PHYSICAL PROPERTIES OF SOLUTIONS Do all the exercises in your study guide. PHYSICAL PROPERTIES OF SOLUTIONS A solution is a homogeneous mixture of a solute and a solvent. A solvent is a substance that dissolves a solute. For an example: water, milk, carbon tetrachloride, etc. A solute is a substance that dissolves in a solvent. For an example: Salt, Sugar, Eno, etc. PHYSICAL PROPERTIES OF SOLUTIONS For substances in the same phase: A solvent is a substance that is in majority in a solution. A solute is a substance that is in minority in a solution. For an example: In a homogeneous mixture of two liters of milk and a cup of water; milk is a solvent and water is a solute. PHYSICAL PROPERTIES OF SOLUTIONS Different types of solution: Liquid in Liquid Solid in Liquid. Gas in Liquid. Gas in Gas These are the units that shows the amount of solute in a solution or in a solvent. Molarity (M) which is more important in stoichiometry cannot be used in most colligative properties (which will be discussed in the coming section) because molarity indicates the number of moles of solute in a liter of a solution. Molarity does not specify the amount of solvent in a solution. The following concentration units can be used in most colligative properties and is going to be discussed in this section: Molality Mole fraction Weight percentage Molality is the number of moles of a solute in a kilogram of a solvent The formula indicating molality is shown on the next slide. 1
Example: Calculate the molality (m) of 20.0g of sodium chloride dissolve in 50.0 g of water. Solution: (a) Convert 20.0g NaCl to mol. (b) Convert 50.0g H 2 O to kg. (c) Apply a formula. (a) = 0.34223 mol NaCl (b) (c) = 0.05 kg H 2 O = 6.8446mol/kg= 6.85m Mole fraction ( ) is the moles of a solute divided by the number of moles of solute plus the number of moles of solvent. Example: Calculate the (a) Mole fraction ( ) of NaCl and the (b) mole fraction ( )of H 2 O, if 20.0g of sodium chloride dissolve in 50.0 g of water. Solution: (a) Convert 20.0g NaCl to mol. (b) Convert 50.0g H 2 O to mol. (c) Apply a formula. 2
(c) =0.34223 mol NaCl =2.753304mol = 0.1112845 =0.111 But the mole fraction must always be less than one (1). The mole fraction of NaCl plus the mole fraction of H 2 O is equals to 1. Therefore: (b) Therefore + = 1 = 1 - Substituting from the above problem, we get: = 1 - = 1-0.1112845 = 0.8887 = 0.889 Weight percentage is the mass of a solute divided by the mass of a solute plus the mass of a solvent multiplied by hundred. Example: Calculate the weight percentage of NaCl if 20.0g of sodium chloride dissolve in 50.0 g of water. Solution: Apply a formula. 3
THE SOLUTION PROCESS This is the process in which the solute dissolves in a solvent to form a solution. There are three types of solutions: = 0.97560976 =0.975 Unsaturated saturated supersaturated THE SOLUTION PROCESS Unsaturated solution is a solution in which a less solute dissolve in solvent. Saturated solution is a solution in which a solute is just enough to dissolve in solvent. In a microscopic level; the number of solute particles dissolved is equal to the number of dissolved particles forming a solute in a solution. That is, one may not notice the formation of solute in a solution with the naked eyes. These are in dynamic equilibrium. THE SOLUTION PROCESS For an example: If NaCl (s) is dissolved in water to form a saturated solution; then the number of Na + (aq) and Cl - (aq) that are formed is equal to the number of NaCl (aq) that are formed as it is indicated below NaCl (s) NaCl (aq) Na + (aq)+ Cl - (aq) THE SOLUTION PROCESS Supersaturated solution is a solution in which a solute is more than enough to dissolve in a solvent. That is, one can notice the formation of solute in a solvent with the naked eyes. Do all the exercises in your study guide. LIQUIDS DISSOLVING IN LIQUIDS This is a process in which two or more liquids are miscible. Example 1: Hydrochloric acid and water Ethanol and water These are miscible with water because they are also polar substances. Like dissolve like 4
LIQUIDS DISSOLVING IN LIQUIDS Example 2: Oil and paraffin Oil and petrol These are miscible with oil because they are both non-polar substances and they can dissolve in a non-polar substance (oil). LIQUIDS DISSOLVING IN LIQUIDS In contrast, immiscible liquids are liquids that cannot mix with each other. Example: oil and water ethane and water These are not miscible with water because they are both non-polar substances and they cannot dissolve in a polar substance (water). LIQUIDS DISSOLVING IN LIQUIDS Like dissolve like They form a heterogeneous mixture and not a homogeneous mixture (solution) SOLIDS DISSOLVING IN WATER This is a process in which a solid substance dissolves in water. Example: Sugar and Water Salt (NaCl) and Water Although most solids are soluble in water, not all the solid substance are soluble in water. For example, the network solids are not soluble in water: Diamond, graphite, etc are not soluble in water. SOLIDS DISSOLVING IN WATER In the case of the network solids it is because their intra-molecular forces (covalent bonds) are so strong that their enthalpy of dissolution is higher than their enthalpy of hydration. ENTHALPY OF SOLUTION This is the sum of the change in lattice enthalpy and the change in enthalpy of solvation. The change in enthalpy of solution ( soln H) must be negative for a solute to dissolve in a solvent. If soln H is negative, then the solute dissolves spontaneously in a solvent. If soln H is positive, then the solute do not dissolves spontaneously in a solvent. 5
ENTHALPY OF SOLUTION Example of the above-mentioned process is the dissolution of KF in H 2 O. In this case, you must have enough energy to separate the ionic bond between potassium and fluorine atom to form K + and F -. This is called the change in lattice enthalpy ( lattice H) of KF. This is an endothermic reaction. ENTHALPY OF SOLUTION Again there should be an energy released when each of these ions is solvated by water molecules. This is called an the change in enthalpy of solvation ( solv H). If the solvent is water the it is: the change in enthalpy of solvation ( hydr H). It is an exothermic reaction. When adding solv H and lattice H you should get a negative soln H for a spontaneous reaction: (+ lattice H) + (- solv H) = - soln H. ENTHALPY OF SOLUTION The above process is shown below: Formation of ions in gaseous form: KF (s) K + (g) + F - (g) -- lattice H = +821 kj/mol ENTHALPY OF SOLUTION Formation of the solvated or hydrated ions. K + (g) + F - (g) K + (aq) + F - (aq) -- solv H = -837kJ/mol Net process: KF (s) K + (aq) + F - (aq) -- soln H = -16 kj/mol ENTHALPY OF SOLUTION:THERMODYNAMIC DATA When the enthalpy of solution is determined under standard conditions (STP) and one molal (m) of the product is formed from its elements at their standard conditions and in their stable form, then the change in enthalpy of solution is called change in standard enthalpy of formation ( f H o ). These values together with the standard enthalpy of solution ( soln H o ) are listed in Table 14.1 and appendix L of your text book, Kotz et al. FACTORS AFFECTING SOLUBITITY There are three factors that influence the solubility: Solute-Solvent Pressure Temperature 6
SOLVENT - SOLUTE INTERACTION Solvent-solute inter-action is one of the main factors that affect the solubility of a solute in a solvent. The stronger the force of attraction between the solute and the solvent molecules, the greater the solubility. In this case, the solubility will increases when the force of attraction between the solute and the solvent increase. SOLVENT - SOLUTE INTERACTION Like dissolve like: Substances with similar intermolecular forces tend to be soluble in one another. For an example: The polar solute tend to be soluble in polar solvent. e.g. HCl can easily dissolve in H 2 O. The non-polar solute tend to be soluble in nonpolar solvent. e.g. ethane can easily dissolve in heptane. SOLVENT SOLUTE INTERACTION Like dissolve like: Substances with different intermolecular forces tend to be insoluble in one another. For an example: The polar solute tend to be insoluble in non-polar solvent. e.g. HCl cannot easily dissolve in octane. The non-polar solute tend to be insoluble in polar solvent. e.g. ethane cannot easily dissolve water. PRESSURE EFFECT The solubility of solids and liquids are not appreciably affected by pressure. The solubility of gas in any liquid is directly proportional to its partial pressure. The solubility of gas in liquid can be easily shown by Henry s law. Do all the exercises in your study guide. DISSOLVING GASES IN LIQUIDS: HENRY S LAW Henry s law state that: The solubility of a gas in a liquid is directly proportional to the gas pressure. Mathematically it is written as: S g α P g Where: S g is the solubility of a gas in mol/kg. P g is the partial pressure of a gaseous solute. DISSOLVING GASES IN LIQUIDS: HENRY S LAW Introducing a constant called Henry s law constant, we get: S g = k H P g Where: S g is the solubility of a gas in mol/kg. P g is the partial pressure of a gaseous solute. K H is Henry s law constant. This allow us to substitute proportionality with the equal sign. 7
DISSOLVING GASES IN LIQUIDS: HENRY S LAW Example of this is the dissolution of CO 2 in a bottled or canned beverages. When any of these is opened, the partial pressure of CO 2 decreases and the gas bubble out of bottle or can or bottle. The solubility of a gas is the concentration of the dissolved gas in equilibrium with the substance in gaseous state. DISSOLVING GASES IN LIQUIDS: HENRY S LAW That is, it is the state at which the number of gas molecules leaving the surface of the solution is equal to the number of molecules reentering the solution. The increase in pressure results in more molecules striking the surface of the liquid and entering the solution in a given time. DISSOLVING GASES IN LIQUIDS: HENRY S LAW The solution will eventually reach a new equilibrium when the concentration of a gas dissolved in the solvent is higher enough that the rate of gas molecules escaping the solution again equals to the rate at which the gas molecules enter the solution. TEMPERATURE EFFECT ON SOLUBILTY: LE CHATELIER S PRINCIPLE The solubility of all gases is inversely proportional to temperature. That is, the solubility of all gases decreases as the temperature increases. Gases that are appreciable soluble in water, they do so in exothermic process as is shown below. H soln <0 gas+ liquid solvent saturated solution + energy TEMPERATURE EFFECT ON SOLUBILTY: LE CHATELIER S PRINCIPLE Gases that are not appreciable soluble in water, they do so in endothermic process. The system absorb energy to increase the temperature; decrease the formation of the product and increase the formation of the reactants This will be the reverse of the above mentioned reaction. TEMPERATURE EFFECT ON SOLUBILTY: LE CHATELIER S PRINCIPLE Loss of dissolved gas molecules from a solution, require energy. At equilibrium the rates of the two processes are the same. Temperature effect is well-understood when using Le Chatelier s principle state that A change in any of the factors determining an equilibrium causes the system to adjust by shifting in the direction that reduces or counteracts the effect of the change 8
TEMPERATURE EFFECT ON SOLUBILTY: LE CHATELIER S PRINCIPLE gas+ liquid solvent Exothermic process H soln <0 Add energy: equilibrium shift to the left saturated solution + energy TEMPERATURE EFFECT ON SOLUBILTY: LE CHATELIER S PRINCIPLE As Le Chatelier s principle have said, one of the factors that determine equilibrium (temperature) is changed to increase the formation of reactant; but the system is counteracting this change. In the above reaction the equilibrium shift to the left as the energy is added to increased the temperature. TEMPERATURE EFFECT ON SOLUBILTY: LE CHATELIER S PRINCIPLE Unlike the dissolution of gases, the solubility of most solids in water is directly proportional to the temperature. Examples of this are: salt (NaCl) in water sugar (C 12 H 22 O 11 ) in water This is not true for all solids, the solubility of some solids decreases as the temperature increases. Example of this is cesium sulphate (Ce 2 SO 4 ) TEMPERATURE EFFECT ON SOLUBILTY: LE CHATELIER S PRINCIPLE Exercise: Do Example 14.3 on page 627 of your text book. COLLIGATIVE PROPERTIES Colligative properties of a solution are properties of a solution that depend on the number of solute particles per solvent molecules and not on the identity the solute. The following colligative properties will be discussed: Change in vapor pressure: Rauolt s Law Boiling point elevation Freezing point depresion Osmotic pressure The vapor pressure (P) of a substance at a particular temperature is the pressure of the vapour when the liquid and the vapor are at equilibrium. The vapor pressure of a solvent above the solution is lower than the vapor pressure of a pure solvent. The vapor pressure of a solvent is directly proportional to the number of solvent molecules in a solution. 9
Rauolt s law states that The vapor pressure of a solvent is directly proportional to the mole fraction. P solvent α solvent Where P solvent is the vapor pressure of a solvent. is the mole fraction of a solvent. PRESSURE: RAOULT S LAW Because solvent vapor pressure is proportional to the number of solvent molecules, we can write the following equation for the equilibrium vapor pressure of the solvent over the solution: P solvent = solvent P o solvent Where P solvent is the vapor pressure of a solvent. P o solvent is the vapor pressure of a pure solvent. is the mole fraction of a solvent. The above equation is called Raoult s law. The ideal solution is the solution in which Raoult s law hold. Remember that no solution is truly ideal like the ideal gas. For Raoult s law to hold the forces of attraction between the solute and the solvent molecules must be the same as those between the solvent molecules in a pure solvent. This is always the case with like dissolve like. PRESSURE: RAOULT S LAW If the solvent-solute interaction is stronger than the solvent-solvent interaction, the actual vapor pressure (P o solvent ) will be lower than the calculated Raoult s law (P solvent ). P o solvent < P solvent If the solvent-solute interaction is weaker than the solvent-solvent interaction, the actual vapor pressure (P o solvent ) will be higher than the calculated Raoult s law (P solvent ). P o solvent > P solvent The answer shows that The actual vapor pressure (P o solvent ) is higher than the calculated Raoult s law (P solvent ). That is P o solvent > P solvent Therefore, the solvent-solute interaction is weaker than the solvent-solvent interaction. To check this lowering actual vapor pressure, we subtract the actual one from the calculated one (Rauolt s law value) P solvent = P solvent - P o solvent = 467 mmhg - 525.8mHg = -58.8mmHg -59mmHg 10
Adding a non-volatile solute (like salt or sugar) to a solvent (like water) lowers the vapor pressure of a solvent. We can therefore calculate P solvent directly from the mole fraction of the solute ( and the actual vapor pressure of the solvent (P solvent ). Let derive this formula from the previous one. solute) P solvent = P solvent but P solvent = Therefore P solvent = - P o solvent solvent x P o solvent solventp o solvent -P o solvent Take out P o solvent as a common factor: P solvent = P o solvent - P o solvent = -P o solvent(- solvent +1) = -P o solvent(+1 - solvent ) = -P o solvent(1 - solvent ) For the solution containing only volatile solvent and the non-volatile solute, the sum of their mole fraction must be one. solute + solvent =1 solute = 1 solvent = - (1 - solvent) P o solvent Substituting 1 solvent by solute in the equation below, we get: P solvent =- (1 - solvent) P o solvent = - solutep o solvent BOILING POINT ELEVATION A boiling point is a temperature at which the vapor pressure of a substance is equal to the atmospheric pressure (1 atm or 760mmHg). When adding the non-volatile solute to a volatile solvent, the vapor pressure of that solvent will decrease. This will result in an increase in a boiling point of that solvent. 11
BOILING POINT ELEVATION That is because the vapor pressure of this solution is lower than the actual vapor pressure, the temperature must be taken higher for that solvent to reach 1 atm or 760 mmhg. This new boiling point or elevated boiling point is directly propotional to the molality of a solution. That is: T bp α m BOILING POINT ELEVATION This elevation in boiling point ( T bp ) is calculate as T bp = k bp m Where k bp is the molal boiling point elevation constant. m is the molality of that solution. See the figure below: BOILING POINT ELEVATION FREEZING POINT DEPRESSION Freezing point is a temperature at which the condensation and melting of that substance are at equilibrium. The freezing points of an ideal solution are affected by a decrease in vapor pressure. The decrease in vapor pressure of a substance results in a decrease in freezing point. The freezing point is directly proportional to the molality of a solute in a solution. T fp α m FREEZING POINT DEPRESSION Like the boiling point elevation; a freezing point depression is calculated by: T fp = k fp m Where k fp is the molal freezing point depression constant in o C/m. m is the molality of that solution. Osmosis is the movement of water molecules through a semi-permeable membrane from the region of higher concentration of water (hypertonic region) to the region of lower concentration of water (hypotonic region) of the same molecules until the two regions have the same concentration of water molecules (isotonic). Examples of this terms are shown on the figure below 12
On the figure (a) above, the blood cell with lower concentration of solute (hypertonic solution) is placed in the in the solution that has higher concentration of solute (hypotonic solution). The water flow from the cell into the solution. The cell will shrivel. this process is called crenation. On Figure (b), The cell with higher concentration of solute (hypotonic solution ) is placed in the solution of lower concentration of solute (hypertonic solution). The water flow from the solution into the cell. This cause the cell to rupture. This process is called hemolysis. If a liquid water and an aqueous solution, like NaCl (aq), C 12 H 22 O 11(aq), etc, are separated by a semipermeable membrane, then water will flow from the water section to the solution section until the concentration of water molecules is the same on both sides of the membrane. That is until the two solutions are isotonic. Examples of this is shown on the figure below At this time the level of the solution will increase to the new level. The different between the new and the initial level is called. This is the pressure caused by osmosis. It is calculated by using the following formula. 13
On the above mentioned formula: Is the osmotic pressure. c is the concentration of the solution in molarity. R is the gas constant (0.082057 L.atm/K) T is the absolute temperature in Kelvin. C, is the same as; This equation can therefore be written as COLLIGATIVE PROPERTIES AND MOLAR MASS CALCULATION Colligative properties can be used to calculate the molar mass of a compound. One can get the molar mass from: Vapor pressure lowering Boing point elevation Freezing depression Osmotic pressure COLLIGATIVE PROPERTIES AND MOLAR MASS CALCULATION Calculation of molar mass from the boiling point elevation is shown on the example below. COLLIGATIVE PROPERTIES AND MOLAR MASS CALCULATION Molar mass can also be calculated from the osmotic pressure. COLLIGATIVE PROPERTIES OF SOLUTIONS CONTAINIG IONS. These are the solutions formed by the solutes that totally dissolve and completely dissociate in the solvent. For an example: the following salts totally dissolve and completely dissociate in water: NaCl Na 2 SO 4 (Al) 2 (SO 4 ) 3 14
COLLIGATIVE PROPERTIES OF SOLUTIONS CONTAINIG IONS. The ions so formed can be used to calculate the molality of the solution and to estimate the freezing point depression of that solution. For an example: COLLIGATIVE PROPERTIES OF SOLUTIONS CONTAINIG IONS. To estimate the freezing point depression for the ionic compound, find the molality of the solution, from the mass and the molar mass of the compound and the mass of the solvent. Then multiply the molality by the number of ion in the formula. COLLIGATIVE PROPERTIES OF SOLUTIONS CONTAINIG IONS. For an example: Two ions for NaCl Three ions for Na 2 SO 4 Five ions for (Al) 2 (SO 4 ) 3 The T fp can be measured or calculated. The measured one is called T fp,measured The calculated one is called T fp,calculated. The ratio (i) between T fp,measured and T fp,calculated. and is called van t Hoff Factor. COLLIGATIVE PROPERTIES OF SOLUTIONS CONTAINIG IONS. That is: N.B. T fp is the change in freezing point before the ions are formed COLLIGATIVE PROPERTIES OF SOLUTIONS CONTAINIG IONS. The above ratio decreases and approaches the whole number of the ion in the compound but do not quite reaches it. For an example it approaches Two for NaCl, Three for Na 2 SO 4, Five ions for (Al) 2 (SO 4 ) 3,etc. This is true for the dilute solution. COLLIGATIVE PROPERTIES OF SOLUTIONS CONTAINIG IONS. The value Can be written as 15
COLLIGATIVE PROPERTIES OF SOLUTIONS CONTAINIG IONS. This shows that, the measured value is: T fp, measured = T fp x mx i 16