solutions.pdf

www.tntextbooks.in ICT Corner Effect of temperature and pressure in an equilibrium process By using this tool, we can P...

www.tntextbooks.in ICT Corner Effect of temperature and pressure in an equilibrium process By using this tool, we can Please go to the URL determine the effect of pressure http://www.freezeray.com/ and temperature in the flashFiles/a equilibrium concentration of mmoniaConditions.htm the components in ammonia (or) Scan the QR code on the synthesis (Haber process) right side Steps Open the Browser and type the URL given (or) Scan the QR Code. The website will show the equilibrium reaction involved in ammonia synthesis and the relative concentration of the components. The visual representation and the actual concentration values are given in the box 1. Now change the pressure or temperature using the corresponding slider indicated in the box2. As you move the slider you will be able to see the change in the equilibrium concentration of reactants and products. Now you can understand that if a stress is applied on the system at equilibrium, the system will adjust itself to nullify the effect of the stress. 29 11th Chemistry Chapter-8 Vol-2 EM.indd 29 09-01-2023 14:38:33  www.tntextbooks.in Unit 9 Solutions Learning Objectives After studying this unit students will be able to describe the formation of different types of solutions express the concentration of a solution in different units prepare solutions of required concentrations by diluting the stock solution state Henry’s and Raoult’s Law François - Marie Raoult François - Marie Raoult explain the deviation of real solutions from was a French chemist who Raoult’s Law conducted research into the behaviour of solutions, correlate colligative properties of solutions especially their physical with the molar masses of their solutes properties. explain the abnormal colligative properties In his first research paper, he described the define Van't Hoff factor and calculate degree of action of solutes in depressing dissociation / association the freezing point of the solutions. He also gave a 9.1 INTRODUCTION relation between the vapour pressure of the solution with There are many chemicals that play an important the molecular wight of the role in our daily life. All these chemicals are in different solute. physical forms, viz solid, liquid and gas. If we do close examination on their composition, we could find that most of them are mixtures and rarely pure substances. One more interesting aspect is that most of the mixtures are homogeneous irrespective of their physical state and such homogeneous mixtures are called as solutions. 30 Unit 9.indd 30 23-12-2021 17:00:49  www.tntextbooks.in Sea water is one of the naturally the solvent,the resultant solution is called as existing solutions which covers more than an aqueous solution. If solvents (Benzene, 70% of the earth’s surface. We cannot imagine CCl4, ether etc.,) other than water is used, life on earth without sea water. It contains then the resultant solution is called as a non- many dissolved solids, mostly NaCl. Another aqueous solution. important naturally occurring solution is air. Air is a homogeneous mixture of nitrogen, The following table illustrates the oxygen, carbon dioxide, and other trace different types of solutions based on the gases. Even solid material such as brass is a physical state of the solute and solvent. homogeneous mixture of copper and zinc. Table 9.1 Types and examples of In the above examples the solutions solutions are in different physical states viz... liquid solution (sea water), gas (air) and solid (alloys), and State of Solvent Solute S. No. one common property of all the above is Examples their homogeneity. The homogeneity implies uniform distribution of their constituents or Air (A mixture of Gas Gas Gaseous solution nitrogen, oxygen and components throughout the mixture. In this other gases) chapter, we learn about the solutions and 1 Solid Liquid Humid oxygen (Oxygen their properties. Gas containing water) 9.2 Types of solutions Gas Camphor in nitrogen gas A solution is a homogeneous mixture Solid Liquid Liquid Liquid of two or more substances, consisting of CO2 dissolved in water Gas (carbonated water) Liquid solutions atoms, ions or molecules. The compound that is present in largest amount in a Liquid Ethanol dissolved in homogeneous mixture is called the solvent, 2 water and others are solutes. For example, when a Solid small amount of NaCl is dissolved in water, Salt water a homogeneous solution is obtained. In this solution, Na+ and Cl- ions are uniformly Solution of H2 in Gas distributed in water. Here water is the solvent palladium Solid solutions as the amount of water is more compared to Liquid Amalgam of potassium Solid the amount of NaCl present in this solution, 3 (used for dental filling) and NaCl is the solute. Gold alloy (of copper Solid Solid used in making The commonly used solutions are the Jewelery) solutions in which a solid solute is dissolved in a liquid solvent. However, solute or 9.3 Expressing concentration of solutions solvent can be in any of the three states of matter (solid, liquid, gas). If water is used as In our life we have come across many solutions of varying strengths or 31 Unit 9.indd 31 23-12-2021 17:00:49  www.tntextbooks.in concentrations such as mouthwash, antiseptic solutions, household disinfectants etc... Have you ever noticed the concentration of the ingredients present in those solutions? For example, chlorhexidine mouthwash solution contains 0.2 % (w/v) chlorhexidine gluconate; The concentration of the commercially available hydrogen peroxide is 3% (w/v). Similarly, other terms such as ppm (TDS of water), molar and normal (laboratory reagents) are used to express the concentration of the solution. The concentration of a solution gives the amount of solute present in a given quantity of solvent. As we have seen, there are different ways of expressing the concentration of a solution. Let us learn the different concentration terms and to prepare a solution of a specific concentration. Table 9.2 Different concentration units and their illustrations Concentration S. No. term Expression Illustration The molality of the solution containing 45 g of glucose dissolved in 2 kg of water Molality (m)  45  Number ofmoles of solute Number ofmoles of solute  180  0..25 1 Mass of the solvent (in kg ) = = = 0.125 m Mass of the solvent (in kg ) 2 2  45  Number ofmoles of solute  180  0..25 = = = 0.125 m Mass of the solvent (in kg ) 2 2 5.845 g of sodium chloride is dissolved in water and the solution was made up to 500 mL using a standard flask. The strength of the Molarity (M) Number ofmoles of solute solution in molarity is 2  5.845  Volume of solution (in L)  58.45  0.1 Number ofmoles of solute = = = 0. 2 M Volume of solution (in L) 0. 5 0. 5  5.845  Number ofmoles of solute  58.45  =  = 0. 1 = 0. 2 M Volume of solution (in L) 0. 5 0. 5 3.15 g of oxalic acid dihydrate, is dissolved in water and the solution was made up to 100 mL using a standard flask. The strength of the solution in normality is Normality (N) Number of gram equivalents of solute = Number of gram equivalents of solute Volume of solution (in L) 3 Volume of solution (in L)  mass of oxalic acid     3.15   Equivalent mass of oxalic acid   63   = = volume of solution (inL) 0. 1 0.05 = = 0. 5 N 0. 1 32 Unit 9.indd 32 23-12-2021 17:00:50  www.tntextbooks.in Concentration S. No. term Expression Illustration 5.85g of sodium chloride is dissolved in water and the solution was made up to 500 mL using a standard flask. The strength of the solution Formality (F) in formality is Number of Formula mass of solute Number of Formula mass of solute 4 Volume of solution (in L) formality = Volume of solution (in L) 5.85 = 58.5 × 0.5L = 0.2 F 0.5 mole of ethanol is mixed with 1.5 moles of Number of moles of the component water. The mole fraction of ethanol in the above solution Total number of moles of all the is compon nents present in solution Mole fraction (of a component) (x) Number of moles of the ethanol = Consider a solution containing Total number of moles of ethanol and watter two components A and B whose = 0.5 = 0.5 = 0.25 mole fractions are xA and xB, 1. 5 + 0. 5 2. 0 respectively. Let the number of The mole fraction of water in the above solution 5 moles of the two components A is and B be nA and nB, respectively. Number of moles of water = nA nB Total number of moles of ethanol and water xA = and x B = nA + nB nA + nB 1..5 = = 0.75 2. 0 Now, nA nB The mole fraction of water can also be calculated xA + xB = + =1 as follows mole fraction of water + mole fraction nA + nB nA + nB of ethanol = 1; mole fraction of water = 1 – mole fraction of ethanol = 1-0.25 = 0.75 Neomycin, aminoglycoside antibiotic cream Mass percentage (% w/w) contains 300 mg of neomycin sulphate the active ingredient, in 30g of ointment base. The mass Mass of the solute (in g) percentage of neomycin is 6 ×100 Mass of solution (in g ) Mass of the neomycin sulphate (in g) 0. 3 g × 100 = × 100 = 1% w / w Mass of solution (in g ) 30 g Mass of the neomycin sulphate (in g) 0. 3 g × 100 = × 100 = 1% w / w Mass of solution (in g ) 30 g 33 Unit 9.indd 33 23-12-2021 17:00:52  www.tntextbooks.in Concentration S. No. term Expression Illustration 50 mL of tincture of benzoin, an antiseptic Volume percentage solution contains 10 mL of benzoin. The volume percentage of benzoin (% v/v) Volume of the solute (in mL) 7 ×100 Volume of the benzoin (in mL) Volume of solution (in mL) 10 = × 100 = × 1000 = 20% v / v Volume of solution (in mL) 50 Volume of the benzoin (in mL) 10 = × 100 = × 1000 = 20% v / v Volume of solution (in mL) 50 A 60 mL of paracetamol paediatric oral suspension contains 3g of paracetamol. The mass percentage percentage (% w/v) Mass by volume of paracetamol is Mass of the solute (in g ) Mass of the paracetamol (in g ) 3 8 ×100 100 100 5% w / v Volume of solution (in mL) Volume of solution (in mL) 60 Mass of the paracetamol (in g ) 3 100 100 5% w / v Volume of solution (in mL) 60 50 g of tap water contains 20 mg of dissolved solids. The TDS value in ppm is Parts per million Number of parts of the component × 106 Mass of the dissolved solids (ppm) Total number of parts of all componentts 9 × 106 Mass of the solute Mass of the water = × 106 Mass of the solution 20 × 10−3 g × 106 = 400 ppm 50 g Evaluate Yourself ? 1) If 5.6 g of KOH is present in (a) 500 mL and (b) 1 litre of solution, calculate the molarity of each of these solutions. 2) 2.82 g of glucose is dissolved in 30 g of water. Calculate the mole fraction of glucose and water. 3) The antiseptic solution of iodopovidone for the use of external application contains 10 % w/v of iodopovidone. Calculate the amount of iodopovidone present in a typical dose of 1.5 mL. 4) A litre of sea water weighing about 1.05 kg contains 5 mg of dissolved oxygen (O2). Express the concentration of dissolved oxygen in ppm. The concentration of a solution is expressed in different units. The choice of unit depends on the type of measurement applied. For example,in complexometric titrations involving 34 Unit 9.indd 34 23-12-2021 17:00:53  www.tntextbooks.in EDTA, the reaction between EDTA and the Where the Cs & Vs are concentration metal ions takes place in the 1:1 mole ratio and volume of the stock solution and Cw and hence molar solutions are used in this & Vw are concentration and volume of the titrations. In the redox and neutralisation working standard, respectively. titrations we use normal solutions. The mole fraction is used to calculate the partial 9.3.2 Advantages of using standard pressure of gases and the vapour pressure solutions: of solutions. The percentage units are used 1. The error in weighing the solute can be to express the active ingredients present in minimised by using concentrated stock therapeutics, and the ppm is used to express solution that requires large quantity of the quantity of solutes present in small solute. amounts in solutions. 9.3.1 Standard solutions and working 2. We can prepare working standards of standards different concentrations by diluting the A standard solution or a stock stock solution, which is more efficient solution is a solution whose concentration since consistency is maintained. is accurately known. A standard solution 3. Some of the concentrated solutions of required concentration can be prepared are more stable and are less likely to by dissolving a required amount of a solute, support microbial growth than working in a suitable amount of solvent. Its done by standards used in the experiments. (i) transforming a known amount of a solute to a standard flask of definite volume. Example Problem 1 (ii) a small amount of water is added to the 1. What volume of 4M HCl and 2M HCl flask and shaken well to dissolve the solute. should be mixed to get 500 mL of 2.5 M (iii) then water is added to the flask HCl? to bring the solution level to the mark indicated at the top end of the flask. Let the volume of 4M HCl required to (iv) the flask is stoppered and shaken well to prepare 500 mL of 2.5 MHCl = x mL make concentration uniform. Therefore, the required volume of 2M HCl At the time of experiment, the = (500 - x) mL solution with required concentration is We know from the equation (9.1) prepared by diluting the stock solution. This diluted solution is usually called C1V1+ C2V2 = C3V3 working standard. A known volume of stock (4x)+2(500-x) = 2.5 × 500 solution is transferred to a new container and brought to the calculated volume. The 4x+1000-2x = 1250 necessary volumes of the stock solution and 2x = 1250 - 1000 final volume can be calculated using the 250 following expression. x = 2 = 125 mL Cs Vs = Cw Vw ----------(9.1) 35 Unit 9.indd 35 23-12-2021 17:00:53  www.tntextbooks.in Hence, volume of 4M HCl required = 125 mL Volume of 2M HCl required = (500 - 125) mL= 375 mL Evaluate Yourself ? 5) Describe how would you prepare the following solution from pure solute and solvent (a) 1 L of aqueous solution of 1.5 M CoCl2. (b) 500 mL of 6.0 % (V/V) aqueous methanol solution. 6) How much volume of 6 M solution of NaOH is required to prepare 500 mL of 0.250 M NaOH solution. 9.4 Solubility of the solutes Solubility of a solute is the maximum amount of solute that can be dissolved in a specific amount of solvent at a specified temperature. When maximum amount of solute is dissolved in a solvent, any more addition of solute will result in precipitation at a given temperature and pressure. Such a solution is called as a saturated solution. The solubility of a substance at a given temperature is defined as the amount of the solute that can be dissolved in 100 g of the solvent at a given temperature to form a saturated solution. 9.4.1 Factors influencing the solubility The solubility of a solute generally depends on the nature of the solute and the solvent in which it is dissolved. It also depends on the temperature and pressure of the solution. Nature of solute and solvent: Sodium chloride, an ionic compound, dissolves readily in a polar solvent such as water, but it does not dissolve in non-polar organic solvents such as benzene or toluene. Many organic compounds dissolve readily in organic solvents and do not dissolve in water. Different gases dissolve in water to different extents: for example, ammonia is more soluble than oxygen in water. Effect of temperature: Solid solute in liquid solvent: Generally, the solubility of a solid solute in a liquid solvent increases with increase in temperature. When the temperature is increased, the average kinetic energy of the molecules of the solute and the solvent increases. The increase in kinetic energy facilitates the solvent molecules to break the intermolecular attractive forces that keep the solute molecules together and hence the solubility increases. 36 Unit 9.indd 36 23-12-2021 17:00:53  www.tntextbooks.in When a solid is added to a solvent, it begins to dissolve. i.e. the solute leaves from the solid state (dissolution). After some time, some of the dissolved solute returns back to the solid state (recrystallisation). If there is excess of solid present, the rate of both these processes becomes equal at a particular stage. At this stage an equilibrium is established between the solid solute molecules and dissolved solute molecules. Solute (solid) ⇆ Solute (dissolved) According to Le-Chatelier principle, if the dissolution process is endothermic, the increase in temperature will shift the equilibrium towards right i.e solubility increases. for an exothermic reaction, the increase in temperature decreases the solubility. The solubilities of ammonium nitrate, calcium chloride, ceric sulphate nano-hydrate and sodium chloride in water at different temperatures are given in the following graph. 0 10 20 30 40 50 60 70 80 90 100 250 225 200 3 NO Solubility (g/100g H2 O) 4 NH 175 150 CaCl 2 125 100 75 50 NaCl 25 Ce2(SO4)3. 9H2O 0 0 10 20 30 40 50 60 70 80 90 100 Temperature (oC) Figure 9. 1 Plot of solubility versus temperature for selective compounds The following conclusions are drawn from the above graph. ▶ The solubility of sodium chloride does not vary appreciable as the maximum solubility is achieved at normal temperature. In fact, there is only 10 % increase in solubility between 0 ˚ to 100 ˚C. ▶ The dissolution process of ammonium nitrate is endothermic, the solubility increases steeply with increase in temperature. ▶ In the case of ceric sulphate, the dissolution is exothermic and the solubility decreases with increase in temperature. ▶ Even though the dissolution of calcium chloride is exothermic, the solubility increases moderately with increase in temperature. Here, the entropy factor also plays a significant role in deciding the position of the equilibrium. 37 Unit 9.indd 37 23-12-2021 17:00:53  www.tntextbooks.in Gaseous solute in liquid solvent: When pressure is increased In the case of gaseous solute in liquid solvent, the solubility decreases with increase in temperature. When a gaseous solute dissolves in a liquid solvent, its molecules interact with solvent molecules with weak intermolecular forces. When the temperature increases, the average Under normal More gas molecules are kinetic energy of the molecules present conditions soluble at higher pressure in the solution also increases. The Figure 9.2 Effect of pressure on solubility increase in kinetic energy breaks the weak intermolecular forces between the gaseous Effect of pressure: solute and liquid solvent which results in Generally the change in pressure the release of the dissolved gas molecules to does not have any significant effect in the the gaseous state. Moreover, the dissolution solubility of solids and liquids as they are of most of the gases in liquid solvents is an not compressible. However, the solubility exothermic process, and in such processes, of gases generally increases with increase of the increase in temperature decreases the pressure. dissolution of gaseous molecules. Consider a saturated solution of a gaseous solute dissolved in a liquid solvent Activity: in a closed container. In such a system, the Open the soda bottle and put a following equilibrium exists. balloon over it. The balloon will inflate with Gas (in gaseous state) ⇆ Gas (in solution) the released carbon dioxide from the soda. Carry out the same experiment by placing According to Le-Chatelier principle, the soda bottle in a container of hot water. the increase in pressure will shift the You will observe the balloon is inflated equilibrium in the direction which will much faster now. This shows the decrease in reduce the pressure. Therefore, more solubility of gases in solution with increase number of gaseous molecules dissolves in in temperature. In the rivers where hot the solvent and the solubility increases. water is discharged from industrial plants, 9.5 Henry's law the aquatic lives are less sustained due to the decreased availability of dissolved oxygen. William Henry investigated the relationship between pressure and solubility of a gaseous solute in a particular solvent. According to him, “the partial pressure of the gas in vapour phase (vapour pressure of 38 Unit 9.indd 38 23-12-2021 17:00:55  www.tntextbooks.in the solute) is directly proportional to the mole fraction(x) of the gaseous solute in the solution at low concentrations”. This statement is known as Henry’s law. Henry’s law can be expressed as, psolute α xsolute in solution psolute = KHxsolute in solution Here, psolute represents the partial pressure of the gas in vapour state which is commonly called as vapour pressure. xsolute in solution represents the mole fraction of solute in the solution. KH is a empirical constant with the dimensions of pressure. The value of ‘KH’ depends on the nature of the gaseous solute and solvent. The above equation is a straight-line in the form of y=mx. The plot partial pressure of the gas against its mole fraction in a solution will give a straight line as shown in fig 9.3. The slope of the straight line gives the value of KH. 1000 Partial pressure of HCl (Torr) 500 0 0.01 0.02 Mole fraction of HCl gas in its solution in cyclohexane Figure 9.3 Solubility of HCl gas in cyclohexane at 293 K. Why the carbonated drinks are stored in a pressurized container? We all know that the carbonated beverages contain carbon dioxide dissolved in them. To dissolve the carbon dioxide in these drinks, the CO2 gas is bubbled through them under high pressure. These containers are sealed to maintain the pressure. When we open these containers at atmospheric pressure, the pressure of the CO2 drops to the atmospheric level and hence bubbles of CO2 rapidly escape from the solution and show effervescence. The burst of bubbles is even more noticeable, if the soda bottle is in warm condition. 39 Unit 9.indd 39 23-12-2021 17:00:55  www.tntextbooks.in Why deep-sea divers use air diluted with helium gas in their air tanks? The professional deep-sea divers carry a compressed air tank for breathing at high pressure under water. The normal compressed air contains nitrogen and oxygen and these gases are not very soluble in blood and other body fluids at normal pressure. As the pressure at that depth is far greater than the surface atmospheric pressure, more nitrogen dissolves in the blood and other body fluids when the diver breathes from tank. When the diver ascends to the surface, the pressure decreases, the dissolved nitrogen comes out of the blood and other body fluids quickly forming bubbles in the blood stream. These bubbles restrict blood flow, affect the transmission of nerve impulses and can even burst the capillaries or block them. This condition is called “the bends”, which are painful and dangerous to life. Now a days, to avoid such dangerous condition, the professional divers, use air diluted with helium gas (about 11.7% Helium, 56.2% Nitrogen and 32.1% Oxygen), because of lower solubility of helium in the blood than nitrogen. Moreover, because of small size of helium atoms they can pass through cell walls without damaging them. The excess oxygen dissolved in the blood is used in metabolism and does not cause the condition of bends 9.5.1 Limitations of Henry’s law Henry’s law is applicable at moderate temperature and pressure only. Only the less soluble gases obeys Henry’s law The gases reacting with the solvent do not obey Henry’s law. For example, ammonia or HCl reacts with water and hence does not obey this law. + – NH3+ H2O ⇆ NH4 + OH The gases obeying Henry’s law should not associate or dissociate while dissolving in the solvent. Example Problem 2: 0.24 g of a gas dissolves in 1 L of water at 1.5 atm pressure. Calculate the amount of dissolved gas when the pressure is raised to 6.0 atm at constant temperature. psolute = KHxsolute in solution At pressure 1.5 atm, 40 Unit 9.indd 40 23-12-2021 17:00:55  www.tntextbooks.in p1 = KH x1------------(1) At pressure 6.0 atm, p2 = KH x2------------(2) Dividing equation (1) by (2) From equation p1/p2 = x1/x2 1.5/6.0 = 0.24/x2 Therefore x2 = 0.24 x 6.0/1.5 = 0.96 g/L Evaluate Yourself ? 7) Calculate the proportion of O2 and N2 dissolved in water at 298 K. When air containing 20% O2 and 80% N2 by volume is in equilibrium with water at 1 atm pressure. Henry’s law constants for two gases are KH(O2) = 4.6 x 104 atm and KH (N2) = 8.5 x 104 atm. 8) Explain why the aquatic species are more comfortable in cold water during winter season rather than warm water during the summer. 9.6 Vapour pressure of liquid Generally, liquids have a tendency to evaporate. If the kinetic energy of molecules in the liquid state overcomes the intermolecular force of attraction between them, then the molecules will escape from the liquid state. This process is called 'evaporation' and it happens on the surface of the liquid. If evaporation is carried out in a closed container then the vapour remains in contact with the surface of the liquid. These vapour molecules are in continuous random motion during which they collide with each other and also with the walls of the container. As the collision is inelastic, they lose their energy and as result the vapour returns back to liquid state. This process is called as 'condensation'. Evaporation and condensation are continuous processes. If the process is carried out in a closed system, a stage is reached when the rate of evaporation becomes equal to the rate of condensation. Thus, an equilibrium is established between liquid and its vapour. The pressure of the vapour in equilibrium with its liquid is called vapour pressure of the liquid at the given temperature. The vapour pressure of a liquid depends on its nature, temperature and the surface area. The following simple apparatus demonstrates the measurement of vapour pressure of a liquid. 41 Unit 9.indd 41 23-12-2021 17:00:55  www.tntextbooks.in Pgas = equlibrium Vapour Pressure Figure 9.4 : a) A closed round bottomed flask in which ethanol is in equilibrium with its vapour. b) In the same setup the vapour is allowed to escape through a U tube filled with mercury. The escaped vapour pushes the mercury in the U tube and the difference in mercury level gives the vapour pressure of ethanol present in the RB flask. 9.7 Vapour pressure of liquid solutions When a solute (of any physical state - solid, liquid or gas ) is dissolved in a liquid solvent the resultant solution is called a liquid solution. The solution which contains only two components (one solvent and one solute) is called a binary solution. We have already discussed the solution of a gaseous solute in liquid solvent under Henry's law. 9.7.1 Vapour pressure of binary solution of liquid in liquids Now, let us consider a binary liquid solution formed by dissolving a liquid solute ‘A’ in a pure solvent ‘B’ in a closed vessel. Both the components A and B present in the solution would evaporate and an equilibrium will be established between the liquid and vapour phases of the components A and B. The French chemist Raoult, proposed a quantitative relationship between the partial 42 Unit 9.indd 42 23-12-2021 17:00:55  www.tntextbooks.in pressures and the mole fractions of two components A & B, which is known as Raoult’s Law. This law states that “in the case of a solution of volatile liquids, the partial vapour pressure of each component (A & B) of the solution is directly proportional to its mole fraction”. According to Raoult’s law, pAα xA ----------(9.3) pA = k xA when xA = 1, k = p°A where p°A is the vapour pressure of pure component ‘A’ at the same temperature. Therefore, pA = p°A xA ---------- (9.4) Similarly, for component ‘B’ pB= p°B xB ----------(9.5) xA and xB are the mole fraction of the components A and B respectively. According to Dalton’s law of partial pressure the total pressure in a closed vessel will be equal to the sum of the partial pressures of the individual components. Hence, Ptotal = pA + pB ----------(9.6) Substituting the values of pA and pB from equations (9.4) and (9.5) in the above equation, Ptotal = xAp°A + xBp°B (9.7) We know that xA + xB = 1 or xA = 1 - xB Therefore, Ptotal = (1 - xB) p°A + xB p°B ----------(9.8) Ptotal= p°A + xB( p°B- p°A) ----------(9.9) The above equation is of the straight-line equation form y = mx+c. The plot of Ptotal versus xB will give a straight line with (p°B- p°A) as slope and p°A as the y intercept. Let us consider the liquid solution containing toluene (solute) in benzene (solvent). 43 Unit 9.indd 43 23-12-2021 17:00:55  www.tntextbooks.in The variation of vapour pressure of pure benzene and toluene with its mole fraction is given in the graph. Xtoluene 1.0 0.8 0.6 0.4 0.2 0 100 100 Vapour Pressure (mmHg) 75 75 74.7 P0benzene on 50 P soluti 50 ene P benz 25 25 P0toluene 22.3 Pto luene 0 0 0.2 0.4 0.6 0.8 1.0 Xbenzene Figure 9.5 Solution of benzene in toluene obeying Raoult’s law. The vapour pressures of pure toluene and pure benzene are 22.3 and 74.7 mmHg, respectively. The above graph shows, the partial vapour pressure of the pure components increases linearly with the increase in the mole fraction of the respective components. The total pressure at any composition of the solute and solvent is given by the following straight line (represented as red line) equation. Psolution = p°toluene+ xbenzene( p°benzene- p°toluene) ---------- (9.10) 9.7.2 Vapour pressure of binary solution of solids in liquids When a nonvolatile solute is dissolved in a pure solvent, the vapour pressure of the pure solvent will decrease. In such solutions, the vapour pressure of the solution will depend only on the solvent molecules as the solute is nonvolatile. For example, when sodium chloride is added to the water, the vapour pressure of the salt solution is lowered. The vapour pressure of the solution is determined by the number of molecules of the solvent present in the surface at any time and is proportional to the mole fraction of the solvent. 44 Unit 9.indd 44 23-12-2021 17:00:55  www.tntextbooks.in Volatile solvent Particles on this expression, Raoult’s Law can also be stated as “the relative lowering of vapour Nonvolatile solute particles pressure of an ideal solution containing the nonvolatile solute is equal to the mole Add Nonvolatile fraction of the solute at a given temperature”. solute Comparison of Raoult’s law and Henry’s law According to Raoult’s law, for a solution containing a nonvolatile solute Equilibrium Rate of vaporization reduced by presence of psolute = p°solutexsolute ---------- (9.17) nonvolatile solute According to Henry’s law: Fig 9.6 Rate of vapourization reduced by presence of nonvolatile solute. psolute = KHxsolute in solution ---------- (9.18) Psolution α xA ---------- (9.11) The difference between the above two expressions is the proportionality Where xA is the mole fraction of the solvent constant p°A (Raoults Law) and KH.(Henry's Law). Henry's law is applicable to solution Psolution = k xA ---------- (9.12) containing gaseous solute in liquid solvent, o while the Raoult's Law is applicable to When xA = 1, K = Psolvent nonvolatile solid solute in a liquid solvent. o If the solute is non volatile then the Henry's (Psolvent is the partial pressure of pure solvent) law constant will become equal to the o vapour pressure of the pure solvent (p°A) and Psolution = Psolvent xA ----------(9.13) thus, Raoult’s law becomes a special case of Psolution Henry’s law. For very dilute solutions the o = xA ----------(9.14) solvent obeys Raoult’s law and the solute Psolvent obeys Henry’s law. Psolution 1– o =1-xA ----------(9.15) 9.8 Ideal and non-ideal solutions Psolvent o Psolvent – Psolution = xB 9.8.1 Ideal Solutions: o Psolvent ---------- (9.16) An ideal solution is a solution in Where xB is the mole fraction of the solute which each component i.e. the solute as well as the solvent obeys the Raoult’s law (∴ xA + xB = 1, xB = 1 - xA) over the entire range of concentration. The above expression gives the Practically no solution is ideal over the relative lowering of vapour pressure. Based entire range of concentration. However, 45 Unit 9.indd 45 23-12-2021 17:00:56  www.tntextbooks.in when the concentration of solute is very intermolecular interactions between solute low, the dilute solution behaves ideally. If (B) and solvent (A). Consider a case in which the two components present in the solution the intermolecular attractive forces between (A and B) are identical in size, structure, A and B are weaker than those between the and having almost similar intermolecular molecules of A (A-A) and molecules of B (B- attractive forces between them (i.e. between B). The molecules present in such a solution A-A, B-B and B-A) and then the solution have a greater tendency to escape from tends to behave like an ideal solution. the solution when compared to the ideal solution formed by A and B, in which the For an ideal solution intermolecular attractive forces (A-A, B-B, A-B) are almost similar. Consequently, the 1. There is no change in the volume vapour pressure of such non-ideal solution on mixing the two components increases and it is greater than the sum of (solute & solvents). (ΔVmixing= 0) the vapour pressure of A and B as predicted 2. There is no exchange of heat when the by the Raoult’s law. This type of deviation is solute is dissolved in solvent (ΔHmixing = 0). called positive deviation. 3. Escaping tendency of the solute Here, pA > p°A xA and pB > p°B xB. and the solvent present in it should be same as in pure liquids. Hence ptotal > p°A xA + p°B xB ----------(9.19) Examples for ideal solutions: (Benzene & Let us understand the positive Toluene) ; (n-hexane & n-heptane) ; (Ethyl deviation by considering a solution of bromide & Ethyl iodide) ; (Chlorobenzene ethyl alcohol and water. In this solution & Bromobenzene). the hydrogen bonding interaction between ethanol and water is weaker than those 9.8.2 Non-ideal solutions hydrogen bonding interactions amongst themselves (ethyl alcohol-ethyl alcohol and The solutions which do not water-water interactions). This results in the obey Raoult’s law over the entire range increased evaporation of both components of concentration, are called non-ideal (H2O and C2H5OH) from the aqueous solutions. For a non-ideal solution, there is solution of ethanol. Consequently, the a change in the volume and enthalpy upon vapour pressure of the solution is greater mixing. i.e. ΔHmixing ≠ 0 & ΔVmixing ≠ 0. The than the vapour pressure predicted by deviation of the non-ideal solutions from Raoult’s law. Here, the mixing process is the Raoult’s law can either be positive or endothermic i.e. ΔHmixing> 0 and there will negative. be a slight increase in volume (ΔVmixing> 0). Non-ideal solutions - positive deviation Examples for non-ideal solutions from Rauolt's Law: showing positive deviations: Ethyl alcohol & cyclohexane, Benzene & acetone, Carbon The nature of the deviation from the tetrachloride & chloroform, Acetone & ethyl Rauolt’s law can be explained in terms of the alcohol, Ethyl alcohol & water. 46 Unit 9.indd 46 23-12-2021 17:00:56  www.tntextbooks.in Mole fraction of water 1.0 0.8 0.6 0.4 0.2 0.0 12 O 10 p C 2H 5OH +H 2 Vapour pressure(in kPa) → 8 6 OH p H5 C2 4 pH 2O 2 0 0.0 0.2 0.4 0.6 0.8 1.0 Mole fraction of ethanol → Vapour pressure diagram showing positive deviation Figure 9.7 Positive deviations from Raoult’s law. The dotted line (-----) is ideal behavior and the solid lines (____) is actual behaviour Non-ideal solutions - negative deviation from Rauolt's Law: Let us consider a case where the attractive forces between solute (A) and solvent (B) are stronger than the intermolecular attractive forces between the individual components (A-A & B-B). Here, the escaping tendency of A and B will be lower when compared with an ideal solution formed by A and B. Hence, the vapour pressure of such solutions will be lower than the sum of the vapour pressure of A and B. This type of deviation is called negative deviation. For the negative deviation pA < p°A xA and pB < p°B xB. Let us consider a solution of phenol and aniline. Both phenol and aniline form hydrogen bonding interactions amongst themselves. However, when mixed with aniline, the phenol molecule forms hydrogen bonding interactions with aniline, which are stronger than the hydrogen bonds formed amongst themselves. Formation of new hydrogen bonds considerably reduce the escaping tendency of phenol and aniline from the solution. As a result, the vapour pressure of the solution is less and there is a slight decrease in volume (ΔVmixing< 0) on mixing. During this process evolution of heat takes place i.e. ΔHmixing< 0 (exothermic) 47 Unit 9.indd 47 23-12-2021 17:00:56  www.tntextbooks.in Examples for non-ideal solutions showing negative deviation: Acetone + chloroform, Chloroform + diethyl ether, Acetone + aniline,Chloroform + Benzene. ← Mole fraction of Acetone 1 0.8 0.6 0.4 0.2 0 330 p CH 300 3 CO C Vapour pressure (in Torr) → H + 3 C HCl 3 p CH 3 CO 200 Cl 3 CH 3 CH p 100 0 0 0.2 0.4 0.6 0.8 1 Mole fraction of chloroform → Vapour pressure diagram showing negative deviation Figure 9.8 Negative deviation from Raoult’s law. The dotted line (-----) is ideal behavior and the solid lines (____) is actual behaviour 9.8.3 Factors responsible for deviation from Raoult’s law The deviation of solution from ideal behavior is attributed to the following factors. i) Solute-solvent interactions For an ideal solution, the interaction between the solvent molecules (A-A),the solute molecules (B-B) and between the solvent & solute molecules (A-B) are expected to be similar. If these interactions are dissimilar, then there will be a deviation from ideal behavior. ii) Dissociation of solute When a solute present in a solution dissociates to give its constituent ions, the resultant ions interact strongly with the solvent and cause deviation from Raoult’s law. For example, a solution of potassium chloride in water deviates from ideal behavior + – because the solute dissociates to give K and Cl ion which form strong ion-dipole interaction with water molecules. 48 Unit 9.indd 48 23-12-2021 17:00:56  www.tntextbooks.in + – KCl (s) + H2O (l) → K (aq)+ Cl (aq) Evaluate Yourself ? iii)Association of solute 9) Calculate the mole fractions of benzene Association of solute molecules can also cause deviation from ideal behaviour. and naphthalene in the vapour phase For example, in solution, acetic acid exists when an ideal liquid solution is formed as a dimer by forming intermolecular hydrogen bonds, and hence deviates from by mixing 128 g of naphthalene with Raoult’s law. 39 g of benzene. It is given that the O H O vapour pressure of pure benzene is 50.71 H3C C C CH3 mmHg and the vapour pressure of pure O H O naphthalene is 32.06 mmHg at 300 K. Fig 9.9 Acetic acid (dimer) iv) Temperature An increase in temperature of the 9.9 Colligative properties solution increases the average kinetic energy of the molecules present in the solution Pure water is tasteless. When you add which causes decrease in the attractive force sugar it becomes sweet, while addition of salt between them. As a result, the solution makes it salty. It implies that the properties deviates from ideal behaviour. of a solution depend on the nature of solute v) Pressure particles present in the solution. However, for an ideal dilute solution, the properties, At high pressure the molecules tend namely, relative lowering of vapour pressure, to stay close to each other and therefore there elevation of boiling point, depression in will be an increase in their intermolecular freezing point and osmotic pressure do not attraction. Thus, a solution deviates from depend on the chemical nature of the solute Raoult’s law at high pressure. but depends only on the number of solute particles (ions/molecules) present in the vi) Concentration solution. These four properties are known as colligative properties. Though the magnitude of If a solution is sufficiently dilute there these properties are small, they have plenty of is no pronounced solvent-solute interaction practical applications. For example the osmotic because the number of solute molecules are pressure is important for some vital biological very low compared to the solvent. When the systems. concentration is increased by adding solute, the solvent-solute interaction becomes significant. This causes deviation from the Raoult’s law. 49 Unit 9.indd 49 23-12-2021 17:00:56  www.tntextbooks.in Relative lowering of vapour pressure The vapour pressure of a solution containing a nonvolatile, non-electrolyte solute is always lower than the vapour pressure of the pure solvent. Consider a closed system in which a pure solvent is in equilibrium with its vapour. At equilibrium the molar Gibbs free energies of solvent in the liquid and gaseous phase are equal (ΔG = 0). When a solute is added to this solvent, the dissolution takes place and its free energy (G) decreases due to increase in entropy. In order to maintain the equilibrium, the free energy of the vapour phase must also decrease. At a given temperature, the only way to lower the free energy of the vapour is to reduce its pressure. Thus the vapour pressure of the solution must decrease to maintain the equilibrium. We know that from the Raoult's law the relative lowering of the vapour pressure is equal to the mole fraction of the solute (equation 9.16) PSolution PSolvent o Solvent alone Solvent + Solute Figure 9.10 Measuring relative lowering of vapour pressure From the above equation,it is clear that the relative lowering of vapour pressure depends only on the mole fraction of the solute (xB) and is independent of its nature. Therefore, relative lowering of vapour pressure is a colligative property. Determination of molar mass from relative lowering of vapour pressure The measurement of relative lowering of vapour pressure can be used to determine the molar mass of a nonvolatile solute. In this method, a known mass of the solute is dissolved in a known quantity of solvent. The relative lowering of vapour pressure is measured experimentally. According to Raoult’s law the relative lowering of vapor pressure is, 50 Unit 9.indd 50 23-12-2021 17:00:57  www.tntextbooks.in 0 Psolvent − Psolution bar at the boiling point of the solvent. What 0 = xB is the molar mass of the solute when PA° is Psolvent 1.013 bar? Let wA and wB be the weights of the solvent and solute respectively and their ΔP WB x MA = corresponding molar masses are MA and PA° MB x WA MB, then the mole fraction of the solute xB is In a 2 % solution weight of the solute is 2 g nB xB = (9.20) and solvent is 98 g n A + nB Here, nA & nB are the moles of the ΔP = PA° – Psolution= 1.013 -1.004 bar = 0.009 solvent and the solute respectively. For dilute bar solutions nA>>nB. Hence nA +nB ≈ nA. Now PA° x WB xMA nB MB = xB = ΔP x WA nA Number of moles of solvent and the solute MB = 2 x 18 x 1.013/(98 x 0.009) are, = 41.3 g mol-1 wA wB =nA = , nB Evaluate Yourself MA MB ? wB 10) Vapour pressure of a pure liquid Therefore, x B = MB A is 10.0 torr at 27°C. The vapour wA pressure is lowered to 9.0 torr on MA ------ (9.21) dissolving one gram of B in 20 g of A. Thus, If the molar mass of A is 200 g mol-1 wB then calculate the molar mass of B. P° solvent – Psolution M ° = B Psolvent wA Elevation of boiling point MA ΔP w B × MA Boiling point is an important =........ ((9.22) ------ 9.22) physical property of a liquid. The boiling PA° w A × MB point of a liquid is the temperature at which From the equation (9.22) the molar its vapour pressure becomes equal to the mass of the solute (MB) can be calculated atmospheric pressure (1 atm). When a using the known values of wA, wB, MA and nonvolatile solute is added to a pure solvent the measured relative lowering of vapour at its boiling point, the vapour pressure of pressure. the solution is lowered below 1 atm. To bring the vapour pressure again to 1 atm, Example Problem3: the temperature of the solution has to be increased. As a result,the solution boils at An aqueous solution of 2 % a higher temperature (Tb) than the boiling nonvolatile solute exerts a pressure of 1.004 point of the pure solvent (Tb°). This increase 51 Unit 9.indd 51 23-12-2021 17:00:58  www.tntextbooks.in in the boiling point is known as elevation of boiling point. A plot of vapour pressure versus temperature for water and an aqueous solution is given below 1 atm Free t zing en o lv lines Liquid phase es pur f eo of so r Solid phase su p res Pressure our lid so Vap ut ion l f so lvent re o p ressu ur Vapo Gas phase ∆Tf ∆Tb Tf Tf0 Temperature Tb0 Tb Figure 9.11 Elevation of boiling point and depression in freezing point