Chapter 13. Physical Properties of Solutions

Transcription

1 Chapter 13 Overview Composition of Matter All matter can be divided into two major classes which are pure substance and mixture. The pure substance could be Element or could be compound. The mixture can be either Homogeneous mixture or heterogeneous as in figure 1. Matter solid, liquid gas Pure Substance constant composition Physical methods Separating mixture Preparing mixture Mixture variable composition Elements Compounds Homogeneous Heterogeneous Metals Non -metals Semimetals (metalloids) Fig. 1 Composition of Matter Homogeneous Mixture: The prefixes "homo"- indicate sameness A homogeneous mixture has the same uniform appearance and composition throughout. Many homogeneous mixtures are commonly referred to as solutions as example dissolving sugar in H 2 O. Heterogeneous Mixture: The prefixes: "hetero"- indicate difference. the individual components of a mixture remain physically separated and can be seen as separate components, for example mixing sand with H 2 O 1

2 What is a Solution? A solution is a homogeneous mixture of two or more substances dissolved in each other. Solutions mainly consist of two components one is called solvent and the other is called solute. Solvent: the substance in which the substances (solute) are dissolves to produce a homogeneous mixture (solution) and it is the substance that present in large amount. Solute: the substances that dissolves in a solvent to produce a homogeneous mixture Although we shall concentrate mainly on liquid solutions, solutions can be solids, liquids or gases. Classification of solution: Solutions can be classified on the basis of their state: solids, liquids or gases. Examples of various ways of preparing a two component solution in each of the three states are summarized in the following table. Solution type solvent solute Example Solid solution Solid Solid Metalloids, copper dissolved in nickel Solid Gas Hydrogen dissolved in palladium Solid Liquid Mercury dissolved in gold Gas solution Gas Gas Oxygen dissolved in Nitrogen (air) Gas Solid Dry ice dissolved in (sublimed) nitrogen Gas Liquid Chloroform dissolved in (evaporated into) nitrogen Liquid solution Liquid Solid Sugar dissolved in water Liquid Gas Carbon dioxide dissolved in water (soft drinks) Liquid Liquid Ethanol dissolved in water 2

3 In order to form a liquid solution in which the solute and solvent are both liquids, the two liquids must be miscible. Miscible liquids mean that the two liquids can dissolve completely in each other at any proportions نسبھ) (با ي as example water and ethanol. In the other hand immiscible means they can t dissolve in each other as example oil and water. Types of Solutions Well, as we have seen in the above section that solution are formed by mixing solute and solvent with each other. In order to such as this process to takes place the molecules of the solute and the solvent must get in direct contact with each other leading to solubility of the solute in the solvent and forming the solution. The ratio of solute and its solubility in the solvent as well as the amount of the solvent used determines the nature (type) of the produced solution. Solubility: solubility is defined as the maximum amount of substance (solute) that will dissolve in a solvent at that temperature. Usually solubility is measured by g/l. Chemist usually classifying solution according to the capacity of solvent to dissolve a solute at a specific temperature; in other words solution can be classified according to the ratio between solute and solvent at a given temperature. This classification led to the following three groups of solution. 1) Saturated solution Contains the maximum amount of solute dissolved in solvent. In other words; when the solvent has dissolved all the solute it can, and left some un-dissolved solute it s a saturated solution. For example; if you add 20 g of sugar to 100mL of water at room temperature, then all of the sugar will dissolved. However if you add 300g to the same amount of water, some of the sugar remains un-dissolved. We said this solution is saturated. In saturated solution there is equilibrium between solution and solute that mean if you for example keep string the solution some of the un-dissolved solute will dissolve and in the same time some of dissolved solute will come out so they are in equilibrium. 2) Unsaturated solution An unsaturated solution contains less of solute than it could hold i.e. solutions can dissolve more solute under normal condition. For example a saturated solution of 3

4 NaCl/H 2 O contains 25g of NaCl in 100mL water; the unsaturated solution of NaCl/H 2 O will contain less than 25g NaCl in 100mL water. In an unsaturated solution there is no equilibrium exists between the solute and the solution because there is no undissolved solute 3) Supersaturated solution: A solution that is temporarily holding more than it can, a seed crystal will make it come out. A saturation solution can be prepared by adding excess solute to a saturated solution. Unsaturated solutions are not stable solutions, for example if just one crystal of solute is added, or the liquid is shaken, the excess solute will crystallize immediately and saturated solution will form. Crystallization: The process in which dissolved solute comes out of solution and form crystals. Crystals: Crystals are solids that form by a regular repeated pattern of atoms or molecules connecting together. In crystal the arrangements of the building blocks (atoms and molecules) lie in an orderly array. In crystals, however, a collection of atoms called the Unit Cell is repeated in exactly the same arrangement over and over throughout the entire material. Diamond the very expensive jewel which is 3D جزیي ( molecule covalently bonded network of carbon atoms. Diamond is a gigantic properties. with crystalline structure which gives it its unique (ضخم 4

5 Aqueous solutions An aqueous solution is a solution in which the solvent is water. It is usually shown in chemical equations as a subscript (aq). The word aqueous means pertaining to, related to, similar to, or dissolved in water. As water is an excellent solvent as well as naturally abundant, it logically has become a ubiquitous solvent in chemistry. Substances that do not dissolve well in water are called hydrophobic ('water fearing') whereas those that do are known as hydrophilic ('water-loving'). An example of a hydrophilic substance would be the sodium chloride (ordinary table salt). H 2 O Na + Cl - Na + Na + Cl - solute H 2 O Cl - A Molecular View of the Solution Process In the previous sections we have discussed the formation of solutions, and you learned that solutions are formed by dissolving a solute in a solvent. Now we need to learn how the dissolving process takes place. To understand this process we need to remember what we have learned about the intermolecular forces and intramolecular forces (Chapter 12), however these forces play very important rolls in interpretation the solubility and solution formation. (Please go back to chapter 12) To make a solution the solute and solvent are mixed together, and in order to obtain a solution the solute molecules will dissolve in the solvent molecules that means the solute molecules will take the positions of some solvent molecules. What happens when you add a solid solute to a liquid solvent? Immediately after the addition of a solid solute to a liquid solvent, the solid state structure begins to disintegrate (یتحلل) little by little. The solvent molecules chip away at the surface of the crystal lattice, prying out solute particles and surrounding them, and finally dispersing them throughout the body of the solution. This process 5

6 (solubility) depends on the relative strength of three attractive forces (intermolecular forces) which are: 1) Solvent-solvent interaction (before dissolution) 2) Solute solute interaction (before dissolving) 3) Solute- solvent interaction (during the dissolving process) When the solution is formed i.e. for the solute to dissolve in the solvent both the solvent solvent intermolecular force and solute solute intermolecular forces will be replaced (تستبدل) by the solute- solvent interaction (see figure 3). From this disscation, we can conclude (نستنتج) that if the solvent solvent intermolecular force and solute solute intermolecular forces are weak, and the solute- solvent interaction is strong the (العكس صحیح) versus. formation of the solution will be easy and vies there is intermolecular force between the solvent molecules solvent -solvent interaction there is intermolecular force between the solvent molecules solute-soluteinteraction In solution, there is intermolecular interaction between solve and solute. The old solvent -solvent interaction and solute-soluteinteraction disappear liquid solvent solid solute solution Fig. 3 Mechanism of dissolving solid solute in liquid solvent Formation of solution & energy changes The solubility process of solid solute in a liquid solvent is companied with changes in the energy of both the solute, solvent and the formed solution. As we have just learned that for a solid solute to dissolve in a liquid solvent the intermolecular force (interaction) between the solvent solvent, and solute solute must be destroyed. In the same time a new interaction between solute- solvent start to take place. In both cases the energy cotenants of the solute and solvent and solution is changed. To make 6

7 this clear and easy; let us assume that the formation of a solution to takes place in three steps as the following. a) Separation of the solvent solvent interaction b) Separation of the solute solute interaction c) Formation of new interaction between solute- solvent In both steps a and b the process need energy in order to overcome the intermolecular force (interaction) between the solvent molecules alone and between the solute molecules alone. Steps a & b are endothermic steps. In any process or a chemical reaction if energy (heat) is absorbed, the process is said to be endothermic process or reaction. In the other hand, in step 3 the solute and solvent are mixed together, in terms of energy; this process can be either endothermic process or can be exothermic process. In any process or a chemical reaction if energy (heat) is released the process is said to be exothermic process or chemical reaction. The total change of energy during the dissolving process is the summation of each change in each step i.e. a and b and c in our example. The total change of heat is known as the heat of solution H or enthalpy H of solution (reaction). For steps a, b and c the H sol = H a + H b + H c If H is positive ( H > zero) that indicates that energy (heat) is absorbed i.e. endothermic process or reaction. If H is negative ( H < zero) that indicates that energy (heat) is released i.e. exothermic process or reaction. When the value of H sol is positive and when it is negative? v If the interaction between solute and solvent is stronger than both the interaction between solvent-solvent molecules and the interaction between solute-solute molecules the value of H sol will be negative ( H sol < zero) and the process is exothermic. In this the solubility process is favourable and the solid solute will dissolve easily (easily here means that there is no great energy deficit) in the liquid solvent to form a solution. v If the interaction between solute and solvent is weaker than both the interaction between solvent-solvent molecules and the interaction between 7

8 solute-solute molecules the value of H sol will be positive ( H sol > zero) and the process is endothermic. In this the solubility process is unfavourable and the solid solute will not dissolve or slightly dissolved in the liquid solvent to form a solution. Figure 5 is explaining the energy changing during the dissolving process. H step d positive value endothermic solubility is unfavourable potential energy step a step b expanded solvent expanded solute step c solvent expanded solute net energy change negative value exothermic H solvent solute solution Fig. 5 The energy changes during the dissolving process Input of energy is required to separate solute-solute (step a) and solvent-solvent (step b) attractions. Energy is released due to solute-solvent attractions (step c). If the amount of energy released in step c is greater than the energy absorbed in steps a and b the process is exothermic ( H sol is negative) and favoured for dissolution, but if the amount of energy in step c is less than the absorbed energy in steps a and b the process is endothermic ( H sol is positive) and un-favoured as in step d in figure 5. 8

9 In addition to the energy factor, there is another factor that effect on formation of solution which is the tendency toward disorder. When solute and solvent molecules are mixed together; to form a solution, there is an increase in randomness or disorder. In the pure state, the solvent and solute have enough degree of order, this order is destroyed when the solute dissolve in the solvent. Therefore, the solution process is accompanied by an increase in disorder. This increasing in disorder of the system is favouring the solubility of any substance. Figure 6 is explaining the above discussion. Does any solute dissolve in any solvent? The direct and simple answer to this question is NO. There are so many solutes that do not dissolve in particular solvent. However; solubility of any solute in a solvent depends mainly on the type of intermolecular force of both solutes and solvents. 9

10 To understand why some solutes do not dissolve in some solvents let us learn a general saying which says like dissolves like. The meaning of this says is that if any substances in this case (solute & solvent) have the same type of intermolecular forces they will likely dissolve in each other. In other words if two or more substances have the same polarity they most likely will dissolve in each other. Solvents and solutes can be broadly classified into polar and non-polar. A molecule is polar if there is some separation of charge in the chemical bonds, so that one part of the molecule has a slight positive charge and the other a slight negative charge. The slight negative and positive charges are arising as a result of the difference in the electronegativities between the elements that form the bond. Polarity can be measured as the dielectric constant or the dipole moment of a compound. Water is a well-known example of a polar molecule, it consist of two H atoms covalently bonded to oxygen O atoms (covalent bond). H 2 O is a polar because there is difference in electronegativities between the H atoms and the O atom (figure 7) Fig. 7 H 2 O is polar molecule The dipole moment (µ) of water is 1.85 deby. The value of µ is calculated using the following equation Where µ is dipole moment, q is the charge and r is the distance between the two charges. Table 1 show some compounds and dielectric constant and dipole moment (µ) for the values. 10

11 Compound Dielectric Constant Dipole Moment (m ) Formamide Water Methanol Benzene Table 1 Please note that benzene has µ equal zero i.e. it is non polar compound. There are no polar bonds in benzene as it contains covalent bond between only H and C which have nearly similar electronegativities (2.2 for H and 2.55 for C). The polarity of a solvent determines what type of compounds it is able to dissolve and with what other solvents or liquid compounds it is miscible. As a rule of thumb, polar solvents dissolve polar compounds BEST and non-polar solvents dissolve nonpolar compounds BEST: "like dissolves like". Strongly polar compounds (ionic compound) like inorganic salts (e.g. table salt NaCl) dissolve only in very polar solvents like water, while strongly non-polar compounds like oils or waxes dissolve only in very non-polar organic solvents like hexane. Similarly, water and hexane are not miscible with each other and will quickly separate into two layers even after being shaken well. Table 12 in the next page has a list of common solvents both polar and non-polar solvent with the values of dielectric constant and dipole moment. You should be able to tell a good thing about the solubility of any solvent with another solvent or the solubility of any solute (you need to know its structure) in any given solvent. A compound, such as water, that is composed of polar molecules. Polar solvents can dissolve ionic compounds or covalent compounds that ionize. Non-polar solvents, such as benzene, will only dissolve non-polar covalent compounds. Now you should be able to expect if a given solute will dissolve in a given solvent or will not. Of course; no need to memorize the values of the dielectric constant and dipole moment and the electronegativities or any other numbers. However all what you need is to use your understanding to the concept of polarity and to understand the classification of the periodic table as this will help you to obtain a good judgment about the solubility of compounds in each other. 11

12 Table 2 Common polar and non-polar Solvents Name Structure bp, o C dipole moment dielectric constant water H-OH methanol CH 3 -OH ethanol CH 3 CH 2 -OH propanol CH 3 CH 2 CH 2 -OH butanol CH 3 CH 2 CH 2 CH 2 - OH formic acid acetic acid formamide acetone tetrahydrofuran (THF) methyl ethyl ketone ethyl acetate acetonitrile N,N-dimethylformamide (DMF) diemthyl sulfoxide (DMSO) hexane CH 3 (CH 2 ) 4 CH benzene diethyl ether CH 3 CH 2 OCH 2 CH methylene chloride CH 2 Cl carbon tetrachloride CCl

13 Now, let us see how ionic compounds like NaCl dissolve in polar solvent like H 2 O? Ionic compounds form ionic solutions. Sodium chloride when dissolve in water forms an ionic solution. Such ionic solutions conduct electricity. The solutions are called electrolytes. Table salt or sodium chloride (NaCl) is an ionic compound. Sodium is in the first group of the periodic table and chlore is in the group 7. The difference in their electronegativities indicates that it is an ionic compound. Ionic compound do not form hydrogen bonding with water. Salt is a solid at room temperature. The sodium and chloride atoms are arranged in a pattern that we called crystal lattice. This process is described on steps as following; 1) When we put salt in water, the sodium and chloride atoms are released from their crystalline lattice and form positive sodium ions (Na + ) & negative chloride ions (Cl - ). The (Na + ) and (Cl - ) are free to move within the water i.e. over come the Intramolecular forces. Fig. Formation of Na + and Cl - in water 13

14 2) Water is a polar solvent, it consist of 2 H atoms and one O atoms. These three atoms are covalently bonded to each other (Intramolecular forces). As a result of the difference in electronegativities between the O and H, a partial negative charge will arise on the O atoms and a partial positive charge will arise on the two H atoms. The partial charges on the O and H give the H 2 O its unique property as polar solvent. However; water molecules are joining to each other via an intermolecular force (hydrogen bonds), this hydrogen bonds are stands behind the high boiling point of water, as well as it makes the H 2 O a good solvent for so many compounds. Fig H bonds (Intermolecular forces) In order for the solute to dissolve in the solvent, the intermolecular forces i.e. the hydrogen bond between the water molecules must be over come, this distortion required energy to over come the interaction between the solvent (water) molecules. As a result of the over comes of the interaction between the solvent molecules; a room are created between the water molecules and enabling the solute to get into the solvent. 14

15 3) Now we have the positive ions Na + and the negative ions Cl - and the H 2 O with its partial positive on H atoms and the partial negative on the O atom. The positive ion Na + will be attracted to the partially negative O atom via ion-dipole interaction, in the same way the partial H atoms on water will be attracted by the Cl - negative ion as well via ion-dipole interaction. Fig. Ion-Dipole interaction between Na + and negative O, and between H and Cl - In the actual situation; so many partial negative O on water will be surrounding the Na + positive ion, and so many H atoms in water surrounding the partial negative Cl - ion via Ion-Dipole interaction and the NaCl H 2 O solution is formed. The process in which an ion or molecules are surrounded by solvent molecules in a specific manner is called solvation. Ion-Dipole interaction and the NaCl H 2 O solution is formed 15

16 Calculating Concentration Units & Dilutions The concentration of a chemical solution refers to the amount of solute that is dissolved in a solvent. We normally think of a solute as a solid that is added to a solvent (e.g., adding table salt to water), but the solute could just as easily exist in another phase. For example, if we add a small amount of ethanol to water, then the ethanol is the solute and the water is the solvent. If we add a smaller amount of water to a larger amount of ethanol, then the water could be the solute! Units of Concentration Once you have identified the solute and solvent in a solution, you are ready to determine its concentration. Concentration may be expressed by several different ways, using percent by mass, mole fraction, molarity, molality, or normality. 1) Percent by Mass (%) Percent by Mass (%) also called the percent by weight or weight percent. This is the mass of the solute divided by the mass of the solution (mass of solute plus mass of solvent), multiplied by 100. Percent by Mass = moles of solute Mass of solution x 100% Percent by Mass has no units because it is a ratio of two similar quantities. Example Determine the percent composition by mass of a 100 g salt solution which contains 20 g salt. Solution: 20 g NaCl / 100 g solution x 100 = 20% NaCl solution 16

17 2) Mole Fraction (X) This is the number of moles of a compound divided by the total number of moles of all chemical species in the solution. Keep in mind, the sum of all mole fractions in a solution always equals 1. For example if we have a solution contains 2 mol of solute A dissolved in 10 mol of solvent B; the mole fraction of the solute A is calculated by using the following formula Mole Fraction A moles of solute A moles of solute A + moles of solute B Mole Fraction A Mole Fraction B / /12 = 1 Example: What are the mole fractions of the components of the solution formed when 92 g glycerol is mixed with 90 g water? (molecular weight water = 18; molecular weight of glycerol = 92) Solution: 90 g water = 90 g x 1 mol / 18 g = 5 mol water 92 g glycerol = 92 g x 1 mol / 92 g = 1 mol glycerol total mol = = 6 mol A water = 5 mol / 6 mol = B glycerol = 1 mol / 6 mol = It's a good idea to check your math by making sure the mole fractions add up to 1: A water + B glycerol = =

18 3) Molarity Molarity is probably the most commonly used unit of concentration. It is the number of moles of solute per liter of solution (not necessarily the same as the volume of solvent!). Molarity = moles of solute litter of solution Example: What is the molarity of a solution made when water is added to 11 g CaCl2 to make 100 ml of solution? Solution: 11 g CaCl2 / (110 g CaCl2 / mol CaCl2) = 0.10 mol CaCl2 100 ml x 1 L / 1000 ml = 0.10 L molarity = 0.10 mol / 0.10 L i.e. molarity = 1.0 M Example 2:.0678 g of NaCl is placed in a 25.0 ml flask full of water. When the NaCl dissolves, what is the molarity of the solution?.0464 M NaCl 4) Molality Molality is the number of moles of solute per kilogram of solvent. Because the density of water at 25 C is about 1 kilogram per liter, molality is approximately equal to molarity for dilute aqueous solutions at this temperature. This is a useful approximation, but remember that it is only an approximation and doesn't apply when the solution is at a different temperature, isn't dilute, or uses a solvent other than water. molality = moles of solute kilogram of solvent 18

19 Example 1:.20 mol of ethylene glycol / 2.0 Kg of solvent =.10 m ethylene glycol Example 2: What is the molality of a solution of 10 g NaOH in 500 g water? Solution: 10 g NaOH / (4 g NaOH / 1 mol NaOH) = 0.25 mol NaOH 500 g water x 1 kg / 1000 g = 0.50 kg water molality = 0.25 mol / 0.50 kg molality = 0.05 M / kg i.e. molality = 0.50 m 5) Normality (N) Normality is equal to the gram equivalent weight of a solute per liter of solution. A gram equivalent weight or equivalent is a measure of the reactive capcity of a given molecule. Normality is the only concentration unit that is reaction dependent. Example: 1 M sulfuric acid (H 2 SO 4 ) is 2 N for acid-base reactions because each mole of surfuric acid provides 2 moles of H + ions. On the other hand, 1 M sulfuric acid is 1 N for sulfate precipitation, since 1 mole of sulfuric acid provides 1 mole of sulfate ions. Dilutions You dilute a solution whenever you add solvent to a solution. Adding solvent results in a solution of lower concentration. You can calculate the concentration of a solution following a dilution by applying this equation: M 1 V 1 = M 2 V 2 where M is molarity, V is volume, and the subscripts 1 and 2 refer to the initial and final values. Example: How many millilitres of 5.5 M NaOH are needed to prepare 300 ml of 1.2 M NaOH? Solution: 5.5 M x V1 = 1.2 M x 0.3 L 19

20 V1 = 1.2 M x 0.3 L / 5.5 M V1 = L V1 = 65 ml So, to prepare the 1.2 M NaOH solution, you pour 65 ml of 5.5 M NaOH into your container and add water to get 300 ml final volume. Factors affecting solubility 1) Effect of temperature a) Solid solute b) Gas solute 2) Effect of pressure 1) Effect of Temperature on Solubility (solid solute) Solubility refers to the ability for a given substance, the solute, to dissolve in a solvent at a specific temperature. Generally, an increase in the temperature of the solution increases the solubility of a solid solute. A few solid solutes, however, are less soluble in warmer solutions. The chart shows solubility curves for some typical inorganic salts (all solids). Many salts behave like barium nitrate and disodium hydrogen arsenate, and show a large increase in solubility with temperature. The solubility increases with temperature 20

21 Some solutes (e.g. NaCl in water) are fairly independent of temperature. A few, such as cerium(iii) sulphate, become less soluble in hot water. There is no direct relationship between the value of H sol (if it is positive or negative) and the variation of solubility with temperature. For example, the solution process of CaCl 2 is exothermic and that for NH 3 NO 3 is endothermic, but the solubility of both of them has increased by increasing the temperature. However, experimental work is the best way to determine the effect of temperature on solubility that because the solubility is depends on the natural of the solute and solvent, for example; only 1 gram of lead (II) chloride can be dissolved in 100 grams of water at room temperature, while 200 grams of zinc chloride can be dissolved at the same temperature. b) Effect of Temperature on Solubility (Gas solute) As the temperature increases, the solubility of a gas decreases as shown by the downward trend in the graph. More gas is present in a solution with a lower temperature compared to a solution with a higher temperature. The reason for this gas solubility relationship with temperature is because the increase in temperature causes an increase in kinetic energy. The higher kinetic energy causes more motion in molecules which break intermolecular bonds and escape from solution. This gas solubility relationship can be remembered if you think about what happens to a bottle of Pepsi as it stands around for awhile at room temperature. The taste of the Pepsi becomes bad since more of the "tangy" carbon dioxide bubbles have escaped. Boiled water also tastes "flat" because all of the oxygen gas has been removed by heating. 21

22 2) Effect of Pressure on Solubility Liquids and solids solutes exhibit practically no change of solubility with changes in pressure. Gases as might be expected, increase in solubility with an increase in pressure. Henry's Law states that: The solubility of a gas in a liquid is directly proportional to the pressure of that gas above the surface of the solution. C α P C = ĸP C = molar concentration of a gas in a liquid P = pressure of the gas over the solution in atmospheres K = Constant that only depends on temperature. Its unit is mol/l.atm If the pressure is increased, the gas molecules are "forced" into the solution since this will best relieve the pressure that has been applied. The number of gas molecules is decreased. The number of gas molecules dissolved in solution has increased as shown in the graphic on the left. Carbonated beverages provide the best example of this phenomena. All carbonated beverages are bottled under pressure to increase the carbon dioxide dissolved in solution. When the bottle is opened, the pressure above the solution decreases. As a result, the solution effervesces and some of the carbon dioxide bubbles off. 22

23 What is Colligative Properties of Solutions? Solutions have different properties than either the solutes or the solvent used to make the solution. Those properties can be divided into two main groups colligative and non-colligative properties. Colligative properties depend only on the number of dissolved particles in solution and not on their identity. Non-colligative properties depend on the identity of the dissolved species and the solvent. A property that depends only on the amount of solute in a solution and not the identity of the solute is called Colligative properties Colligative properties are 1. Boiling point elevation, 2. Freezing point lowering, and 3. Osmotic pressure. 4. Vapour pressure lowering, K f and K b K f and K b are the freezing point depression constant and boiling point elevation constant respectively. When a solute is added to a solvent, the boiling point of the solution is always greater than the boiling point of the pure solvent. Adding a solute also lowers the freezing point. Solvent Formula Melting Point ( o C) Boiling Point( o C) K f ( o C/m) K b ( o C/m) Benzene C 6 H Ethanol C 2 H 5 OH Water H To determine the amount of change in boiling point you will need this equation: T b = K b * m T b change in boiling point 23

24 K b boiling point elevation constant m molality of solution Then T b is added to the normal boiling point of the pure solvent. Note that the identity of the solute is not important, just its concentration (expressed in molality). Therefore, boiling point elevation is a colligative property. To determine the freezing point of a solution, you need to calculate the decrease in freezing point caused by the addition of a solute to the solvent. Use the equation: T f = K f * m T f change in freezing point K f freezing point depression constant m molality of solution Then the T f is subtracted from the normal freezing point of the pure solvent. Freezing point depression is also a colligative property. Example: Calculate the boiling point and freezing point of a solution of.30 g of glycerol (C 3 H 8 O 3 ) in 20.0 g of water. moles glycerol = (.30 g) (1 mole / 92 g) =.0033 moles molality of solution =.0033 moles /.020 kg =.16 m T b = (.521 o C/m)(.16 m) =.083 o C Boiling point = = o C T f = (1.858 o C/m)(.16m) =.30 o C Freezing point = 0.00 o C -.30 o C = -.30 o C In an ionic solution, the total concentration of ions is important. Therefore, another factor (i) is included in the equations. 24

25 T b = K b * m * i T f = K f * m * i "i" is the number of ions from each formula unit. In the previous example, if NaCl had been the solute your change in boiling point and freezing point would have needed to be multiplied by 2. (i=2 because NaCl consists of a Na + ion and a Cl - ion To explain the difference between the two sets of solution properties, we will compare the properties of a 1.0 M aqueous sugar solution to a 0.5 M solution of table salt (NaCl) in water. Despite the concentration of sodium chloride being half of the sucrose concentration, both solutions have precisely the same number of dissolved particles because each sodium chloride unit creates two particles upon dissolution a sodium ion, Na +, and a chloride ion, Cl -. Therefore, any difference in the properties of those two solutions is due to a non-colligative property. Both solutions have the same freezing point, boiling point, vapour pressure, and osmotic pressure because those colligative properties of a solution only depend on the number of dissolved particles. The taste of the two solutions, however, is markedly different. The sugar solution is sweet and the salt solution tastes salty. Therefore, the taste of the solution is not a colligative property. Another non-colligative property is the colour of a solution. A 0.5 M solution of CuSO4 is bright blue in contrast to the colourless salt and sugar solutions. Other non-colligative properties include viscosity, surface tension, and solubility. 25

26 Making solutions 1) Pour in a small amount of solvent 2) Then add the solute and dissolve it 3) Then fill to final volume. M x L = moles Example1: How many moles of NaCl are needed to make 6.0 L of a 0.75 M NaCl solution? Example 2: How many grams of CaCl 2 are needed to make 625 ml of a 2.0 M solution? Example 3: 10.3 g of NaCl are dissolved in a small amount of water then diluted to 250 ml. What is the concentration? Example 4: How many grams of sugar are needed to make 125 ml of a 0.50 M C6H12O6 solution? 26

27 Dilution Dilution is the addition of water to a solution or (solvents) that depends on the type of solution you are making. The number of moles of solute doesn t change if you add more solvent. The moles before = the moles after M1 x V1 = M2 x V2 M1 and V1 are the starting concentration and volume. M2 and V2 are the starting concentration and volume. Stock solutions are pre-made to known M Practice 2.0 L of a 0.88 M solution are diluted to 3.8 L. What is the new molarity? You have 150 ml of 6.0 M HCl. What volume of 1.3 M HCl can you make? Need 450 ml of 0.15 M NaOH. All you have available is a 2.0 M stock solution of NaOH. How do you make the required solution? 27