Chemistry 1028/1029

CHAPTER 13

PROPERTIES OF SOLUTIONS (Please note that you have to attend lectures to complete this set of notes) Dr A Hart 1

SOLUTIONS • Solutions are two or more pure substances.

of

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

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FORCES IN SOLUTIONS • In a solution there are intermolecular forces between solute and solvent particles.

• In order for a solution to form, the intermolecular forces between solute and solvent particles must be strong enough to compete with:  interactions between solute particles (solute-solute interactions) and  interactions between solvent particles (solvent-solvent interactions). • The interaction between solvent and solute particles is called If water is the solvent this interaction is called

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EXAMPLES OF SOLUTIONS State of Solution Gas Liquid

State of Solvent Gas Liquid

State of Solute Gas Gas

Liquid Liquid Solid

Liquid Liquid Solid

Liquid Solid Gas

Solid Solid

Solid Solid

Liquid Solid

Example

Hydrogen in palladium Mercury in silver Silver in gold (alloy)

We are most concerned with liquid solutions.

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TYPES OF SOLUTIONS •  

• 

solution Solvent holds as much solute as is possible at that temperature. Dissolved solute is in dynamic equilibrium with solid solute particles.

solution Less than the maximum amount of solute is dissolved in the solvent for that temperature. 5

• Supersaturated  Solvent holds more solute than is normally possible at that temperature.  These solutions are unstable and crystallization can usually be stimulated by adding a “seed crystal” or scratching the side of the flask.

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EXAMPLE • The solubility of NaCl in water at 0 °C is 35.7 g NaCl per 100 ml of water.

• Therefore in 100 ml of water at 0 °C: 30 g NaCl 35.7 g NaCl 36 g NaCl

solution solution solution (i.e. unstable and at some point NaCl will precipitate) 7

FACTORS AFFECTING SOLUBILITY Three factors affect solubility: • Solute-solvent interactions • Pressure • Temperature

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SOLUTE-SOLVENT INTERACTIONS • The stronger the attractions between solute and solvent molecules, the the solubility. • Therefore substances with similar intermolecular forces tend to be soluble in each other.

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EXAMPLE Ethanol is soluble in water since they both have between their molecules.

ethanol

ethanol

ethanol

water 10

“LIKE DISSOLVES LIKE” Chemists use the axiom “like dissolves like”: • Polar substances tend to dissolve in solvents. Example: Ethanol (polar) dissolves in water (polar) but not in hexane (non-polar). • Non-polar substances tend to dissolve in solvents. Example: Hexane (non-polar) dissolves in oil (non-polar). • Ionic solutes are more likely to be soluble in 11 polar solvents.

MISCIBLE AND IMMISCIBLE LIQUIDS • Miscible liquids dissolve in all proportions. Example: Acetone or ethanol are miscible in water since they dissolve in all proportions. • Immiscible liquids do not dissolve significantly. Example: Oil is immiscible in water. Two layers form when oil is added to water. 12

EFFECT OF PRESSURE ON SOLUBILITY • The solubility of liquids and solids in liquid solvents is not greatly affected by pressure. • The solubility of a gas in a liquid depends on

Low Pressure (less gas dissolved)

High Pressure (more gas dissolved)

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HENRY’S LAW The solubility of a gas in a liquid is directly proportional to the of the gas above the liquid. Sg = kPg where • Sg is the solubility of the gas i.e. concentration of that gas in the solvent(molarity). • k is the Henry’s law constant for that gas in that solvent at a specific temperature. • Pg is the partial pressure of the gas above the liquid. 14

EXAMPLE • The solubility of pure nitrogen gas in water at 25 °C and 1 atm is 6.8 x 10-4 mol L-1. What is the concentration of nitrogen dissolve in water under atmospheric conditions when the partial pressure of nitrogen gas in the atmosphere is 0.78 atm?

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ANSWER • k is not given, so we have to work it out. Sg = kPg k = Sg = 6.8 x 10-4 mol L-1 = Pg 1 atm Therefore the solubility of N2 at 0.78 atm: Sg = (6.8 x 10-4 mol L-1atm -1 ) x (0.78 atm) =

Therefore the concentration of gas dissolved decreased with a decrease in pressure.

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PRACTICAL USE OF HENRY’S LAW • Bottlers use the effect of pressure on solubility in producing carbonated drinks such as beer and soft drinks. These are bottled under a CO2 pressure that is greater than 1 atm. • When the carbonated drink is opened, CO2 bubbles out of solution, because the CO2 partial pressure above the liquid is reduced and hence the solubility of CO2 decreases. 17

HENRY’S LAW AND DEEP-SEA DIVING • As a diver descends the pressure increases to 3 to 4 times that of atmospheric pressure and the solubility of gases in the diver’s blood increases. • If the diver ascends too fast, the solubility of the gases decreases so rapidly that they bubble in the blood vessels and blood flow is blocked. This is known as “the bends” or decompression sickness. It is characterised by locked joints, deafness, paralysis and even death. 18

• Nitrogen is the main problem in deep-sea diving because it has the highest partial pressure in air. • Deep-sea divers sometimes substitute helium for nitrogen in their tanks because helium has a much lower solubility than nitrogen.

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• The ideal partial pressure for oxygen is 0.2 atm (the normal partial pressure for oxygen in air at 1 atm). • Deep-sea divers use a gas mixture with less oxygen than there is in air so that at great depths the oxygen partial pressure is about 0.2 atm. • If the partial pressure of oxygen becomes too great, the urge to breath is reduced and carbon dioxide is not removed from the body. 20

LAKE NYOS DISASTER (CAMEROON) • 2000 people and many livestock were asphyxiated by carbon dioxide. • Lake Nyos is thermally stratisfied with layers of warm water on top of cold water layers, which are near the bottom. Very little mixing occurs normally. • CO2 has seeped into the lower cold water layers over thousands of years and has dissolved because of high pressure (Henry’s Law). 21

• Some unusual cooling took place and the lower layers became supersaturated with CO2. • A disturbance (possibly a landslide) caused the lower layers of water to rise to the surface, releasing a tremendous amount of CO2, suffocating humans and livestock. • This is an illustration of Henry’s Law that led to tragedy. 22

Temperature and Solubility of Solids Generally, the solubility of solid solutes in liquid solvents with increasing temperature.

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TEMPERATURE AND SOLUBILITY OF GASES In contrast, the solubility of gases in water decreases with an increase in temperature.  Carbonated soft drinks are more “bubbly” (more CO2 dissolves) if stored in the refrigerator.

 Warm lakes have less O2 dissolved in them than cool lakes. Poor dissolution of oxygen leads to fish suffocating. 24

WAYS OF EXPRESSING CONCENTRATIONS OF SOLUTIONS Concentration of a solution can be expressed: • Qualitatively Dilute Concentrated Small amount of solute

Large amount of solute

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• Quantitaively

• • • • • • •

Mass percentage Parts per million/billion (ppm, ppb) Mole fraction Molarity Molality Equivalents Osmolarity (see later) 26

Mass Percentage Mass % = mass of component (g) total mass (g) Examples 1) A solution of HCl that is 36% by mass is HCl in every of solution.

100

of

2) 0.25 g CaCO3 in a 1.5 g tablet: mass % CaCO3 = 0.25 g x 100 = 1.5 g 27

3) 3 g NaCl in 150 mL solution: Mass % NaCl = 3 g x 100 = 150 mL

4) 3 mL of ethanol in 150 mL solution: % ethanol = 3 mL x 100 = 150 mL Therefore % is parts per 100.

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PARTS PER MILLION AND PARTS PER BILLION • Parts per million (ppm) ppm =

mass of component (g) x 106 total mass of solution (g)

• Parts per billion (ppb) ppb = mass of component (g) x 109 total mass of solution (g) 29

EXAMPLE • 1 ppm = 1 mg solute 106 mg of solution But the density of water = 1 g mL-1 106 mg = 1 kg = 1000 g = 1000 mL = 1L 1 ppm = 1 mg solute 1 L of solution • Ppm and ppb are very useful when concentrations are very small. e.g. Arsenic is only allowed to be 0.01 ppm in water. i.e. 0.01 mg arsenic 1 L water 30

EXAMPLE Express the concentration of F– as ppm for a solution containing 0.35 g F– in 500 L of water.

density of water = 1 g mL-1 500 L = 500 kg = 500 x 103 g Conc F– = 0.35 g x 106 500 x 103 g + 0.35 g = 31

MOLE FRACTION (X)

X=

moles of component total moles of all components

• Mole fraction has no units.

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MOLARITY (M) moles of solute Molarity (M) = litres of solution • Molarity is concentration: c = n V • Units are : mol L-1, mol dm-3 or M

• Because volume is temperature dependent, molarity can change with temperature.

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MOLALITY (M) moles of solute Molality(m) = kg of solvent

• Units are mol kg-1 or molal (m) • Because both moles and mass do not change with temperature, molality (unlike molarity) is not temperature dependent. 34

EQUIVALENT (Eq) • 1 equivalent (Eq) = 1 mol of charges (positive or negative) • Example:

1 mol Na+ = 1 mol Mg2+ = 1 mol HCO3– = • Equivalent is the unit often used by doctors when referring to ionic components in blood (e.g. K +, Na+, PO43–). • Since the concentration of ions is low the mEq (1 x 10-3 Eq) is often used. Concentration is reported as mEq L-1. 35

EXAMPLE What is the mass percentage of iodine (I2) in a solution containing 0.045 mol I2 in 115 g CCl4? mass % = n=m M

mass solute x 100 total mass of solution mI = nM = 0.045 mol x 253.8 g mol-1 2 =

mass % =

11.421 g x 100 (11.421 + 115) g = (2 sig figs)

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EXAMPLE • What is the molality of a solution formed by dissolving 1.50 mol of KCl in 16.0 mol water? Molality = moles of solute kg of solvent n=m m H2O = nM = 16.0 mol x 18.02 g mol-1 M = Molality = 1.50 mol KCl 288.3 x 10-3 kg

= (

of KCl ) 37

EXAMPLE Vitamin C, ascorbic acid, (C6H8O6) is a water soluble vitamin. A solution containing 80.5 g ascorbic acid dissolved in 210 g of water has a density of 1.22 g mL-1 at 55 °C. Calculate: a) the mole fraction b) the molarity of ascorbic acid a)

nC6H8O6 = m = 80.5 g = M 176.1 g mol-1

n H2O = m = 210 g = M 18.02 g mol-1

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mole fraction: Xvit C = 0.4571 mol = (0.4571 + 11.7) mol vit C

water

b) Molarity = c = n V density = mass volume = mass = (210 + 80.5) g volume density 1.22 g mL-1 volume = 238.1 mL =

molarityVit C = 0.4571 mol 0.2381 L

= 39

CONVERSION OF CONCENTRATION UNITS • Sometimes the concentration of a given solution needs to be known in several different concentration units. • It is possible to interconvert concentration units. • See sample exercises in text book Chapter 13. 40

COLLIGATIVE PROPERTIES • A colligative property of a solution is a property that depends on the number (concentration) of solute particles present regardless of the nature of the solute. • A colligative property depends on the collective effect of all the solute particles. • Examples:  NaCl(s) Na+(aq) + Cl-(aq)  CaCl2(s) Ca2+(aq) + 2Cl-(aq)  Molecular solids, e.g. glucose, result in 41

WHAT ARE COLLIGATIVE PROPERTIES? • Colligative properties include:  Boiling point elevation.  Melting point depression.  Osmotic pressure.

Vapor pressure lowering (Not important for Chem 1028/29.)

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BOILING POINT ELEVATION The greater the number of solute particles in a solution, the the boiling point of the solution. The increase in boiling point is directly proportional to the (concentration) of the solute particles in the solution.

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FREEZING POINT DEPRESSION • The greater the number of solute particles in a solution, the lower the freezing point of the solution. • The decrease in freezing point is directly proportional to the molality (concentration) of the solute particles in the solution.

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EXAMPLES OF BOILING POINT ELEVATION AND FREEZING POINT DEPRESSION • Antifreeze (ethylene glycol, a nonelectrolyte) added to a car’s radiator lowers the freezing point and raises the boiling point. • Salt (NaCl) or CaCl2 is spread on roads in very cold countries to lower the freezing point of water and thus prevent ice from forming. 45

EXAMPLE List the following aqueous solutions in order of increasing boiling point: 0.120 M glucose, 0.050 M LiBr and 0.050 M Zn(NO3)2. For dilute solutions such as these, molarity is essentially the same as molality as the density of the solution is about the same as that of water (1 g mL-1).

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The more solute particles, the higher the bpt. • 0.120 M glucose - nonelectrolyte - 1 particle particle concentration = • O.050 M LiBr LiBr(s) Li+(aq) + Br-(aq) 0.050 M

0.050 M

0.050 M

particle concentration = • 0.050 M Zn(NO3)2 Zn(NO3)2(s) Zn2+(aq) + 2NO3-(aq) 0.050 M

0.050 M

(2 x 0.050 M)

particle concentration =

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boiling point of: O.050 M LiBr < 0.120 M glucose < 0.050 Zn(NO3)2 lowest bpt highest bpt

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OSMOSIS • In living systems, the most important colligative property is osmosis. • Osmosis is the movement of solvent molecules through a membrane from a less concentrated to a more concentrated solution i.e. from lower solute concentration to higher solute concentration. • A semipermeable membrane allows water to pass through it but NOT solute particles such as Na+, Cl-, glucose etc.

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OSMOSIS

Osmosis occurs and there is a net movement of solvent from the dilute solution to the concentrated solution.

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OSMOSIS • Osmosis stops when the column of solution on the left becomes high enough to exert sufficient pressure at the membrane to stop the net movement of solvent. • This pressure is the , . 51

OSMOTIC PRESSURE ( ) • The pressure required to stop osmosis, is known as osmotic pressure, Since, PV = nRT n =( V

)RT

n P = ( V )RT and

= MRT

where, M is the molarity of the solution (c = n/V). R is ideal gas constant. T is the temperature in kelvin. •

is a colligative property, dependent on the concentration of solute particles, not their nature.

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EXAMPLES OF OSMOSIS • Osmosis is used by trees to get water to their leaves. • Trees lose water to the atmosphere through their leaves. • The solute concentration in the leaves increases. • This generates an osmotic pressure within the tree that forces water from the soil up to the leaves.

• Making pickles. • Brine (salt solution) is used to make pickles. • Water flows out of the cucumbers into the more concentrated brine and the cucumbers shrivel. 53

TYPES OF SOLUTIONS When dealing with osmosis, there are three types of solutions we have to consider: • • •

solutions solutions solutions

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HYPERTONIC SOLUTIONS • A hypertonic solution is one that has a solute concentration than a solution on the other side of a membrane. Example: • If the solute concentration of a solution outside a red blood cell is greater than that inside the cell, the solution is hypertonic. • Water will flow out of the cell, and crenation results. 55

HYPOTONIC SOLUTIONS • A hypotonic solution is one that has a solute concentration than a solution on the other side of a membrane. Example • If the solute concentration of a solution outside a red blood cell is less than that inside the cell, the solution is hypotonic. • Water will flow into the cell and haemolysis (rupture of the cell) results.

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ISOTONIC SOLUTIONS • Solutions are isotonic if the osmotic pressure is the same on both sides of a membrane. • It means the concentrations of both solutions are the • Osmosis does not occur i.e. there is no net flow of water into or out of the cell. • Example: The concentration of salts in normal blood plasma is isotonic to 0.9% NaCl i.e. saline solution. 57

USES OF HYPOTONIC AND HYPERTONIC SOLUTIONS • Hypotonic solutions can be used for oral hydration. e.g. 0.45 % NaCl. Solute concentration is lower than in normal blood plasma. • Hypertonic solutions can be used on patients with excessive water retension. e.g. 5 – 10 % dextrose or 5 % dextrose in 0.45 % NaCl. Solute concentration is higher than in normal blood plasma. 58

DIALYSIS • Most animal cell membranes are dialysing membranes. • The following pass through these membranes:  water  ions (Na+, Cl-, K +)  nutrients  waste products • Larger particles like proteins, cells or colloids (blood) do not pass through them. 59

EXAMPLE OF DIALYSIS • A dialysis machine is used on patients with kidney damage. (5-7 h, once a week). blood out

blood in (colloid)

Waste products diffuse out at a greater rate than in, resulting in purification

dialysis tube (cellophane) isotonic solution

Isotonic solutes diffuse in/out at same rate

(isotonic for all components that must remain in the blood 60

Osmolarity • Osmolarity is the measure of the number of moles of osmotically active solute particles per unit volume of solution, i.e. osmoles of solute per litre of solution (osmol L-1). • The osmolarity of a solution is usually expressed as Osm (pronounced "osmolar"). osmolarity = molarity x number of particles • The unit milliosmole, mosmol, is also used.

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Osmolarity Examples • 2 M glucose osmolarity = 2 M x 1 particle = • 2 M NaCl osmolarity = 2 M x 2 particles =

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EXAMPLE What is the osmolarity of a 0.9% (w/v) NaCl solution?

0.9% (w/v) NaCl = nNaCl =

0.9 g NaCl 100 mL solution =

0.9 g 56.4 g mol-1 molarity = n = 1.5 x 10-2 mol = V 100 x 10-3 L NaCl Na+ + Closmolarity = 0.15 M x 2 particles = = 300 mosmol L-1 Note: Human blood has an osmolarity of 300 mosmol L-1. 63

COLLOIDS • Colloids are suspensions of particles larger than individual ions or molecules, but too small to be settled out by gravity. • Thus colloids are mixtures of particles that are larger than those forming solutions but smaller than the particles forming suspensions. 64

TYPES OF COLLOIDS Colloids are classified by the solvent (dispersing medium) and by the colloidal matter (dispersed medium). They can be gas liquid or solid Phase of Colloid

Dispersing Dispersed Colloid Type (solventlike) (solutelike) substance substance

Example

Gas

Gas

Gas

-

None (all are solutions)

Gas

Gas

Liquid

Aerosol

Gas

Gas

Solid

Aerosol

Smoke

Liquid

Liquid

Gas

Foam

Whipped cream

Liquid

Liquid

Liquid

Emulsion

Liquid

Liquid

Solid

Sol

Paint, Blood

Solid

Solid

Gas

Solid Foam

Marshmallow

Solid

Solid

Liquid

Solid Emulsion

Solid

Solid

Solid

Solid sol

Ruby glass

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PROPERTIES OF COLLOIDS • Colloidal particles range in size from 10 – 2000 Å. (1 Å = 1 x 10-10 m) • Colloids cannot be separated by ordinary filter paper. • Colloids exhibit Brownian motion i.e. the random movement of particles caused by their bombardment by solvent molecules. • Colloids scatter light. This is known as the Tyndal effect. Example: Light streaming through a forest - colloidal particles (dust) in the air scatter the light.

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HYDROPHILIC AND HYDROPHOBIC COLLOIDS • The most important colloids are those in which the dispersing medium is water. • Some colloidal molecules have a polar, a hydrophilic (water-loving) end and a nonpolar, hydrophobic (water-hating) end.

hydrophilic end

hydrophobic end

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HYDROPHILIC (WATER-LOVING) COLLOIDS

• In biological systems these are often enzymes and antibodies that must not settle out of solution • These molecules fold in such a way that the hydrophilic, polar groups are on the surface. And can interact with water by hydrogen bonding. • Blood and gelatin are hydrophilic colloids. hydrophilic groups

O H H water

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HYDROPHOBIC (WATER-HATING) COLLOIDS • These colloids are not stable in water. • They can be stabilised by the adsorption (adhering) of ions to their surface. • These adsorbed ions can interact with water and stabilise the colloid.

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SURFACTANTS • Another way to keep hydrophobic particles, like grease or oil, in solution is to use a surfactant like a soap or detergent, e.g. sodium stearate. • Surfactants have a hydrophobic end and a hydrophilic end. • Stabilisation results from the interaction of the end with the oil droplet and the hydrophilic end with 70

BIOLOGICAL APPLICATION • When fats reach the small intestine, the gall bladder secretes bile, which contains components very similar to sodium stearate. • These compounds have a hydrophilic (polar) end and a hydrophobic (nonpolar) end. • They emulsify the fats that are present in the intestine and therefore facilitate digestion and absorption of fat soluble vitamins. • Emulsification is the formation of a suspension of one liquid in another. • End of Section

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