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Prentice Hall © 2003Chapter 13 Hlanganani Tutu, C403 School of Chemistry

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Presentation on theme: "Prentice Hall © 2003Chapter 13 Hlanganani Tutu, C403 School of Chemistry"— Presentation transcript:

1 Prentice Hall © 2003Chapter 13 Hlanganani Tutu, C403 School of Chemistry Email: hlanganani.tutu@wits.ac.za

2 Prentice Hall © 2003Chapter 13 Solution - homogeneous mixture of solute and solvent. In solutions, intermolecular forces become rearranged. The Solution Process

3 Prentice Hall © 2003Chapter 13 gas in gas – e.g. air gas in liquid -- e.g. soda gas in solid -- e.g. gas on solid, catalyst liquid in liquid liquid in solid -- e.g. mercury amalgam solid in liquid solid in solid -- e.g. 14-karat gold, brass Examples of solutions

4 Prentice Hall © 2003Chapter 13 Consider NaCl (solute) dissolving in water (solvent): –Interruption of water H-bonds, –NaCl→Na + + Cl -, –ion-dipole forces form: Na + …  -OH 2 and Cl - …  +H 2 O. –If water is the solvent, we say the ions are hydrated.

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7 Prentice Hall © 2003Chapter 13 Energy Changes and Solution Formation 3 energy steps in forming a solution: separation of solute molecules (  H1), separation of solvent molecules (  H2), and formation of solute-solvent interactions (  H3).

8 Prentice Hall © 2003Chapter 13  H soln =  H 1 +  H 2 +  H 3.  H soln can be +ve or -ve depending on the intermolecular forces.

9 Prentice Hall © 2003Chapter 13

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11 Prentice Hall © 2003Chapter 13 “Rule”: polar solvents dissolve……………………….? Non-polar solvents dissolve..............................................?

12 Prentice Hall © 2003Chapter 13 Exercise: Why doesn’t gasoline dissolve NaCl? Exercise: Why doesn’t water and octane mix well (immiscible)? Remember: the resultant solution’s interactions must be stronger than the interactions in the original substance

13 Prentice Hall © 2003Chapter 13 Solution Formation, Spontaneity, and Disorder When the energy of the system decreases (e.g. dropping a book and allowing it to fall to a lower potential energy), the process is spontaneous.

14 Prentice Hall © 2003Chapter 13 Example: a mixture of CCl 4 and C 6 H 14 is less ordered than the two separate liquids. Therefore, they spontaneously mix

15 Prentice Hall © 2003Chapter 13

16 Prentice Hall © 2003Chapter 13 There are solutions that form by physical processes and those by chemical processes.

17 Prentice Hall © 2003Chapter 13 Consider: Ni(s) + 2HCl(aq)  NiCl 2 (aq) + H 2 (g). When all the water is removed from the solution, no Ni is found only NiCl 2 ·6H 2 O. Therefore, Ni dissolution in HCl is a chemical process.

18 Prentice Hall © 2003Chapter 13 Consider: NaCl(s) + H 2 O (l)  Na + (aq) + Cl - (aq). When the water is removed from the solution, NaCl is found. Therefore, NaCl dissolution is a physical process.

19 Prentice Hall © 2003Chapter 13 Dissolve: solute + solvent  solution. Crystallization: solution  solute + solvent. Saturation: crystallization and dissolution are in equilibrium. Saturated Solutions and Solubility

20 Prentice Hall © 2003Chapter 13 Solubility: amount of solute required to form a saturated solution. Supersaturation: reached when more solute is dissolved than in a saturated solution.

21 Prentice Hall © 2003Chapter 13

22 Prentice Hall © 2003Chapter 13 Solute-Solvent Interaction Miscible liquids: mix in any proportions. Immiscible liquids: do not mix. Intermolecular forces are important The more C atoms, the less the solubility in water. Factors Affecting Solubility

23 Prentice Hall © 2003Chapter 13 The -OH groups in a molecule increase solubility in water….“like dissolves like”

24 Prentice Hall © 2003Chapter 13

25 Prentice Hall © 2003Chapter 13 Which of these two would be more soluble in water?

26 Prentice Hall © 2003Chapter 13 Network solids do not dissolve. Why?

27 Prentice Hall © 2003Chapter 13 Pressure Effects Solubility of a gas in a liquid is a function of the pressure of the gas.

28 Prentice Hall © 2003Chapter 13 Pressure Effects

29 Prentice Hall © 2003Chapter 13 The higher the pressure, the more molecules of gas are close to the solvent

30 Prentice Hall © 2003Chapter 13 where: S g - solubility of a gas, k is a constant, and P g is the partial pressure of a gas Henry’s Law gives:

31 Prentice Hall © 2003Chapter 13 Example 27g of acetylene, C 2 H 2, dissolves in 1L of acetone at 1.0 atm pressure. If the partial pressure of acetylene is increased to 12 atm, what is the solubility in acetone? Solution: S 1 = kP 1 …………(1) S 2 = kP 2 …………(2) Ans: 3.2 x 10 2 g

32 Prentice Hall © 2003Chapter 13 Carbonated beverages are bottled with a partial pressure of CO 2 > 1 atm. What happens when a bottle is opened?

33 Prentice Hall © 2003Chapter 13 Temperature Effects As temperature increases, solubility of solids generally increases, e.g. sugar in warm water Sometimes, solubility decreases as temperature increases (e.g. Ce 2 (SO 4 ) 3 ).

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35 Prentice Hall © 2003Chapter 13 Temperature Effects Gases - less soluble at high temperature Thermal pollution in dams and rivers – loss of O 2

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37 Prentice Hall © 2003Chapter 13 Mass Percentage, ppm, and ppb Definitions: Ways of Expressing Concentration

38 Prentice Hall © 2003Chapter 13 Example: How would you prepare 425 g of an aqueous solution containing 2.40% by mass of sodium acetate, NaC 2 H 3 O 3 ? Ans: Mass of NaC 2 H 3 O 3 = 10.2 g Mass of H 2 O = mass of solution - mass of NaC 2 H 3 O 3 = 415 g

39 Prentice Hall © 2003Chapter 13

40 Prentice Hall © 2003Chapter 13

41 Prentice Hall © 2003Chapter 13 Mole Fraction, Molarity, and Molality

42 Prentice Hall © 2003Chapter 13 Converting between molarity (M) and molality (m) requires density. Exercise: 0.2 mol of ethylene glycol is dissolved in 2000 g of water. Calculate the molality

43 Prentice Hall © 2003Chapter 13 Example: What is the molality of a solution containing 5.67 g of glucose, C 6 H 12 O 6 (M r = 180.2 g), dissolved in 25.2 g of water? (Calc. the mole fractions of the components as well). Solution:  Think about the solute!................glucose (express in moles)  Think about the solvent!...............water (express in kilograms) Ans: 1.25 m

44 Prentice Hall © 2003Chapter 13 Example: Converting molarity to molality An aqueous solution is 0.907M Pb(NO 3 ) 2. What is the molality of lead nitrate, Pb(NO 3 ) 2, in this solution? The density of the solution is 1.252 g/mL. (Molar mass of Pb(NO 3 ) 2 = 331.2 g) Ans: 0.953 m Pb(NO 3 ) 2


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