No. 6 Disoluciones

Solutions

2009, Prentice-Hall, Inc.

Chapter 13Properties of Solutions

John D. Bookstaver

St. Charles Community College

Cottleville, MO

Chemistry, The Central Science, 11th editionTheodore L. Brown, H. Eugene LeMay, Jr.,

and Bruce E. Bursten

Solutions


Solutions

• Solutions are homogeneous mixtures of two or more pure substances.

• In a solution, the solute is dispersed uniformly throughout the solvent.

Solutions


Solutions

The intermolecular forces between solute and solvent particles must be strong enough to compete with those between solute particles and those between solvent particles.

Solutions


How Does a Solution Form?

As a solution forms, the solvent pulls solute particles apart and surrounds, or solvates, them.

Solutions


How Does a Solution Form

If an ionic salt is soluble in water, it is because the ion-dipole interactions are strong enough to overcome the lattice energy of the salt crystal.

Solutions


Student, Beware!

Just because a substance disappears when it comes in contact with a solvent, it doesn’t mean the substance dissolved.

Solutions


Student, Beware!

• Dissolution is a physical change — you can get back the original solute by evaporating the solvent.

• If you can’t, the substance didn’t dissolve, it reacted.

Solutions


Types of Solutions

• Saturated– In a saturated solution,

the solvent holds as much solute as is possible at that temperature.

– Dissolved solute is in dynamic equilibrium with solid solute particles.

Solutions


Types of Solutions

• Unsaturated– If a solution is

unsaturated, less solute than can dissolve in the solvent at that temperature is dissolved in the solvent.

Solutions


Types of Solutions

• Supersaturated– In supersaturated solutions, the solvent holds

more solute than is normally possible at that temperature.

– These solutions are unstable; crystallization can usually be stimulated by adding a “seed crystal” or scratching the side of the flask.

Solutions


Factors Affecting Solubility

• Chemists use the axiom “like dissolves like."– Polar substances tend to dissolve in polar solvents.– Nonpolar substances tend to dissolve in nonpolar

solvents.

Solutions



The more similar the intermolecular attractions, the more likely one substance is to be soluble in another.

Solutions



Glucose (which has hydrogen bonding) is very soluble in water, while cyclohexane (which only has dispersion forces) is not.

Solutions



• Vitamin A is soluble in nonpolar compounds (like fats).

• Vitamin C is soluble in water.

Solutions


Gases in Solution

• In general, the solubility of gases in water increases with increasing mass.

• Larger molecules have stronger dispersion forces.

Solutions


Gases in Solution

• The solubility of liquids and solids does not change appreciably with pressure.

• The solubility of a gas in a liquid is directly proportional to its pressure.

Solutions


Henry’s Law

Sg = kPg

where

• Sg is the solubility of the gas,

• k is the Henry’s Law constant for that gas in that solvent, and

• Pg is the partial pressure of the gas above the liquid.

Solutions


Temperature

Generally, the solubility of solid solutes in liquid solvents increases with increasing temperature.

Solutions


Temperature

• The opposite is true of gases.– Carbonated soft

drinks are more “bubbly” if stored in the refrigerator.

– Warm lakes have less O2 dissolved in them than cool lakes.

Solutions

Sample Exercise 13.2 Predicting Solubility Patterns

Predict whether each of the following substances is more likely to dissolve in the nonpolar solvent carbon tetrachloride (CCl4) or in water: C7H16, Na2SO4, HCl, and I2.SolutionAnalyze: We are given two solvents, one that is nonpolar (CCl4) and the other that is polar (H2O), and asked to determine which will be the best solvent for each solute listed.Plan: By examining the formulas of the solutes, we can predict whether they are ionic or molecular. For those that are molecular, we can predict whether they are polar or nonpolar. We can then apply the idea that the nonpolar solvent will be best for the nonpolar solutes, whereas the polar solvent will be best for the ionic and polar solutes.Solve: C7H16 is a hydrocarbon, so it is molecular and nonpolar. Na2SO4, a compound containing a metal and nonmetals, is ionic. HCl, a diatomic molecule containing two nonmetals that differ in electronegativity, is polar. I2, a diatomic molecule with atoms of equal electronegativity, is nonpolar. We would therefore predict that C7H16 and I2 (the nonpolar solutes) would be more soluble in the nonpolar CCl4 than in polar H2O, whereas water would be the better solvent for Na2SO4 and HCl (the ionic and polar covalent solutes).

Solutions

Sample Exercise 13.2 Predicting Solubility Patterns

Arrange the following substances in order of increasing solubility in water:

Answer: C5H12 < C5H11Cl < C5H11OH < C5H10(OH)2 (in order of increasing polarity and hydrogen-bonding ability)

Practice Exercise

Solutions

The curve describes the solubility properties of

• a liquid.

• an ionic solid.

• a covalent solid.

• a gas.

• a gas or liquid. 0

1.0

2.0

So

lub

ilit

y m

M

15 30 45 60Temperature (° C)

0

Solutions

Considering the substances given below, predict which will be soluble in water.

• 1 • 2 • 3 • 2, 3• 1, 2, 3

CH3CH2CH2CH2CH2CH2CH2CH2OH

NH3 CH3C CH3

O

1

2 3

Solutions


Ways of Expressing

Concentrations of Solutions

Solutions


Mass Percentage

Mass % of A =mass of A in solutiontotal mass of solution

100

Solutions


Parts per Million andParts per Billion

ppm =mass of A in solutiontotal mass of solution

106

Parts per Million (ppm)

Parts per Billion (ppb)

ppb =mass of A in solutiontotal mass of solution

109

Solutions


moles of Atotal moles in solution

XA =

Mole Fraction (X)

• In some applications, one needs the mole fraction of solvent, not solute — make sure you find the quantity you need!

Solutions


mol of soluteL of solution

M =

Molarity (M)

• You will recall this concentration measure from Chapter 4.

• Since volume is temperature-dependent, molarity can change with temperature.

Solutions

When concentration is expressed in units of ________ , changing temperature will affect the

solution concentration.

• molality• molarity• ppm• molarity and molality• molarity, molality, and ppm

Solutions


mol of solutekg of solvent

m =

Molality (m)

Since both moles and mass do not change with temperature, molality (unlike molarity) is not temperature-dependent.

Solutions


Changing Molarity to Molality

If we know the density of the solution, we can calculate the molality from the molarity and vice versa.

Copyright ©2009 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458

All rights reserved.

Chemistry: The Central Science, Eleventh EditionBy Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. MurphyWith contributions from Patrick Woodward

Sample Exercise 13.4 Calculation of Mass-Related Concentrations

(a) Calculate the mass percentage of NaCl in a solution containing 1.50 g of NaCl in 50.0 g of water. (b) A commercial bleaching solution contains 3.62 mass % sodium hypochlorite, NaOCl. What is the mass of NaOCl in a bottle containing 2.50 kg of bleaching solution?Answer: (a) 2.91%, (b) 90.5 g of NaOCl

Practice Exercise

(a) A solution is made by dissolving 13.5 g of glucose (C6H12O6) in 0.100 kg of water. What is the mass percentage of solute in this solution? (b) A 2.5-g sample of groundwater was found to contain 5.4 µg of Zn2+. What is the concentration of Zn2+ in parts per million?

Solution

(a) Analyze: We are given the number of grams of solute (13.5 g) and the number of grams of solvent (0.100 kg = 100 g). From this we must calculate the mass percentage of solute.Plan: We can calculate the mass percentage by using Equation 13.5. The mass of the solution is the sum of the mass of solute (glucose) and the mass of solvent (water).

Comment: The mass percentage of water in this solution is (100 – 11.9)% = 88.1%.(b) Analyze: In this case we are given the number of micrograms of solute. Because 1 µg is 1 × 10–6 g, 5.4 µg = 5.4 × 10–6 g.Plan: We calculate the parts per million using Equation 13.6.




Sample Exercise 13.5 Calculation of Molality

A solution is made by dissolving 4.35 g glucose (C6H12O6) in 25.0 mL of water at 25 °C. Calculate the molality of glucose in the solution. Water has a density of 1.00 g/mL.

What is the molality of a solution made by dissolving 36.5 g of naphthalene (C10H8) in 425 g of toluene (C7H8)?Answer: 0.670 m

Practice Exercise

Solution

Analyze: We are asked to calculate a molality. To do this, we must determine the number of moles of solute (glucose) and the number of kilograms of solvent (water).Plan: We use the molar mass of C6H12O6 to convert grams to moles. We use the density of water to convert milliliters to kilograms. The molality equals the number of moles of solute divided by the number of kilograms of solvent (Equation 13.9).

Solve: Use the molar mass of glucose, 180.2 g/mL, to convert grams to moles:

Because water has a density of 1.00 g/mL, the mass of the solvent is

Finally, use Equation 13.9 to obtain the molality:




Sample Exercise 13.6 Calculation of Mole Fraction and Molality

An aqueous solution of hydrochloric acid contains 36% HCl by mass. (a) Calculate the mole fraction of HCl in the solution. (b) Calculate the molality of HCl in the solution.

Solution

Analyze: We are asked to calculate the concentration of the solute, HCl, in two related concentration units, given only the percentage by mass of the solute in the solution.Plan: In converting concentration units based on the mass or moles of solute and solvent (mass percentage, mole fraction, and molality), it is useful to assume a certain total mass of solution. Let’s assume that there is exactly 100 g of solution. Because the solution is 36% HCl, it contains 36 g of HCl and (100 – 36) g = 64 g of H2O. We must convert grams of solute (HCl) to moles to calculate either mole fraction or molality. We must convert grams of solvent (H2O) to moles to calculate mole fractions, and to kilograms to calculate molality.

Solve: (a) To calculate the mole fraction of HCl, weconvert the masses of HCl and H2O to moles andthen use Equation 13.7:

(b) To calculate the molality of HCl in the solution,we use Equation 13.9. We calculated the number ofmoles of HCl in part (a), and the mass of solvent is 64 g = 0.064 kg:




Sample Exercise 13.6 Calculation of Mole Fraction and Molality

Acommercial bleach solution contains 3.62 mass % NaOCl in water. Calculate (a) the mole fraction and (b) the molality of NaOCl in the solution.Answer: (a) 9.00 × 10–3, (b) 0.505 m.

Practice Exercise




Sample Exercise 13.7 Calculation of Molality Using the Density of a Solution

A solution with a density of 0.876 g/mL contains 5.0 g of toluene (C7H8) and 225 g of benzene. Calculate the molarity of the solution.

The units for our answer (mol/L) are correct, and the answer, 0.21 M, has two significant figures, corresponding to the number of significant figures in the mass of solute (2).

Solution

Analyze: Our goal is to calculate the molarity of a solution, given the masses of solute (5.0 g) and solvent (225 g) and the density of the solution (0.876 g/mL).Plan: The molarity of a solution is the number of moles of solute divided by the number of liters of solution (Equation 13.8). The number of moles of solute (C7H8) is calculated from the number of grams of solute and its molar mass. The volume of the solution is obtained from the mass of the solution (mass of solute + mass of solvent = 5.0 g + 225 g = 230 g) and its density.Solve: The number of moles of solute is

The density of the solution is used to convert the mass of the solution to its volume:

Molarity is moles of solute per liter of solution:

Check: The magnitude of our answer is reasonable. Rounding moles to 0.05 and liters to 0.25 gives a molarity of





A solution containing equal masses of glycerol (C3H8O3) and water has a density of 1.10 g/mL. Calculate (a) the molality of glycerol, (b) the mole fraction of glycerol, (c) the molarity of glycerol in the solution.Answer: (a) 10.9 m, (b) = 0.163, (c) 5.97 M

Practice Exercise

Solution (continued)

Comment: Because the mass of the solvent (0.225 kg) and the volume of the solution (0.263 L)are similar in magnitude, the molarity and molalityare also similar in magnitude:

Solutions


Colligative Properties

• Changes in colligative properties depend only on the number of solute particles present, not on the identity of the solute particles.

• Among colligative properties are– Vapor pressure lowering – Boiling point elevation– Melting point depression– Osmotic pressure

Solutions


Vapor Pressure

Because of solute-solvent intermolecular attraction, higher concentrations of nonvolatile solutes make it harder for solvent to escape to the vapor phase.

Solutions


Vapor Pressure

Therefore, the vapor pressure of a solution is lower than that of the pure solvent.

Solutions


Raoult’s Law

PA = XAPAwhere– XA is the mole fraction of compound A, and

– PA is the normal vapor pressure of A at that temperature.

NOTE: This is one of those times when you want to make sure you have the vapor pressure of the solvent.

Solutions


Boiling Point Elevation and Freezing Point Depression

Nonvolatile solute-solvent interactions also cause solutions to have higher boiling points and lower freezing points than the pure solvent.

Solutions


Boiling Point Elevation

• The change in boiling point is proportional to the molality of the solution:

Tb = Kb m

where Kb is the molal boiling point elevation constant, a property of the solvent.Tb is added to the normal

boiling point of the solvent.

Solutions


Boiling Point Elevation

• The change in freezing point can be found similarly:

Tf = Kf m

• Here Kf is the molal freezing point depression constant of the solvent.

Tf is subtracted from the normal boiling point of the solvent.

Solutions


Boiling Point Elevation and Freezing Point Depression

Note that in both equations, T does not depend on what the solute is, but only on how many particles are dissolved.

Tb = Kb m

Tf = Kf m

Solutions


Colligative Properties of Electrolytes

Since these properties depend on the number of particles dissolved, solutions of electrolytes (which dissociate in solution) should show greater changes than those of nonelectrolytes.

Solutions


Colligative Properties of Electrolytes

However, a 1M solution of NaCl does not show twice the change in freezing point that a 1M solution of methanol does.

Solutions


van’t Hoff Factor

One mole of NaCl in water does not really give rise to two moles of ions.

Solutions


van’t Hoff Factor

Some Na+ and Cl- reassociate for a short time, so the true concentration of particles is somewhat less than two times the concentration of NaCl.

Solutions


van’t Hoff Factor

• Reassociation is more likely at higher concentration.

• Therefore, the number of particles present is concentration-dependent.

Solutions


van’t Hoff Factor

• We modify the previous equations by multiplying by the van’t Hoff factor, i.

Tf = Kf m i





A solution containing equal masses of glycerol (C3H8O3) and water has a density of 1.10 g/mL. Calculate (a) the molality of glycerol, (b) the mole fraction of glycerol, (c) the molarity of glycerol in the solution.Answer: (a) 10.9 m, (b) = 0.163, (c) 5.97 M

Practice Exercise


Comment: Because the mass of the solvent (0.225 kg) and the volume of the solution (0.263 L)are similar in magnitude, the molarity and molalityare also similar in magnitude:




Sample Exercise 13.8 Calculation of Vapor-Pressure Lowering

Glycerin (C3H8O3) is a nonvolatile nonelectrolyte with a density of 1.26 g/mL at 25 °C. Calculate the vapor pressure at 25 °C of a solution made by adding 50.0 mL of glycerin to 500.0 mL of water. The vapor pressure of pure water at 25 °C is 23.8 torr (Appendix B), and its density is 1.00 g/mL.Solution

Analyze: Our goal is to calculate the vapor pressure of a solution, given the volumes of solute and solvent and the density of the solute.Plan: We can use Raoult’s law (Equation 13.10) to calculate the vapor pressure of a solution. The mole fraction of the solvent in the solution, XA, is the ratio of the number of moles of solvent (H2O) to total moles of solution (moles C3H8O3 + moles H2O).

Solve: To calculate the mole fraction of water in the solution, we must determine the number of moles of C3H8O3 and H2O:

We now use Raoult’s law to calculate thevapor pressure of water for the solution:

The vapor pressure of the solution has been lowered by 0.6 torr relative to that of pure water.




Sample Exercise 13.8 Calculation of Vapor-Pressure Lowering

The vapor pressure of pure water at 110 °C is 1070 torr. A solution of ethylene glycol and water has a vapor pressure of 1.00 atm at 110 °C. Assuming that Raoult’s law is obeyed, what is the mole fraction of ethylene glycol in the solution?Answer: 0.290

Practice Exercise




Sample Exercise 13.9 Calculation of Boiling-Point Elevation and Freezing-Point Lowering

Automotive antifreeze consists of ethylene glycol, CH2(OH)CH2(OH), a nonvolatile nonelectrolyte. Calculate the boiling point and freezing point of a 25.0 mass % solution of ethylene glycol in water.

Solution

Analyze: We are given that a solution contains 25.0 mass % of a nonvolatile, nonelectrolyte solute and asked to calculate the boiling and freezing points of the solution. To do this, we need to calculate the boiling-point elevation and freezing-point depression.Plan: To calculate the boiling-point elevation and the freezing-point depression using Equations 13.11 and 12, we must express the concentration of the solution as molality. Let’s assume for convenience that we have 1000 g of solution. Because the solution is 25.0 mass % ethylene glycol, the masses of ethylene glycol and water in the solution are 250 and 750 g, respectively. Using these quantities, we can calculate the molality of the solution, which we use with the molal boiling-point-elevation and freezing-point depression constants (Table 13.4) to calculate ΔTb and ΔTf. We add ΔTb to the boiling point and subtract ΔTf from the freezing point of the solvent to obtain the boiling point and freezing point of the solution.Solve: The molality of the solution is calculated as follows:

We can now use Equations 13.11 and 13.12 to calculate the changes in the boiling and freezing points:




Sample Exercise 13.9 Calculation of Boiling-Point Elevation and Freezing-Point Lowering

Calculate the freezing point of a solution containing 0.600 kg of CHCl3 and 42.0 g of eucalyptol (C10H18O), a fragrant substance found in the leaves of eucalyptus trees. (See Table 13.4.)Answer: –65.6 °C

Practice Exercise


Hence, the boiling and freezing points of the solution are

Comment: Notice that the solution is a liquid over a larger temperature range than the pure solvent.




Sample Exercise 13.10 Freezing-Point Depression in Aqueous Solutions

List the following aqueous solutions in order of their expected freezing point: 0.050 m CaCl2, 0.15 m NaCl, 0.10 m HCl, 0.050 m CH3COOH, 0.10 m C12H22O11.

Solution

Analyze: We must order five aqueous solutions according to expected freezing points, based on molalities and the solute formulas.Plan: The lowest freezing point will correspond to the solution with the greatest concentration of solute particles. To determine the total concentration of solute particles in each case, we must determine whether the substance is a nonelectrolyte or an electrolyte and consider the number of ions formed when an electrolyte ionizes.Solve: CaCl2, NaCl, and HCl are strong electrolytes, CH3COOH (acetic acid) is a weak electrolyte, and C12H22O11 is a nonelectrolyte. The molality of each solution in total particles is as follows:

Because the freezing points depend on the total molality of particles in solution, the expected ordering is 0.15 m NaCl (lowest freezing point), 0.10 m HCl, 0.050 m CaCl2, 0.10 m C12H22O11, and 0.050 m CH3COOH (highest freezing point).




Sample Exercise 13.10 Freezing-Point Depression in Aqueous Solutions

Which of the following solutes will produce the largest increase in boiling point upon addition to 1 kg of water: 1 mol of Co(NO3)2, 2 mol of KCl, 3 mol of ethylene glycol (C2H6O2)?Answer: 2 mol of KCl because it contains the highest concentration of particles, 2 m K+ and 2 m Cl–, giving 4 m in all

Practice Exercise

Solutions

Predict which aqueous solution will have the lowest freezing point.

• 0.25 m C2H5OH• 0.15 m CaCl2• 0.20 m NaCl• 0.15 m NH4NO3

• 0.15 m Na3PO4

Solutions

Arrange the aqueous solutions according to increasing boiling point.

• AlCl3 < KNO3 < Na2SO4

• Na2SO4 < AlCl3 < KNO3 • Na2SO4 < KNO3 < AlCl3 • KNO3 < AlCl3 < Na2SO4

• KNO3 < Na2SO4 < AlCl3

0.10 m Na2SO4

0.15 m AlCl3

0.20 m KNO3

Solutions


Colloids

Suspensions of particles larger than individual ions or molecules, but too small to be settled out by gravity are called colloids.

Solutions


Tyndall Effect

• Colloidal suspensions can scatter rays of light.

• This phenomenon is known as the Tyndall effect.

Solutions


Colloids in Biological Systems

Some molecules have a polar, hydrophilic (water-loving) end and a non-polar, hydrophobic (water-hating) end.

Solutions



Sodium stearate is one example of such a molecule.

Solutions



These molecules can aid in the emulsification of fats and oils in aqueous solutions.

No. 6 Disoluciones

Documents

Transcript of No. 6 Disoluciones