chapter 13 properties of solutions

Chapter 13Properties of Solutions

13.1 The Solution Process

Solutions

Solutions are homogeneous mixtures of two or more pure substances.

In a solution, the solute is dispersed uniformly throughout the solvent (present in greatest amount).

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.

How Does a Solution Form?

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

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.

Energy Changes in Solution Simply put, three

processes affect the energetics of solution: separation of solute

particles, separation of solvent

particles, new interactions between

solute and solvent.

Energy Changes in Solution

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

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.

Why Do Endothermic Processes Occur?

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

Enthalpy Is Only Part of the Picture

The reason is that increasing the disorder or randomness (known as entropy) of a system tends to lower the energy of the system.

Enthalpy Is Only Part of the Picture

So even though enthalpy may increase, the overall energy of the system can still decrease if the system becomes more disordered.

Student, Beware!

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

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.

13.1 GIST

Why doesn’t NaCl dissolve in nonpolar solvents such as hexane, C6H14?

Label the following processes as exothermic or endothermic: Forming solvent-solute interactions Breaking solvent-solvent interactions

Silver chloride, AgCl, is essentially insoluble in water. Would you expect a significant change in entropy of the system when 10 g of AgCl is added to 500 mL of water?

Sample Exercise 13.1

1) Water vapor reacts with excess solid sodium sulfate to form a hydrate:

Na2SO4(s) + 10 H2O(g) → Na2SO4●10H2O(s)

a) Does the system become more or less ordered in the process?

b) Does the entropy of the system increase or decrease?

Practice

2) Does the entropy increase or decrease when the stopcock is opened allowing mixing of the two gases?

13.2 Saturated Solutions and Solubility

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.

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.

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.

p.535 GIST: Is a supersaturated solution of sodium acetate a stable equilibrium solution?

13.3 Factors Affecting Solubility

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.

Factors Affecting Solubility

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

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

p.538 GIST: Suppose the hydrogens on the OH groups in glucose were replaced with methyl groups, CH3. Would you expect the solubility of the resulting molecule to be higher than, lower than, or about the same as the solubility of glucose?

Factors Affecting Solubility

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

Vitamin C is soluble in water.

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.

Sample Exercise 13.2 Predicting Solubility Patterns

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

Practice Exercise

Gases in Solution

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

Larger molecules have stronger dispersion forces.

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.

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.

Exercise 13.13

1) 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 25ºC. The Henry’s law constant for CO2 in water at this temperature is 3.1 x 10-2 mol/L-atm.

2) Calculate the concentration of CO2 in a soft drink after the bottle is opened and equilibrates at 25ºC under a CO2 partial pressure of 3.0 x 10-4 atm.

Temperature

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

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.

13.4 Ways of Expressing Concentration

Mass Percentage

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

100

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

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?

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?

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!

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.

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.

Changing Molarity to Molality

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

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.

Practice

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

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.

Practice

1) A commercial bleach contains 3.62 mass % NaOCl in water. Calculate: A) the molality B) the mole fraction of NaOCl

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.

Practice

1) 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

13.5 Colligative Properties

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

Vapor Pressure

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

Vapor Pressure

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

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.

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.

Practice

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

P.547 GIST: Adding 1 mol of NaCl to 1 kg of water reduces the vapor pressure of water more than adding 1 mol of C6H12O6. Explain.

Ideal Solutions Ideal solutions obey Raoult’s Law Real liquid-liquid solutions best approximate

ideal behavior when the solute concentration is low and when the solute and solvent have similar molecular sizes and similar types of intermolecular attractions.

If a solution is not ideal, the observed vapor pressure will either by higher or lower than predicted using Raoult’s Law

Deviation from Raoult’s Law negative deviation

ex. acetone and water when solvent has special attraction to solute observed vapor pressure will be lower than

predicted

Deviation from Raoult’s Law positive deviation

ex. ethanol and hexane when solvent has especially weak interactions

with solute observed vapor pressure will be higher than

predicted

Volatile Solutions when both components are

volatile

where PA and PB are partial pressures

PA° and PB

° are vapor pressure of pure A and B

BBAABAtotal PPPPP

A solution is prepared by mixing 5.81 g of acetone (C3H6O, molar mass = 58.1 g/mol) and 11.9 g of chloroform (HCCl3, molar mass = 119.4 g/mol). At 35oC, this solution has a total pressure of 260. torr. Is this an ideal solution? The vapor pressure of pure acetone and pure chloroform at 35oC are 345 torr and 293 torr respectively.

Boiling Point Elevation a nonvolatile solute lowers the vapor pressure a higher T is required to reach the 1 atm of

pressure which defines boiling point a nonvolatile solute elevates the boiling point

of the solvent the amount of the elevation depends on

concentration of the solute

soluteb mKT p.549 GIST: An unknown solute dissolved in water causes the boiling point to increase by 0.51°C. Does this mean that the concentration of the solute is 1.0m?

Freezing Point Depression vapor pressure of solid and liquid are equal at

freezing point nonvolatile solute lowers the vapor pressure so a

lower T is needed to decrease the vapor pressure to that of the solid

a nonvolatile solute depresses the freezing point of the solvent

the amount of the depression depends on concentration of the solute

solutef mKT

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

Example 1

18.00 g of glucose are added to 150.0 g of water. The boiling point constant is 0.51 C*kg/mol.

What is the boiling point of the solution?

Example 2

What mass of C2H6O2 (M=62.1 g/mol) needs to be added to 10.0 L H2O to make a solution that freezes at -23.3°C?

density is 1.00 g/mL boiling point constant is

1.86°C*kg/mol

Exercise 13.9

1) Automotive antifreeze consists of ethylene glycol (C2H6O2), a nonvolative nonelectrolyte. Calculate the boiling point and freezing point of a 25.0 mass % solution of ethylene glycol in water.

2) 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.

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.

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.

van’t Hoff Factor

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

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.

van’t Hoff Factor

Reassociation is more likely at higher concentration.

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

van’t Hoff Factor

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

Tf = Kf m i

Exercise 13.10

1) List the following aqueous solutions in order of their expected freezing point: 0.050m CaCl2, 0.15m NaCl, 0.10m HCl, 0.050m HC2H3O2, and 0.10m C12H22O11.

2) Which of the following solutes will produce the largest increase in boiling point upon addition ot 1 kg of water: 1 mol Co(NO3)2, 2 mol KCl, or 3 mol of ethylene glycol (C2H6O2)?

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.

Semi-permeable Membrane a partition that allows solvent particles to pass

through but not solute particles separates a solution and a pure solvent

Osmosis the flow of solvent through a semi-

permeable membrane when the system has reached

equilibrium, the water levels are different since there is greater hydrostatic pressure in a solution

Osmosis

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

Osmotic Pressure

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

nV

= ( )RT = MRT

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.

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 crenation results.

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 hemolysis results.

Sample Exercise 13.11 Calculations 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?

Practice

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

P.533 GIST: Of two KBr solutions, one is 0.5m and the other is 0.20m, which is hypotonic with respect to the other?

Example 3

When 1.00x10-3 g of a protein is mixed with water to make 1.00 mL of solution, the osmotic pressure is 1.12 torr at 25.0°C.

Find the molar mass of this protein.

Sample Exercise 13.12 Molar Mass for Freezing-Point Depression

A solution of an unknown nonvolatile nonelectrolyte was prepared by dissolving 0.250 g of the substance in 40.0 g of CCl4. The boiling point of the resultant solution was 0.357 °C higher than that of the pure solvent. Calculate the molar mass of the solute.

Sample Exercise 13.12 Molar Mass for Freezing-Point Depression

Camphor (C10H16O) melts at 179.8 °C, and it has a particularly large freezing-point-depression constant, Kf = 40.0 °C/m. When 0.186 g of an organic substance of unknown molar mass is dissolved in 22.01 g of liquid camphor, the freezing point of the mixture is found to be 176.7 °C. What is the molar mass of the solute?

Practice Exercise

Sample Exercise 13.13 Molar Mass from Osmotic Pressure

The osmotic pressure of an aqueous solution of a certain protein was measured to determine the protein’s molar mass. The solution contained 3.50 mg of protein dissolved in sufficient water to form 5.00 mL of solution. The osmotic pressure of the solution at 25 °C was found to be 1.54 torr. Treating the protein as a nonelectrolyte, calculate its molar mass.

Practice

A sample of 2.05 g of polystyrene of uniform polymer chain length was dissolved in enough toluene to form 0.100 L of solution. The osmotic pressure of this solution was found to be 1.21 kPa at 25ºC. Calculate the molar mass of polystyrene.

13.6 Colloids

Colloids

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

Tyndall Effect

Colloidal suspensions can scatter rays of light.

This phenomenon is known as the Tyndall effect.

Colloids in Biological Systems

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

Colloids in Biological Systems

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

Sample Integrative Exercise Putting Concepts Together

A 0.100-L solution is made by dissolving 0.441 g of CaCl2(s) in water. (a) Calculate the osmotic pressure of this solution at 27 °C, assuming that it is completely dissociated into its component ions. (b) The measured osmotic pressure of this solution is 2.56 atm at 27 °C. Explain why it is less than the value calculated in (a), and calculate the van’t Hoff factor, i, for the solute in this solution. (See the “A Closer Look” box on Colligative Properties of Electrolyte Solutions in Section 13.5.) (c) The enthalpy of solution for CaCl2 is ΔH = –81.3 kJ/mol. If the final temperature of the solution is 27 °C, what was its initial temperature? (Assume that the density of the solution is 1.00 g/mL, that its specific heat is 4.18 J/g-K, and that the solution loses no heat to its surroundings.)