Chapter 13 Lecture on Solutions & Colligative Properties

Chapter 11- Properties of Solutions Sections 13.1 - 13.3 Dissolving Solubility Read pg 529 – 543 pg 564 #1, 2, 3, 4, 6, 12, 13, 17, 19, 22, 25, 27, 29, 31, 84

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.

From weakest to strongest, rank the following solutions in terms of solvent – solute interactions: NaCl in water, butane (C 4 H 10 ) in benzene (C 6 H 6 ), water in ethanol. NaCl in water < C 4 H 10 in C 6 H 6 < water in ethanol Water in ethanol < NaCl in water < C 4 H 10 in C 6 H 6 C 4 H 10 in C 6 H 6 < water in ethanol < NaCl in water

Correct Answer: Butane in benzene will have only weak dispersion force interactions. Water in ethanol will exhibit much stronger hydrogen-bonding interactions. However, NaCl in water will show ion – dipole interactions because NaCl will dissolve into ions. NaCl in water < C 4 H 10 in C 6 H 6 < water in ethanol Water in ethanol < NaCl in water < C 4 H 10 in C 6 H 6 C 4 H 10 in C 6 H 6 < water in ethanol < NaCl in water

Energy Changes in Solution Simply put, three processes affect the energetics of the process: 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.

Water vapor reacts with excess solid sodium sulfate to form the hydrated form of the salt. The chemical reaction is Does the entropy increase or decrease? SAMPLE EXERCISE 13.1 Assessing Entropy Change

Answer: The entropy increases because each gas eventually becomes dispersed in twice the volume it originally occupied. The water vapor becomes less dispersed (more ordered). When a system becomes more ordered, its entropy is decreased . PRACTICE EXERCISE Does the entropy of the system increase or decrease when the stopcock is opened to allow mixing of the two gases in this apparatus?

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

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.

Types of Solutions Saturated 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 Less than the maximum amount of solute for that temperature is dissolved in the solvent.

Types of Solutions Supersaturated 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.

Solution C 7 H 16 is a hydrocarbon, so it is molecular and nonpolar. Na 2 SO 4 , a compound containing a metal and nonmetals, is ionic; HCl, a diatomic molecule containing two nonmetals that differ in electronegativity, is polar; and I 2 , a diatomic molecule with atoms of equal electronegativity, is nonpolar. We would therefore predict that C 7 H 16 and I 2 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. SAMPLE EXERCISE 13.2 Predicting Solubility Patterns Predict whether each of the following substances is more likely to dissolve in carbon tetrachloride (CCl 4 ) or in water: C 7 H 16 , Na 2 SO 4 , HCl, and I 2 .

Answer: C 5 H 12 < C 5 H 11 Cl < C 5 H 11 OH < C 5 H 10 (OH) 2 (in order of increasing polarity and hydrogen-bonding ability) SAMPLE EXERCISE 13.2 continued PRACTICE EXERCISE Arrange the following substances in order of increasing solubility in water :

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 S g = kP g where S g is the solubility of the gas; k is the Henry’s law constant for that gas in that solvent; P g is the partial pressure of the gas above the liquid.

At a certain temperature, the Henry’s law constant for N 2 is 6.0  10  4 M /atm. If N 2 is present at 3.0 atm, what is the solubility of N 2 ? 6.0  10  4 M 1.8  10  3 M 2.0  10  4 M 5.0  10  5 M

Correct Answer: Henry’s law, S g = kP g S g = ( 6.0  10  4 M /atm)(3.0 atm) S g = 1.8  10  3 M 6.0  10  4 M 1.8  10  3 M 2.0  10  4 M 5.0  10  5 M

SAMPLE EXERCISE 13.3 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 3.1  10 –2 mol/L-atm. Solve: Check: The units are correct for solubility, and the answer has two significant figures consistent with both the partial pressure of CO 2 and the value of Henry’s constant. 2

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.

Chapter 11- Properties of Solutions Section 13.4 Ways of Expressing Concentrations of Solutions Read pg 529 – 543 pg 564 #1, 2, 3, 4, 6, 12, 13, 17, 19, 22, 25, 27, 29, 31, 84

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

Determine the mass percentage of hexane in a solution containing 11 g of butane in 110 g of hexane. 9.0 % 10. % 90.% 91 %

Correct Answer: Thus, 110 g (110 g + 11 g) 100 solution of mass total solution in component of mass component of % mass    100 = 91% 9.0 % 10. % 90.% 91 %

Parts per Million and Parts per Billion ppm =  10 6 Parts per Million (ppm) Parts per Billion (ppb) ppb =  10 9 mass of A in solution total mass of solution mass of A in solution total mass of solution

If 3.6 mg of Na + is detected in a 200. g sample of water from Lake Erie, what is its concentration in ppm? 7.2 ppm 1.8 ppm 18 ppm 72 ppm

Correct Answer: 6 10 solution of mass total solution in component of mass component of ppm   ppm 18 10 g 200. g 0.0036 g 200. mg 3.6 6    7.2 ppm 1.8 ppm 18 ppm 72 ppm

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

Molarity ( M ) You will recall this concentration measure from Chapter 4. Because volume is temperature dependent, molarity can change with temperature. mol of solute L of solution M =

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

What is the molality of 6.4 g of methanol (CH 3 OH) dissolved in 50. moles of water? 0.040 m 0.22 m 0.064 m 0.11 m

Correct Answer: 0.040 m 0.22 m 0.064 m 0.11 m

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 (C 6 H 12 O 6 ) in 25.0 mL of water at 25°C. Calculate the molality of glucose in the solution. Solution molar mass of glucose, 180.2 g/mol water has a density of 1.00 g/mL, so the mass of the solvent is

Chapter 11- Properties of Solutions Section 13.5 & 13.6 Colligative Properties & Colloids Read pg 529 – 543 pg 564 #1, 2, 3, 4, 6, 12, 13, 17, 19, 22, 25, 27, 29, 31, 84

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 P A = X A P  A where X A is the mole fraction of compound A 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 .

At a certain temperature, water has a vapor pressure of 90.0 torr. Calculate the vapor pressure of a water solution containing 0.080 mole sucrose and 0.72 mole water. 9.0 torr 10. torr 80. torr 81. torr 90. torr

Correct Answer: total i P P i   P i = X i P total P i = (0.72 mol/[0.72 + 0.080 mol])(90.0 torr) P i = (0.90)(90.0 torr) = 81. torr 9.0 torr 10. torr 80. torr 81. torr 90. torr

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.

Boiling Point Elevation The change in boiling point is proportional to the molality of the solution:  T b = K b  m where K b is the molal boiling point elevation constant, a property of the solvent.  T b is added to the normal boiling point of the solvent.

Ethanol normally boils at 78.4 ° C. The boiling point elevation constant for ethanol is 1.22 ° C/ m . What is the boiling point of a 1.0 m solution of CaCl 2 in ethanol? 77.2 ° C 79.6 ° C 80.8 ° C 82.1 ° C 83.3 ° C

Correct Answer: The increase in boiling point is determined by the molality of total particles in the solution. Thus, a 1.0 m solution of CaCl 2 contains 1.0 m Ca 2+ and 2.0 m Cl  for a total of 3.0 m . Thus, the boiling point is elevated 3.7 °C, so it is 78.4°C + 3.7°C = 82.1°C. m K T b b   77.2 ° C 79.6 ° C 80.8 ° C 82.1 ° C 83.3 ° C

Freezing Point Depression The change in freezing point can be found similarly:  T f = K f  m Here K f is the molal freezing point depression constant of the solvent.  T f is subtracted from the normal freezing point of the solvent.

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.  T b = K b  m  T f = K f  m

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 1 M solution of NaCl does not show twice the change in freezing point that a 1 M 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 a little less than two times the concentration of NaCl.

The van’t Hoff Factor Reassociation is more likely at higher concentration. Therefore, the number of particles present is concentration dependent.

The van’t Hoff Factor We modify the previous equations by multiplying by the van’t Hoff factor, i  T f = K f  m  i

PRACTICE EXERCISE Which of the following solutes will produce the largest increase in boiling point upon addition to 1 kg of water: 1 mol of Co(NO 3 ) 2 , 2 mol of KCl, 3 mol of ethylene glycol (C 2 H 6 O 2 )? 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

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 are of lower solvent concentration ( higher solute concentration ).

Osmotic Pressure The pressure required to stop osmosis, known as osmotic pressure ,  , is 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 . n V  = ( ) RT = MRT

At 300 K, the osmotic pressure of a solution is 0.246 atm. What is its concentration of the solute? 1.0 M 0.50 M 0.25 M 0.10 M

Correct Answer: M =  / RT M = (0.246 atm)/[(0.0821 L-atm/mol-K)(300K)] M = (0.246 atm)/(0.2463 L-atm/mol) = 1.0 M MRT RT V n    1.0 M 0.50 M 0.25 M 0.10 M ( )

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.

PRACTICE EXERCISE What is the osmotic pressure at 20°C of a 0.0020 M sucrose (C 12 H 22 O 11 ) solution? Answer: 0.048 atm, or 37 torr SAMPLE EXERCISE 13.11 Calculations Involving Osmotic Pressure The average osmotic pressure of blood is 7.7 atm at 25°C. What concentration of glucose (C 6 H 12 O 6 ) will be isotonic with blood? Solution Plan: Given the osmotic pressure and temperature, we can solve for the concentration, using Equation 13.13. Solve:

Molar Mass from Colligative Properties We can use the effects of a colligative property such as osmotic pressure to determine the molar mass of a compound.

Solution Plan: K b for the solvent (CCl 4 )= 5.02ºC/ m .  T b = K b m . SAMPLE EXERCISE 13.12 Molar Mass from Freezing-Point Depression A solution of an unknown nonvolatile electrolyte 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. Solve: The solution contains 0.0711 mol of solute per kilogram of solvent. The solution was prepared using 40.0 g = 0.0400 kg of solvent (CCl 4 ). The number of moles of solute in the solution is therefore

PRACTICE EXERCISE Camphor (C 10 H 16 O) melts at 179.8°C, and it has a particularly large freezing-point-depression constant, K f = 40.0ºC/ 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? Answer: 110 g/mol

SAMPLE EXERCISE 13.13 Molar Mass from Osmotic Pressure The osmotic pressure of an aqueous solution of a certain protein was measured in order 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. Calculate the molar mass of the protein. Solution Plan: The temperature ( T = 25ºC) and osmotic pressure (  = 1.54 torr) are given, and we know the value of R so we can use Equation 13.13 to calculate the molarity of the solution, M . In doing so, we must convert temperature from °C to K and the osmotic pressure from torr to atm. We then use the molarity and the volume of the solution (5.00 mL) to determine the number of moles of solute. Finally, we obtain the molar mass by dividing the mass of the solute (3.50 mg) by the number of moles of solute. Solve: Solving Equation 13.13 for molarity gives Because the volume of the solution is 5.00 ml = 5.00  10 –3 L, the number of moles of protein must be

Comment: Because small pressures can be measured easily and accurately, osmotic pressure measurements provide a useful way to determine the molar masses of large molecules. PRACTICE EXERCISE 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 25°C. Calculate the molar mass of the polystyrene. Answer: 4.20  10 4 g/mol SAMPLE EXERCISE 13.13 continued The molar mass is the number of grams per mole of the substance. The sample has a mass of 3.50 mg = 3.50  10 –3 g. The molar mass is the number of grams divided by the number of moles:

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

Tyndall Effect Colloidal suspensions can scatter rays of light. This phenomenon is known as the Tyndall effect.

Which of the following is not an example of a colloid? Fog Smoke Paint Milk Carbonated water

Correct Answer: Carbonated water is a solution; all the other substances in the list are excellent examples of colloids. Fog Smoke Paint Milk Carbonated water

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

Colloids in Biological Systems Sodium stearate is one example of such a molecule.

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

Chapter 13 Lecture on Solutions & Colligative Properties

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Chapter 13 Lecture on Solutions & Colligative Properties