution. Then, using the density of water, the mass of 50 mL (50 vol %) of H2O is 50 g. The mass of 50 mL (50 vol %) of ethylene glycol, using the density given above is 55.5 g.

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Chapter 13: Physical Properties of Solutions

m=

mol C2 H 6O 2 kg H2O 1 mol = 0.895 mol 62.0 g

mol C2H6O2 = 55.5 g 

m=

0.895 mol C2 H6 O2

= 17.9 mol/kg = 17.9 m

0.050 kg

∆Tb = Kbm = (0.52 °C/m)  17.9 m = 9.3 °C Therefore the boiling point of the solution is 9.3 °C above the normal boiling point of water. Boiling point = 109.3 °C. Freezing-Point Depression In a similar way, the freezing point of a solution will always be lower than that of the pure solvent. The freezingpoint depression, ∆Tf, is given by: ∆Tf = Kf m where Kf is the molal freezing-point depression constant and has the units °C/m. As is the case of K b, Kf is different for each solvent . The equation applies to all nonelectrolyte solutes, even volatile ones, because freezing points are rather insensitive to vapor pressure. EXAMPLE 13.5C: FREEZING-POINT DEPRESSION Estimate the freezing point of a solution made by mixing 20.0 mL of isopropanol, (CH3)2CHOH (density = 0.786 g/mL), with 250. g of water. The Kf for water is 1.86°C/m and its normal freezing point is 0.0°C. Solution First, we need to calculate the molality of the ...