Physical Properties of Solution

2 Types of Solutions We can distinguish six types of solutions: ExampleSolutionSolventSolute AirGas Soda waterLiquid Gas H 2 in PdSolid Gas VinegarLiquid NaCl solnLiquid Solid AlloysSolid

3 A saturated solution: A solution that contains the maximum amount of a solute that will dissolve in a given solvent, at a specific temperature. An unsaturated solution: A solution that contains less solute than the solvent has the capacity to dissolve. A supersaturated solution: A solution that contains more solute than is present in a saturated solution.

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5 Molecular view of solution process When a solute dissolves in a solvent, particles of the solute disperse throughout the solvent. The ease with which a solute particle replaces a solvent molecule depends on the strength of: 1.solvent - solvent interaction. 2.solute - solute interaction. 3.solvent - solute interaction

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7 Intermolecular Attractions and Solubility (((LIKE DISSOLVES LIKE))) Benzene does not dissolve in water while methanol is freely soluble in water. Why is this behavior? To answer this and similar questions, we should consult our knowledge about intermolecular forces. First, let us have a close look at the solubility process:

8 § Solvent molecules should be expanded so that a space for solutes is created. In order to expand solvent molecules, we need to input energy in order to overcome the intermolecular forces § Solute molecules should also be expanded and we need to input energy to overcome intermolecular forces between solute molecules

9 § The solubility of expanded solute molecules in solvent will release energy. The released energy should, at least, be of the same magnitude as the input energy for expanding both solute and solvent molecules so that the solution process is an ideal one.

10 Now, consider the case of benzene and why it does not dissolve in water. A high input energy is needed to expand water molecules since there are relatively strong hydrogen bonds. These hydrogen bonds will be broken so that dispersion forces are formed between water and benzene molecules. This is an unfavorable process since we need to input much higher energy for solute expansion but will gain very little energy as dispersion forces are formed. Therefore, benzene is not expected to dissolve in water and the two solutions are said to be immiscible.

11 On the other hand, methanol dissolves freely in water since water and methanol have same type of intermolecular attractions (hydrogen bonding and dipole-dipole) and when hydrogen bonds in the solvent are broken, new hydrogen bonds with the solute are formed. This is a feasible process and such solutions are very stable.

12 When the hydrocarbon chain becomes longer and longer, the alcohol becomes less soluble (miscible) in water since the alcohol size becomes larger which requires breaking several hydrogen bonds to accommodate one solute molecule; an unfavorable process. Methanol, ethanol and propanol are all completely miscible with water while octanol is not. On the other hand miscibility decreases from butanol to heptanol

13 It could be fairly concluded that polar solutes will dissolve in polar solvents and vice versa. In another statement, one can say that the intermolecular forces in both solute and solvent should be alike for the solute to dissolve in the solvent. A non polar solute like I 2 has only dispersion forces as the intermolecular forces between I 2 molecules and will thus dissolve in a non polar solvent (has dispersion forces) like CCl 4 but will not dissolve appreciably in water.

14 Solubility of ionic compounds Ionic solids are highly polar and will only dissolve in very polar solvents like water. Even polar solvents like ethanol do not have enough polarity to dissolve most ionic solutes. Ions in water become hydrated where water molecules form a layer of water around each ion; thus decreasing the available charge, which adjacent ions can feel.

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16 Heats of Solutions (  H soln ) When solutes dissolve in solvents, the solution process either releases or absorbs energy. The amount of heat that is absorbed or released when a solute is dissolved in a solvent is called the heat of solution (  H soln ).  H soln is negative when the solution process releases heat (exothermic, like solubility of LiCl, LiI, AlCl 3, and Al 2 (SO 4 ) 3 in water) and is a positive value when the solution absorbs heat (as for the solution of KCl, KBr, NH 4 Cl, and NH 4 NO 3 in water).

17  H soln = 0 in the case where no heat is absorbed or released like the case of solubility of benzene in carbon tetrachloride). The magnitude of  H soln provides information about relative intermolecular forces of solute, solvent, and solution. When  H soln = 0, the solution is referred to as an ideal solution.

18 The solubility process of liquids or nonionized solids in liquid solvents involves the following steps: § Expanding solute molecules requires an input of energy equals  H solute to overcome potential energy. § Expanding solvent molecules requires an input of energy equals  H solvent to overcome potential energy. § Combining expanded solute and solvent molecules will release energy equals  H combination, due to decrease in potential energy (attraction occurs)

19 Three points could be made: 1.The solution process will be exothermic when the intermolecular forces between solute-solvent molecules are stronger than solute-solute or solvent-solvent molecules. 2.An endothermic solution process is observed when the intermolecular forces between solute-solute and solvent-solvent molecules are stronger than solute-solvent molecules.  H soln = 0 for ideal solutions

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21 For most solids and liquids dissolved in water, the solubility increases as the temperature is increased. This means that the equilibrium shifts towards forming more concentrated solutions at higher temperatures. At any temperature, a solute dissolves to the extent where the equilibrium at that definite temperature is reached. Solubility and Temperature

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23 In the case of the solubility of gas solutes in liquids, usually the solubility decreases as the temperature is increased. This is because the solubility of gases in liquids is almost always exothermic. According to Le Chatelier principle solubility will thus be inversely related to temperature.

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26 Concentration Units Molarity is defined as the number of moles of solute dissolved in one liter of solution Molality the number of moles of solute dissolved in one kg of solvent Mole Fraction The number of moles of a particular component to the overall number of moles of everything in solution is called the mole fraction of that component.

27 For example, a solution having 1.0 mole of acetone, 2.0 moles of ethanol, 3.0 moles of methanol, and 15.0 moles of water will have: X acetone = 1.0/( ) = 0.05 X ethanol = 2.0/( ) = 0.10 X methanol = 3.0/( ) = 0.15 X water = 14.0/( ) = 0.70 X T = = 1.0

28 Mole Percent The mole fraction multiplied by 100 is called the mole percent. In the above example: Mol % acetone = 0.05 x 100% = 5% Mol % ethanol = 0.10 x 100% = 10% Mol % methanol = 0.15 x 100% = 15% Mol % water = 0.70 x 100% = 70%

29 Weight Fraction The weight of a particular component to the total number of grams of solution is called the weight fraction of that component. W component = g component/g solution For example, a solution containing 12.0 g methanol and 38.0 g water has: w methanol = 12.0/( ) = 0.24 w water = 38.0/( ) = 0.76

30 Weight Percent The weight fraction multiplied by 100 is called the weight percent. In the example above: Wt% Methanol = 0.24 x 100% = 24% Wt% Water = 0.76 x 100% = 76% Note that summation of mole or weight percent of all components will add up to 100%. Also, summation of mole or weight fractions of all components will add up to 1.0.

31 Molality Molality is defined as the number of moles of solute dissolved in one kilogram of solvent. Therefore, the volume of one molal solution can be more or less than 1.0 L. Molality is not concerned with volume. You should be able to contrast molality and molarity and be able to use molality in calculations as well as conversion to other concentration units as mentioned earlier.

32 Example A certain aqueous solution contains 7% ethanol (C 2 H 5 OH) by mass. Calculate the mole fraction, mole percent, and the molality Solution 7% by mass would mean that we have 7g ethanol in a 100g sample (and, therefore, 93g H 2 O) Molar mass of ethanol is (2 x 12.0) + (6 x 1.01) + (1 x 16.0) = 46.1g/mole Molar mass of H 2 O is (2 x 1.01) + (1 x 16.0) = 18.0g/mole

33 Number of moles ethanol = 7/46.1 = mol Number of moles of water = 93/18.0 = 5.17 mol Total moles in sample is ( ) = 5.32 X ethanol = 0.152/5.32 = X water = 1 – = Mol% ethanol = x100% = 2.9% Mol% water = x100% = 97.1% Molality: In the same 100g sample above, the ethanol mol and the H 2 O weighs 93g (or kg), therefore: molality = 0.152mol/0.093 kg = 1.63 m