13 Solutions [호환 모드] - 연세대학교 이과대학 화학과chem.yonsei.ac.kr/~mhmoon/pdf/GenChem_Brown/13_S… · · 2013-09-23Solution SCO 2 = kPCO 2 = (3.4 10 2 mol/L-atm)

Chapter 13

Properties of Solutions

Lecture Presentation

Yonsei University

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Solutions• homogeneous mixtures of two or more pure

substances: Solutions may be gases, liquids, or solids

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

© 2012 Pearson Education, Inc.

The intermolecular forces between solute and solvent particles must be strong enough to compete with those between solute particles and those between solvent particles.– Solute-solute, Solvent-solvent,

Solvent-solute interactions

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Solution Process

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How Does a Solution Form?As a solution forms, the solvent pulls solute particles apart and surrounds, or solvates, them.– Solvation, hydration


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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.


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Energetics of Solution Formation

• Simply put, three processes affect the energetics of solution:– Separation of solute particles,

– Separation of solvent particles,

– New interactions between solute and solvent.

• The enthalpy change of the overall process depends on H for each of these steps.


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Why Do Endothermic Processes Occur?

• Things do not tend to occur spontaneously (i.e., without outside intervention) unless the energy of the system is lowered.

• Yet we know that in some processes, like the dissolution of NH4NO3 in water, heat is absorbed, not released.

– MgSO4 added to water has ∆Hsoln = –91.2 kJ/mol.

– NH4NO3 added to water has ∆Hsoln = + 26.4 kJ/mol.

• entropy


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Solution Formation and Chemical Reactions

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

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

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

Ni(s) + 2HCl(aq) NiCl2(aq) + H2(g)

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Types of Solutions

• Saturated – In a saturated

solution, dissolved solute is in dynamic equilibrium with solid solute particles (T).

• Solubility

• Unsaturated solution


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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.


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Factors Affecting Solubility

• The tendency of a substance to dissolve in another depends on:– the nature of the solute.

– the nature of the solvent.

– the temperature.

– the pressure (for gases).


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Nature of Solvent & Solute• Chemists use the axiom “like dissolves like.”

– Solute-Solvent Interactions

– Polar substances tend to dissolve in polar solvents.

– Nonpolar substances tend to dissolve in nonpolar solvents.


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Nature of Solvent & Solute

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


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Glucose

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


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Solubility of alcohols in water• The number of carbon atoms in a chain

affects solubility. – The greater the number of carbons in the chain,

the more the molecule behaves like a hydrocarbon.

– Thus, the more C atoms in the alcohol, the lower its solubility in water.

• Increasing the number of –OH groups within a molecule increases its solubility in water.– The greater the number of –OH groups along the

chain, the more solute-water H-bonding is possible.


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Vitamins

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

• Vitamin C is soluble in water.


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Sample Exercise 13.1 Predicting Solubility Patterns

SolutionC7H16 and I2 (the nonpolar solutes) would be more soluble in the nonpolar CCl4than in polar H2O, whereas water would be the better solvent for Na2SO4 and HCl (the ionic and polar covalent solutes).

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.

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Sample Exercise 13.1 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

Continued

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Gases in Solution

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

• Larger molecules have stronger dispersion forces.


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Pressure Effects

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

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


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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.


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Sample Exercise 13.2 A Henry’s Law Calculation

SolutionSCO2

= kPCO2= (3.4 102 mol/L-atm)(4.0 atm) = 0.14 mol/L = 0.14 M

Practice ExerciseCalculate the concentration of CO2 in a soft drink after the bottle is opened and equilibrates at 25 C under a CO2

partial pressure of 3.0 104 atm.Answer: 1.0 105 M

Calculate the concentration of CO2 in a soft drink that is bottled with a partial pressure of CO2 of 4.0 atm over the liquid at 25 C. The Henry’s law constant for CO2 in water at this temperature is 3.4 102 mol/L-atm.

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

• The opposite for gases.– Carbonated soft drinks are more

“bubbly” if stored in the refrigerator.

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

• Thermal pollution

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

100


Ways of Expressing Concentrations of Solutions

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

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ppm, ppb, ppt in dilute aqueous solution

6

9

12

1 1

10

1 1

10

1 1

10

g solute mg soluteppm

g of solution L solution

g solute g soluteppb


g solute ng soluteppt


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moles of Atotal moles of all componentsXA =

Mole Fraction (X)


moles of soluteliters of solution

M =Molarity (M)

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

moles of solutekilograms of solventm =Molality (m)

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

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Changing Molarity to Molality

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


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Sample Problem-Conversions between units-

ex) What is the molality of a 0.200 M aluminum nitrate

solution (d = 1.012g/mL)?

– Work with 1 liter of solution. mass = 1012 g

– mass Al(NO3)3 = 0.200 mol × 213.01 g/mol = 42.6 g ;

– mass water = 1012 g -43 g = 969 g

0.2000.206 /

0.969

molMolality mol kg

kg

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Sample Exercise 13.4 Calculation of Molality

Solution

A solution is made by dissolving 4.35 g glucose (C6H12O6, MM=180.2g/mol) 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.

(25.0 mL)(1.00 g/mL) = 25.0 g = 0.0250 kg

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

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Sample Exercise 13.5 Calculation of Mole Fraction and MolarityAn 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(a)

(b)

Practice ExerciseA commercial 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) 9.00 103, (b) 0.505 m

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Sample Exercise 13.6 Calculation of Molarity Using the Density of the 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.

Solution

Practice ExerciseA 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) XC3H8O3 = 0.163, (c) 5.97 M

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Colligative Properties

• Changes in colligative propertiesdepend 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


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Vapor Pressure

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


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

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Raoult’s Law

where– XA is the mole fraction of compound A, and

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

– The vapor-pressure lowering, ∆P, is

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


osolventsolute PXP

osolution solvent solventP P

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Sample Exercise 13.7 Calculation of Vapor Pressure of a Solution

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 ofpure water at 25 C is 23.8 torr (Appendix B), and its density is 1.00 g/mL.

Solution

Practice ExerciseThe 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

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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.


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Boiling-Point Elevation

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

Tb = Kb mwhere Kb is the molal boiling-point elevation constant, a property of the solvent.

Tb is added to the normal boiling point of the solvent.


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• 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 freezing point of the solvent.


Freezing-Point Depression

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Colligative Properties of Electrolytes

• Solutions of electrolytes (which dissociate in solution) should show greater changes than those of nonelectrolytes.


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

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van’t Hoff FactorOne mole of NaCl in water does not really give rise to two moles of ions.


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.

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van’t Hoff Factor• Reassociation is more

likely at higherconcentration.

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


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

Tf = Kf m i

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Sample Exercise 13.8 Calculation Of Boiling-Point Elevation and Freezing-Point Depression

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

Tb = Kbm = (0.51 C/m)(5.37 m) = 2.7 CTf = Kfm = (1.86 C/m)(5.37 m) = 10.0 CHence, the boiling and freezing points of the solution are

Boiling point = (normal bp of solvent) + Tb

= 100.0 C + 2.7 C = 102.7 CFreezing point = (normal fp of solvent) Tf

= 0.0 C 10.0 C = 10.0 C

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Sample Exercise 13.9 Freezing-Point Depression in Aqueous SolutionsList 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

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).

Practice ExerciseWhich 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

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Osmosis

• Some substances form semipermeable membranes, allowing some smaller particles to pass through, but blocking other larger particles.

• In biological systems, most semipermeable membranes allow water to pass through, but solutes are not free to do so.


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Osmosis

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Osmosis

In osmosis, there is net movement of solvent from the area of higher solvent concentration (lower solute concentration) to the area of lower solvent concentration(higher solute concentration).


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Osmotic Pressure

The pressure required to stop osmosis, known as osmotic pressure, , is

= ( )RT = MRTnV

where M is the molarity of the solution.

If the osmotic pressure is the same on both sides of a membrane (i.e., the concentrations are the same), the solutions are isotonic.


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Sample Exercise 13.10 Calculating Involving Osmotic Pressure

The average osmotic pressure of blood is 7.7 atm at 25 C. What molarity of glucose (C6H12O6) will be isotonic with blood?

Solution

Practice ExerciseWhat is the osmotic pressure at 20 C of a 0.0020 M sucrose (C12H22O11) solution?

Answer: 0.048 atm, or 37 torr

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Osmosis in Blood Cells

• If the solute concentration outside the cell is greater than that inside the cell, the solution is hypertonic.

• Water will flow out of the cell, and crenationresults.


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Osmosis in Cells

• If the solute concentration outside the cell is less than that inside the cell, the solution is hypotonic.

• Water will flow into the cell, and hemolysisresults.


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

• Particle size: 5 to 1000 nm.

• A colloid particle may consist of a single giant molecule.

• Ex) hemoglobin has molecular dimensions of 6.5 5.5 5.0 nm and a molar mass of 64,500 g/mol.


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Tyndall Effect

• Colloidal suspensions can scatter rays of light.

• This phenomenon is known as the Tyndall effect.


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Colloids in Biological Systems

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


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Colloids in Biological Systems

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


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10, 16, 38, 60, 80, 86, 108, 110

Problems

Documents

13 Solutions [호환 모드] - 연세대학교 이과대학 화학과chem.yonsei.ac.kr/~mhmoon/pdf/GenChem_Brown/13_S… · · 2013-09-23Solution SCO 2 = kPCO 2 = (3.4 10 2 mol/L-atm)