1. Solutions

2. Topics  Solution process  Saturated, unsaturated, supersaturated  Miscibility, solubility  Hydrophobic, hydrophilic  Hydration, solvation  Henry’s law  Review Molarity  Concentrations, ppm, ppb  Colligative properties  Colloids, surfactants

3. A Solution is a Homogeneous mixture of two or more substances. Some examples of solutions are; Air 70 % isopropyl alcohol sugar water salt water 5 M HCl, etc.

4. The Solution Process For one substance to dissolve in another, they must have similar intermolecular forces. That is, polar substances dissolve other polar substances and non polar substances dissolve other non polar substances.

5. Like Dissolves Like This is a common phrase used in chemistry to describe which substances will dissolve in certain solvents. C 2 H 6 will dissolve in CCl 4 Why ? HCl will dissolve in H 2 O Why ?

6. The alkali metal salts, with the exception of Lithium compounds, are ionic and will dissolve in water. Some Ionic substances will dissolve in water which is polar.

7. Solubility Solubility of a substance is a relative concentration of the amount of the substance dissolved in a solute at a given temperature.

8. Sample problem: What is the concentration of a sulfuric acid solution if 20.0 mL of it are neutralized by 150. mL of 0.150 M Ca(OH) 2 ?

9. How much 2.0 M NaOH will it take to bring 200. mL of 0.51 M H 3 PO 4 to the endpoint?

10. Other ways to describe solution concentration are: Normality : N = equivalents Liters of solution Mole fraction :    n A / (n A + n B +…) Molality : m = moles of solute Kg of solvent Mass % : mass % = mass of solute * 100 mass of solution

11. Some helpful explanations: Mole fraction is a ratio of moles of a specific substance in a solution, to the total # of moles of all substances in the solution. Equivalents are either protons, H +, or hydroxides, OH -, or moles of electrons (as in a redox equation). For example; H 2 SO 4 has 2 equivalents per mole. HCl has 1 equivalent per mole. Ca(OH) 2 has 2 equivalents per mole.

12. Sample problem: In a solution prepared from 2.00 g of methanol plus 200. g of water, making 202 mL of solution, calculate the following: M, m, mass %,  methanol

13. Sample problem: A solution contains 12.0 g of glucose, C 6 H 12 O 6 In 95.0 g of water. Calculate the molality of the glucose.

14. Sample problem: Concentrated HCl is 36 % by mass, (This means 36 g HCl / 100 g solution). Calculate the mole fraction of HCl.

15. Sample problem: Calculate the molarity, M, of HCl in a 15.4 m solution if the density of the solution is D=1180g/L (for another example like this see example 11.2 in your text).

16. THE EFFECT OF TEMPERATURE ON SOLUBILITY: When solids or gases are dissolved in solution, an equilibrium is established: NaCl (s) Na + (aq) + Cl - (aq) We can predict the affect of temperature change on solubility equilibria by applying the following principle: An increase in temp. favors an endothermic process. In other words, if the solution process is one that absorbs heat, (endothermic), the increase in temp. will increase solubility. Conversely, if the process is exothermic, an increase in temp. will decrease solubility.

17. MOST SOLIDS DISSOLVING INTO SOLUTION ARE ENDOTHERMIC PROCESSES BECAUSE IT TAKES ENERGY TO BREAK DOWN THE CRYSTAL LATTICE. Solubility g /100g water Temperature o C 50 100 o C KNO 3 NaCl small +  H=small increase in solubility KClO 3

18. Dissolving a gas in solution is usually exothermic, that is, it evolves heat. Therefore an increase in temperature makes a gas less soluble. This is one cause of thermal pollution. Power plants discharge large amounts of warm water and there is thus, less dissolved oxygen, which is needed for the fish to survive.

19. THE EFFECTS OF PRESSURE ON SOLUBILITY: (This is only in appreciable amounts for gas – liquid solutions). Increase in pressure goes toward the area of lower moles of gas. Gas solubility is directly proportional to pressure. This is “Henry’s Law”. C g = kP g (continued)…

20. C g = k P g Where P = partial pressure of gas over solution, C g is the molarity of gas, and k is a constant specific for a particular gas in liquid solution which is equal to: K=Molarity of gas pressure of gas This occurs because an increase in pressure raises the concentration of gas molecules. To maintain equilibrium, more gas molecules enter the solution, thus increasing the Molarity of the gas in solution. M P

21. Sample problem: The solubility of O 2 in water at 20. O C and 1.00 atm is 1.38 M. Calculate the concentration of O 2 at 20 o C and a partial pressure of 0.21 atm. (you can first calculate the Henry’s law constant, K.) ( To see another example like this look at 11.4 in your text). This is why divers use a Helium mixture when they go on deep dives. Helium is insoluble in aqueous solution and therefore at high pressure, deep dives, less gas will dissolve in the bloodstream. When the diver surfaces, there will be less dissolved gasses to form insoluble bubbles in the blood and tissues which causes tremendous pain.

22. ENERGIES OF SOLUTION: The dissolution process is dependant on two things: The  H (soln) for the process The Entropy, S (soln) If the  H of solution is large, the solution process will not occur even though there is an increase in entropy. If the  H of solution is small, the solution process will occur due to the large increase in entropy to satisfy the 2 nd law of thermodynamics.

23. There are three steps to the solution process which either absorb or expel energy: 1.The energy required to separate the solute particles (  H 1 ). 2.The energy required to separate solvent molecules (  H 2 ). 3.The energy released in the solvation process (the solvent surrounding the solute), (  H 3 ). Note: the first two steps absorb energy (+  H), and the last step releases energy (-  H). REMEMBER THIS!

24. Many solids that dissolve well in water are ionic. They have the form of a crystal lattice (positive and negative particles lined up next to one another in an orderly fashion due to electrostatic attractions between them). The stronger these ionic particles are held together, the more energy is required to separate them (  H 1 is large). Generally, as charge increases and size of ion decreases, the charge density becomes GREATER and more energy is required to separate the ions. ex. NaCl has a larger charge density than Cs I. Molecular solids have smaller hydration energies (  H values), because the dipole or dispersion forces that hold them together are not as strong as the electrostatic forces that hold ionic particles together.

25. For a solvent such as water (very polar, therefore strong dipoles and H-bonding), the (  H 2 ) is also large. That is, it takes a lot of energy to separate the water molecules from one another. (  H 3 ) depends on the interactions between the solvent and the solute. Polar or ionic interactions will be great and therefore release a large amount of energy (large -  H 3 ). Non-polar mixed with polar or ionic substances have very little interactions so there would be a very small, if any release of energy (small –  H 3 ). Non-polar mixed with Non-polar substances have weak interactions (such as dispersion forces) and will therefore also release a small amount of energy (small -  H 3 ).

26. Some examples of how it works: An ionic solid in water : Large, +  H 1 + Large,+  H 2 + Large, –  H 3 = Small,+  H (soln) therefore solution process does occur. Non-polar solid in water : Small, +  H 1 + Large, +  H 2 + Small, -  H 3 = Large, +  H(soln) therefore no solution process occurs. Non-polar solid in Non-polar solvent : Small, +  H 1 + Small, +  H 2 + Small, -  H 3 =Small, +  H (soln) therefore solution process does occur.

27. WHY DOES THE SOLUTION PROCESS OCCUR IF THERE IS A +  H (Soln), which means an increase in energy (which opposes the first law of thermodynamics: the energy of the universe is constant) The answer is that the Large increase in entropy, S, that occurs when the solid particles come apart and go into solution, satisfies the second law of thermodynamics: the entropy of the universe is increasing. The conditions of a process will dictate whether the first or second law of thermodynamics will be satisfied. If the energy increase were large, then the entropy increase would not be sufficient to overcome the first law and therefore the solution process would not occur.