Chapter 13 Properties of Solutions Solutions Solutions are

Chapter 13 Properties of Solutions

Solutions • Solutions are homogeneous mixtures of two or more pure substances. • In a solution, the solute is dispersed uniformly throughout the solvent.

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 that in some processes, like the dissolution of NH 4 NO 3 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. It may have reacted.

Student, Beware! • 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.

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.

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

Factors Affecting Solubility • Vitamin A is soluble in nonpolar compounds (like fats). • Vitamin C is soluble in water.

Sample Exercise 13. 1 Predicting Solubility Patterns Predict whether each of the following substances is more likely to dissolve in the nonpolar solvent carbon tetrachloride (CCl 4) or in water: C 7 H 16, Na 2 SO 4, HCl, and I 2. Solve: C 7 H 16 is nonpolar. Na 2 SO 4, is ionic. HCl, is polar. I 2, is nonpolar. We would therefore predict that C 7 H 16 and I 2 (the nonpolar solutes) would be more soluble in the nonpolar CCl 4 than in polar H 2 O, whereas water would be the better solvent for Na 2 SO 4 and HCl (the ionic and polar covalent solutes).

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

Henry’s Law Sg = k. Pg 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.

Sample Exercise 13. 2 A Henry’s Law Calculation Calculate the concentration of CO 2 in a soft drink that is bottled with a partial pressure of CO 2 of 4. 0 atm over the liquid at 25 C. The Henry’s law constant for CO 2 in water at this temperature is 3. 4 10 2 mol/L-atm. Solution SCO 2 = k. PCO 2 = (3. 4 10 2 mol/L-atm)(4. 0 atm) = 0. 14 mol/L = 0. 14 M

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 O 2 dissolved in them than cool lakes.

Ways of Expressing Concentrations of Solutions

Mass Percentage mass of A in solution 100 Mass % of A = total mass of solution

Parts per Million and Parts per Billion Parts per million (ppm) mass of A in solution 106 ppm = total mass of solution Parts per billion (ppb) mass of A in solution 109 ppb = total mass of solution

Sample Exercise 13. 3 Calculation of Mass-Related Concentrations (a) A solution is made by dissolving 13. 5 g of glucose (C 6 H 12 O 6) 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 g of Zn 2+. What is the concentration of Zn 2+ in parts per million? Solution:

Mole Fraction (X ) moles of A XA = total moles of all components • In some applications, one needs the mole fraction of solvent, not solute—make sure you find the quantity you need!

Molarity (M ) moles of solute M= liters of solution • You will recall this concentration measure from Chapter 4. • Since volume is temperature-dependent, molarity can change with temperature.

Molality (m) moles of solute m= kilograms of solvent Since both moles and mass do not change with temperature, molality (unlike molarity) is not temperature-dependent.

Sample Exercise 13. 4 Calculation of Molality A solution is made by dissolving 4. 35 g glucose (C 6 H 12 O 6) in 25. 0 m. L of water at 25 C. Calculate the molality of glucose in the solution. Water has a density of 1. 00 g/m. L. Solution: Because water has a density of 1. 00 g/m. L, the mass of the solvent is: (25. 0 m. L)(1. 00 g/m. L) = 25. 0 g = 0. 0250 kg

Sample Exercise 13. 5 Calculation of Mole Fraction and Molarity 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.

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. 6 Calculation of Molarity Using the Density of the Solution A solution with a density of 0. 876 g/m. L contains 5. 0 g of toluene (C 7 H 8) and 225 g of benzene. Calculate the molarity of the solution. Solution:

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 = XAP A where – XA is the mole fraction of compound A, and – P A 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.

Ideal Solution PA = XA P 0 A PB = XB P 0 B PT = PA + PB PT = XA P A 0 + XB P 0 B

PT is greater than predicted by Raoults’s law PT is less than predicted by Raoults’s law Force < A-A & B-B A-B Force > A-A & B-B A-B

Sample Exercise 13. 7 Calculation of Vapor Pressure of a Solution Glycerin (C 3 H 8 O 3) is a nonvolatile nonelectrolyte with a density of 1. 26 g/m. L at 25 C. Calculate the vapor pressure at 25 C of a solution made by adding 50. 0 m. L of glycerin to 500. 0 m. L of water. The vapor pressure of pure water at 25 C is 23. 8 torr (Appendix B), and its density is 1. 00 g/m. L.

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.

Freezing-Point Depression Tf = Kf m Here Kf is the molal freezingpoint depression constant of the solvent (0 C/m). Tf = T 0 f – Tf T 0 f Tf 0 Tf is subtracted from the normal freezing point of the solvent. is the freezing point of the pure solvent is the freezing point of the solution T f > Tf > 0 m is the molality of the solution

Boiling-Point Elevation Tb = Kb m where Kb is the molal boilingpoint elevation constant (0 C/m). Tb = Tb – T b 0 is the boiling point of the pure solvent. T b is the boiling point of the solution. 0 Tb is added to the normal boiling point of the solvent. Tb > T b Tb > 0 m is the molality of the solution

Boiling-Point Elevation and Freezing-Point Depression Note that in both equations, T T = K m b b does not depend on what the solute is, but only on how many Tf = Kf m particles are dissolved.

Sample Exercise 13. 8 Calculating Of Boiling-Point Elevation and Freezing-Point Depression Automotive antifreeze consists of ethylene glycol, CH 2(OH), a nonvolatile nonelectrolyte. Calculate the boiling point and freezing point of a 25. 0 mass % solution of ethylene glycol in water. Tb = Kbm = (0. 51 C/m)(5. 37 m) = 2. 7 C Tf = Kfm = (1. 86 C/m)(5. 37 m) = 10. 0 C Boiling point = (normal bp of solvent) + Tb = 100. 0 C + 2. 7 C = 102. 7 C Freezing point = (normal fp of solvent) Tf = 0. 0 C 10. 0 C = 10. 0 C

Colligative Properties of Electrolytes Since the colligative properties of electrolytes 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 1 M solution of Na. Cl does not show twice the change in freezing point that a 1 M solution of methanol does.

van’t Hoff Factor One mole of Na. Cl 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 Na. Cl.

van’t Hoff Factor • Reassociation is more likely at higher concentration. • 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

Sample Exercise 13. 9 Freezing-Point Depression in Aqueous Solutions List the following aqueous solutions in order of their expected freezing point: 0. 050 m Ca. Cl 2, 0. 15 m Na. Cl, 0. 10 m HCl, 0. 050 m CH 3 COOH, 0. 10 m C 12 H 22 O 11. Because the freezing points depend on the total molality of particles in solution, the expected ordering is 0. 15 m Na. Cl (lowest freezing point), 0. 10 m HCl, 0. 050 m Ca. Cl 2, 0. 10 m C 12 H 22 O 11, and 0. 050 m CH 3 COOH (highest freezing point).

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.

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).

Osmotic Pressure The pressure required to stop osmosis, known as osmotic pressure, P, is P=( n V )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. 10 Calculating Involving Osmotic Pressure The average osmotic pressure of blood is 7. 7 atm at 25 C. What molarity of glucose (C 6 H 12 O 6) will be isotonic with blood? Solution:

Sample Exercise 13. 11 Molar Mass from 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 CCl 4. 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 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 m. L 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. Solution:

Colligative Properties of Ionic Solutions For ionic solutions we must take into account the number of ions present i = van’t Hoff factor or the number of ions present For vapor pressure lowering: P = i Xsolute. P 0 solvent For boiling point elevation: T b = i Kb m For freezing point depression: For osmotic pressure: im = conc. of particles T f = i Kf m = i MRT

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 a nonpolar, hydrophobic (waterhating) end.

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