# CHAPTER 11 - CHAPTER 11 PROPERTIES OF SOLUTIONS 11.1...

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CHAPTER 11 PROPERTIES OF SOLUTIONS 11.1 Solution composition: Molarity ( M ) = (moles of solute)/(liters of solution) = n / V Molality ( m ) = (moles of solute)/(kilograms of solvent) = n / kg Mass percent = (mass of solute)/(mass of solution) x 100 Mole fraction ( X ) = (moles of components)/(total moles) e.g : For components A and B: 1 = + + = + = B A B A B B B A A A X X n n n X n n n X Normality ( N ) : It is defined as the number of equivalents per liter of solution.

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Normality ( N ) is the number of equivalents per liter of solution: 1) In acid-base reaction : The equivalent is the mass of acid or base that can take or give exactly 1 mole of protons . 2) In redox reaction : The equivalent is the quantity of oxidizing or reducing agent that can take or give 1 mole of electrons . Table 11.2 : Acid or base Molar mass Equivalent mass Relation between ( M ) and ( N ) HCl 36.5 36.5 1 M = 1 N H 2 SO 4 98 98/2 = 49 1 M = 2 N NaOH 40 40 1 M = 1 N Ca(OH) 2 74 74/2 = 37 1 M = 2 N 2
11.2 The Energies of Solution Formation: The formation of a liquid solution takes place in three distinct steps : 1) and 2) Separating each of the solute and solvent particles. These two expansion steps are endothermic (overcoming intermolecular forces). 3) Interaction between solute and solvent to form solution . This step is often exothermic . The enthalpy change associated with the formation of the solution is called enthalpy ( or heat ) of solution Δ H soln and is the sum of the ΔH values of the three steps: Δ H soln = Δ H 1 + Δ H 2 + Δ H 3 In general : Like dissolves like” Three factors affect solubility : 1) Structural 2) Pressure 3) Temperature 3

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11.3 Factors Affecting Solubility: 1) Structural effects : Solubility is favored if the solute and solvent have similar polarity . e.g : Vitamins can be divided into two classes: 1. Fat soluble ( e.g : vitamin A) 2. Water soluble ( e.g : vitamin C) Vitamin A compassed mostly of carbon and hydrogen (i.e. non - polar ) Vitamin C has many OH and CO bonds (i.e. polar ) 2) Pressure effects : While pressure has little effect on the solubility of solids or liquids , it does significantly increase the solubility of gases : For a gas: P α C (concentration of dissolved gas) = k C ( Henry’s Law ) 3) Temperature effect : ( for aqueous solutions ) Predicting the temperature dependence of solubility is very difficult. Solubility of most solids in water increases with increasing the temperature ( see Figure 11.6 ) For some substance such as sodium and cerium sulfates [Na 2 SO 4 and Ce 2 (SO 4 ) 3 ] the solubility decreases with increasing temperature. 4
Example 11.4 : A certain soft drink is bottled so that a bottle at 25 o C contains CO 2 gas at a pressure of 5.0 atm over the liquid. Assuming that the partial pressure of CO 2 in the atmosphere is 4.0 x 10 -4 atm, calculate the equilibrium concentration of CO 2 in the soda both before and after the bottle is opened. The Henry’s law constant for CO 2 in aqueous solution is 32 L.atm/mol at

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CHAPTER 11 - CHAPTER 11 PROPERTIES OF SOLUTIONS 11.1...

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