XII Chemistry Notes PDF - Solutions

Join Now: https://join.hsslive.in Downloaded from https://www.hsslive.in ® 1. SOLUTIONS Solutions are homogeneous mixtures containing two or more components. Generally, the component that is present in la...

Join Now: https://join.hsslive.in Downloaded from https://www.hsslive.in ® 1. SOLUTIONS Solutions are homogeneous mixtures containing two or more components. Generally, the component that is present in larger quantity is called solvent. Solvent determines the physical state of the solution. One or more components present in the solution other than solvent are called solutes. [Or, the substance which is dissolved is called solute and the substance in which solute is dissolved is called solvent]. Solutions containing only two components are called binary solutions. Here each component may be solid, liquid or in gaseous state. Based on this, solutions are of the following types: Types of Solution Solute Solvent Examples Gas Gas Mixture of O2 and CO2 Gaseous solutions Liquid Gas Chloroform mixed with nitrogen gas, water-vapour in air Solid Gas Camphor in nitrogen gas, naphthalene in air Gas Liquid Oxygen dissolved in water, soda water Liquid solutions Liquid Liquid Alcohol dissolved in water, dilute acids and alkalies Solid Liquid Salt in water, glucose in water Gas Solid Hydrogen in Pd, Pt, Ni etc Solid solutions Liquid Solid Amalgam of mercury with sodium Solid Solid Gold ornaments, alloys of metals Concentration of Solutions Composition of a solution can be expressed in terms of concentration. Concentration is defined as the number of moles of solute present per litre of the solution. The concentration of a solution can be expressed by the following ways: (i) Mass percentage (w/w): It is defined as the mass of the component present in 100g of the solution. i.e. Mass % of a component = Mass of the component in the solution × 100 Total mass of the solution For e.g. 10% aqueous solution of glucose by mass means that 10 g of glucose is dissolved in 90 g of water resulting in a 100 g solution. Concentration described by mass percentage is commonly used in industrial chemical applications. (ii) Volume percentage (v/v): It is defined as the volume of a component present in 100 mL of the solution. i.e. Volume % of a component = Volume of the component ×100 Total volume of solution For example, 10% ethanol solution in water means that 10 mL of ethanol is dissolved in 90 mL of water such that the total volume of the solution is 100 mL. Concentration of solutions containing liquids is commonly expressed in this unit. (iii) Mass by volume percentage (w/v): It is the mass of solute dissolved in 100 mL of the solution. It is commonly used in medicine and pharmacy. Mass/volume % of a component = Mass of the component in the solution × 100 Total volume of the solution (iv) Parts per million (ppm): When a solute is present in trace quantities (i.e. very small amounts), its concentration is expressed in parts per million (ppm). It is defined as the number of parts of a particular component in million parts of the solution. i.e. Parts per million (ppm) = Number of parts of the component × 106 Total number of parts of all the components of the solution Concentration in parts per million can be expressed as mass to mass, volume to volume and mass to volume. +2 Chemistry Notes - Prepared by ANIL KUMAR K L, APHSS ADICHACHANALLOOR Page 3  Join Now: https://join.hsslive.in Downloaded from https://www.hsslive.in ® The concentration of pollutants in water or atmosphere is expressed in terms of μg mL –1 or ppm. [μg is microgram] (v) Mole fraction (χ): It is defined as the ratio of the number of moles of a particular component to the total number of moles of the solution. Mole fraction of a component = Number of moles of the component Total number of moles of the solution For example, in a binary solution, if the number of moles of A and B are n A and nB respectively, then the mole fraction of A (χA) = nA (nA + nB) and that of the component B (χB) = nB (nA + nB) χA + χB = 1 i.e. in a given solution sum of the mole fractions of all the components is unity. If there are i components, then χ1 + χ2 + χ3 +.................. + χi = 1 Mole fraction is useful in describing the calculations involving gas mixtures. (vi) Molarity (M): It is defined as the number of moles of solute dissolved per litre of solution. i.e. Molarity (M) = Number of moles of solute (n) Volume of solution in litre (V) For example, 1 M (molar) NaOH solution means that 1 mol (40g) of NaOH is dissolved in one litre of solution. (vii) Molality (m): It is defined as the number of moles of the solute present per kilogram (kg) of the solvent. i.e. Molality (m) = Number of moles of solute Mass of solvent in kg For example, 1 molal (m) solution of KCl means that 1 mol (74.5 g) of KCl is dissolved in 1 kg of water. Among the different methods for expressing the concentration of solution, mass percentage, ppm (in terms of mass), mole fraction and molality are independent of temperature; whereas molarity, mass by volume percentage and volume percentage are depend on temperature. This is because volume depends on temperature and the mass does not. SOLUBILITY Solubility of a substance is its maximum amount that can be dissolved in a specified amount of solvent at a particular temperature. It depends upon the nature of solute, nature of the solvent, temperature and pressure. Solubility of a Solid in a Liquid It is observed that polar solutes dissolve in polar solvents and non-polar solutes in non-polar solvents. In general, a solute dissolve in a solvent if the intermolecular interactions are similar in the two or the general principle related to solubility is that “like dissolves like”. Saturated and Unsaturated solutions When a solid solute is added to the solvent, some solute dissolves and its concentration increases in solution. This process is known as dissolution. Some solute particles in solution collide with the solid solute particles and get separated out of solution. This process is known as crystallisation. After sometime, the rate of dissolution and crystallization becomes equal and a dynamic equilibrium is reached. At this stage the concentration of solute in the solution remain constant and such a solution is called saturated solution. A solution in which no more solute can be dissolved at the same temperature and pressure is called a saturated solution. For such a solution, the concentration of the solution is equal to its solubility. A solution in which more solute can be dissolved at the same temperature is called an unsaturated solution. Effect of temperature on the solubility of a solid in a liquid The solubility of a solid in a liquid mainly depends on temperature. Since the dissolution of a solid in a liquid is an equilibrium process, it should follow Le Chateliers Principle. In general, if in a nearly saturated +2 Chemistry Notes - Prepared by ANIL KUMAR K L, APHSS ADICHACHANALLOOR Page 4  Join Now: https://join.hsslive.in Downloaded from https://www.hsslive.in ® solution, the dissolution process is endothermic (Δsol H > 0), the solubility should increase with rise in temperature and if it is exothermic (Δsol H > 0) the solubility should decrease with temperature. Effect of pressure on the solubility of a solid in a liquid Since solids and liquids are highly incompressible, pressure does not have any significant effect on solubility of solids in liquids. Solubility of a Gas in a Liquid Solubility of gases in liquids is greatly affected by pressure and temperature. The solubility of a gas increases with increase of pressure. A quantitative relation between pressure and solubility of a gas in a liquid was first given by Henry, which is known as Henry’s law. “The law states that at a constant temperature, the solubility of a gas in a liquid is directly proportional to the pressure of the gas”. Or, “the partial pressure of the gas in vapour phase (p) is proportional to the mole fraction of the gas (χ) in the solution” and is expressed as: p = KH χ Here KH is the Henry’s law constant. The value of KH depends on the nature of the gas and temperature. As the value of KH increases, the solubility of the gas in the liquid decreases. A graph of partial pressure (p) of the gas against mole fraction (χ) of the gas in solution is a straight line as follows. The slope of the graph gives the value of KH. Vapour pressure (p) Mole fraction (χ) As the temperature increases solubility of a gas in a liquid decreases. It is due to this reason that aquatic species are more comfortable in cold waters rather than in warm waters. Applications of Henry’s law 1. To increase the solubility of CO2 in soft drinks and soda water, the bottle is sealed under high pressure. 2. A medical condition known as bends in scuba divers. To avoid bends the cylinders used by scuba divers are filled with air diluted with helium (The composition of the air in the cylinders used by scuba divers is 32.1% oxygen, 56.2% nitrogen and 11.7% helium). 3. A medical condition known as anoxia in people living at high altitudes and in mountaineers or climbers. Effect of Temperature on the solubility of a gas in a liquid Solubility of gases in liquids decreases with rise in temperature. When dissolved, the gas molecules are present in liquid phase. So the process of dissolution can be considered similar to condensation, which is exothermic. Hence solubility decreases with increase of temperature. Vapour Pressure of a liquid In a liquid, the molecules with higher energy are escaped to vapour phase. This process is called evaporation. As the density of the vapour increases, the molecules collide with each other and so their energy decreases and returns to the liquid state. This process is called condensation. +2 Chemistry Notes - Prepared by ANIL KUMAR K L, APHSS ADICHACHANALLOOR Page 5  Join Now: https://join.hsslive.in Downloaded from https://www.hsslive.in ® After some time, the rate of evaporation becomes equal to rate of condensation and the two processes attain equilibrium. At this condition, the pressure exerted by the vapour is called vapour pressure. So vapour pressure is defined as the pressure exerted by the vapour in equilibrium with its own liquid. It depends on the nature of the liquid and the temperature. As the temperature increases, the vapour pressure also increases. Vapour Pressure of Liquid Solutions In liquid solutions, the solvent is always a liquid. The solute can be a gas, a liquid or a solid. Generally, the liquid solvent is volatile. The solute may or may not be volatile. Based on the volatility of solute, the vapour pressure of the solution is greater or less than that of the solvent. Vapour Pressure of Liquid-Liquid Solutions – Raoult’s Law A quantitative relationship between the vapour pressure and mole fraction of solute in a solution was first given by a French chemist F.M Raoult and it is known as Raoult’s Law. It states that for a solution of volatile liquids, the partial vapour pressure of each component in the solution is directly proportional to its mole fraction. Consider a binary solution of two volatile liquids 1 and 2. Let p 1 and p2 be the partial vapour pressures of the two components 1 and 2 respectively and ptotal be the total vapour pressure. Let χ1 and χ2 be the mole fractions of the two components 1 and 2 respectively. Then according to Raoult’s law, for component 1, p1 ∝ χ1 or, p1 = p10χ1 and for component 2, p2 ∝ χ2 or, p2 = p20χ2 Where p10 and p20are the vapour pressures of the pure components 1 & 2 respectively. According to Dalton’s law of partial pressures, the total pressure (p total) will be the sum of the partial pressures of the components of the solution. So, ptotal = p1 + p2 Substituting the values of p1 and p2, we get ptotal = χ1 p10 + χ2 p20 = (1 – χ2) p10 + χ2 p20 Or, ptotal = p10 + (p20 – p10) χ2 Plots of p1 or p2 against the mole fractions χ1 and χ2 give straight lines (I and II). Similarly the plot of ptotal versus χ2 (line III) is also linear. The composition of vapour phase in equilibrium with the solution is determined from the partial pressures of the components. If y1 and y2 are the mole fractions of the components 1 and 2 respectively in the vapour phase then, using Dalton’s law of partial pressures: p1 = y1 ptotal and p2 = y2 ptotal In general, pi = yi ptotal +2 Chemistry Notes - Prepared by ANIL KUMAR K L, APHSS ADICHACHANALLOOR Page 6  Join Now: https://join.hsslive.in Downloaded from https://www.hsslive.in ® Raoult’s Law as a special case of Henry’s Law According to Raoult’s law, the vapour pressure of a volatile component in a solution is given by P1 = χ1 p10. According to Henry’s law, solubility of a gas in a liquid is given by p = KH χ. If we compare the equations for Raoult’s law and Henry’s law, we can see that the partial pressure of the volatile component or gas is directly proportional to its mole fraction in solution. Only the proportionality constant KH differs from p10. Thus, Raoult’s law becomes a special case of Henry’s law in which KH becomes equal to p10. Ideal and non-ideal solutions Liquid – liquid solutions can be classified into ideal and non-ideal solutions on the basis of Raoult’s law. 1. Ideal solutions: These are solutions which obey Raoult’s law over the entire range of concentration. For such solutions, the vapour pressure of a component in the solution is equal to the vapour pressure of the pure component multiplied by its mole fraction in the solution. i.e. p1 = p10 χ1 and p2 = p20 χ2 For such solutions, no heat change occurs during the mixing of the pure components [i.e. ∆mixH = 0]. Also, the volume of the solution is equal to the sum of the volumes of the components. Or, there is no change in the volume of the solution during mixing the components [i.e. ∆mixV = 0] Thus for an ideal solution, p1 = p10 χ1 , p2 = p20 χ2 , ∆mixH = 0 and ∆mixV = 0 Ideal behaviour can be explained by considering two components A and B. In pure components, the inter molecular attractive interactions will be of types A-A and B-B. In solution, in addition to these two interactions, A-B type of interaction will also be present. If the A-A and B-B interactions are nearly equal to the A-B interaction, the solution behaves ideally. i.e. solute-solute interactions and solvent-solvent interactions are nearly equal to solute-solvent interaction. A perfectly ideal solution is rare. But some solutions are nearly ideal in behaviour. E.g. solutions of n-hexane and n-heptane, bromoethane and chloroethane, benzene and toluene etc. 2. Non-ideal solutions: These are solutions which do not obey Raoult’s law over the entire range of concentration. The vapour pressure of such a solution is either higher or lower than that predicted by Raoult’s law. If it is higher, the solution shows positive deviation from Raoult’s law and if it is lower, it shows negative deviation from Raoult’s law. (i) Solutions which show positive deviation from Raoult’s law: For such solutions, p1 > p10 χ1 , p2 > p20 χ2 , ∆mixH > 0 and ∆mixV > 0 Here A-B interactions are weaker than A-A and B-B interactions. i.e., in this case solute-solvent interactions are weaker than solute-solute and solvent-solvent interactions. So more molecules are escaped to vapour phase and hence the vapour pressure of the solution increases. E.g. solutions of ethanol and acetone, acetone and CS2, acetone and CCl4 etc. (ii) Solutions which show negative deviation from Raoult’s law: For such solutions, p1 < p10 χ1 , p2 < p20 χ2 , ∆mixH < 0 and ∆mixV < 0 Here A-B interactions are stronger than A-A and B-B interactions. i.e. solute-solvent interactions are stronger than solute-solute interaction and solvent-solvent interaction. So number of molecules escaped to vapour phase decreases and hence the vapour pressure of the solution decreases. E.g. solution of phenol and aniline, chloroform and acetone etc. The plots of vapour pressure curves for ideal solution, solutions which show positive deviation and negative deviation from Raoult’s law are as follows: +2 Chemistry Notes - Prepared by ANIL KUMAR K L, APHSS ADICHACHANALLOOR Page 7  Join Now: https://join.hsslive.in Downloaded from https://www.hsslive.in ® Solutions showing Positive Solutions showing Negative Ideal Solutions deviation from Raoult’s law deviation from Raoult’s law Azeotropes [Constant Boiling Mixtures] These are binary mixtures having the same composition in liquid and vapour phase and boil at a constant temperature. For such solutions, it is not possible to separate the components by fractional distillation. There are two types of azeotropes: minimum boiling azeotrope and maximum boiling azeotrope. The solutions which show a large positive deviation from Raoult’s law form minimum boiling azeotrope at a particular composition. E.g. 95% aqueous ethanol solution by volume at 78.20C [351.35 K]. The solutions which show large negative deviation from Raoult’s law form maximum boiling azeotrope at a particular composition. E.g. a mixture of 68% Nitric acid and 32% water by mass forms a maximum boiling azeotrope at 393.5 K. Vapour Pressure of Solutions of Solids in Liquids The vapour pressure of a liquid is the pressure exerted by the vapour in equilibrium with its own liquid. If a non-volatile solute is added to a pure solvent, the vapour pressure of the resulting solution is always lower than that of the pure solvent. This is because in a pure solvent, there are only solvent molecules, which can vapourise. But when a non-volatile solute is added to the solvent, a fraction of the surface is occupied by solute molecules. So the number of solute molecules passing to the vapour phase decreases and hence the vapour pressure also decreases. The decrease in the vapour pressure of solvent depends on the quantity of non-volatile solute present in the solution and not on its nature. For such a solution the Raoult’s law can be stated as, for any solution, the partial vapour pressure of each volatile component in the solution is directly proportional to its mole fraction in solution. Consider a binary solution containing a solvent 1 and solute 2. Since the solute is non-volatile, only the solvent molecules are present in vapour phase and contribute to vapour pressure. Let p1 be the vapour pressure of the solvent, χ1 be its mole fraction, p10 be its vapour pressure in the pure state. Then according to Raoult’s law, p1 ∝ χ1 or, p1 = p10χ1 A graph between the vapour pressure and the mole fraction of the solvent is linear as follows: +2 Chemistry Notes - Prepared by ANIL KUMAR K L, APHSS ADICHACHANALLOOR Page 8  Join Now: https://join.hsslive.in Downloaded from https://www.hsslive.in ® COLLIGATIVE PROPERTIES AND DETERMINATION OF MOLARMASS The properties which depend only on the number of solute particles and not on their nature are called Colligative properties. The important colligative properties are: Relative lowering of Vapour pressure, Elevation of Boiling point, Depression of Freezing point and Osmotic Pressure. 1. Relative lowering of Vapour Pressure When a non-volatile solute is added to a pure solvent, the vapour pressure (V.P) of the resulting solution is lower than that of the pure solvent. The difference between the vapour pressure of pure solvent and that of the solution is called lowering of vapour pressure (∆p). Consider a binary solution containing a non-volatile solute 2 dissolved in a solvent 1. Let p10 be the vapour pressure of pure solvent 1 and p1 be the vapour pressure of solution [V.P of solution = V.P of the solvent in the solution, since the solute is non-volatile]. Then according to Raoult’s law, p1 = p10.χ1 The lowering of vapour pressure of the solvent (∆p) = p10 – p1 = p 10 – p 10 χ 1 Or, ∆p = p10 (1-χ1) But χ1 + χ2 = 1. Therefore 1-χ1 = χ2 So ∆p = p10.χ2 Or, ∆p = χ2 , the mole fraction of the solute. p 10 Where ∆p/p10 is called relative lowering of vapour pressure. It is defined as the ratio of the lowering of vapour pressure to the vapour pressure pure solvent. But x2 = n2 n1 + n2 Where n1 and n2 are the number of moles of solvent and solute respectively. For dilute solutions, n2 1. Thus for NaCl, i =2, for K2SO4, i = 3, for CaCl2, i = 3 and for acetic acid in benzene, i = ½ Inclusion of van’t Hoff factor modifies the equations for colligative properties as follows: 1. Relative lowering of vapour pressure, ∆P = i x w2 x M1 P 10 w 1 x M 2 2. Elevation of Boiling point (∆Tb) = i.Kb.m 3. Depression of freezing point (∆Tf) = i.Kf.m 4. Osmotic Pressure (π) = i.CRT ********************************************************************************* +2 Chemistry Notes - Prepared by ANIL KUMAR K L, APHSS ADICHACHANALLOOR Page 12