1. Chemistry Solutions

2. I. Introduction: Terminology Solute: substance to be dissolved -- could be solid, liquid, or gas Solvent: substance doing the dissolving Solution: Solute + Solvent

3. II. How Solutes Dissolve in Liquid Solvents (Like dissolves like) Ionic Solutes: A. Non-polar solvents -- do not dissolve B. Polar solvents -- Break into ions in polar solvent (-) ions attracted to (+) end of molecule (+) ions attracted to (-) end of molecule

5. Molecular (covalent) solutes: A. Non-polar solutes -- only non-polar solvents B. Polar solutes Dissolve in polar solvents (like H 2 O) e.g.. PH 3 (-) portion (P) is attracted to (+) H’s in water (+) portion (H’s –in PH 3 ) are attracted to (-) O’s in water

6. PH 3 in H 2 O

7. How Solubility Works (when the solvent is liquid) Solid solute – solvent must be able to get in between solute molecules (or ions) and pull them apart Liquid solute – solvent just needs to get in between molecules (or ions). Easier because molecules are already further apart Gas solute – solvent has to pull gas molecules closer together

8. III. Solubility: The maximum amount of solute that can dissolve in a given amount of solvent The maximum amount of solute that can dissolve in a given amount of solvent -- Units = grams of solute 100.0 grams of solvent Saturated Solution: a solution in which the maximum amount of solute has been dissolved

9. IV. Endothermic v. Exothermic Most solid solutes require NRG to dissolve e.g., NaCl and H 2 O --NRG is required for H 2 O molecules to get in between the Na + and Cl - ions --NRG is taken from H 2 O molecules and surroundings (outside) --H 2 O molecules slow down (lose NRG) and the temperature of water decreases

10. Endothermic: A process that absorbs heat (NRG) -- the (KE) NRG is stored (PE) in the dissolving process -- the solution is colder Like… NaCl + H 2 O + NRG → Na + + Cl - + H 2 O

11. Some solid solutes release NRG when they are dissolved Exothermic: Process that results in a release of heat Like… NaOH + H 2 O → Na + + OH - + H 2 O + NRG (Remember: most solutes dissolve endothermically)

12. 4 Rules for solubility Rule 1: Identity – Solubility of any solute depends both on the identity of the solute and the identity of the solvent

13. Rule 2: Temperature – -Increasing the temperature usually (endothermic processes) increases solubility for solids -Increasing the temperature usually (endothermic processes) increases solubility for solids -Increasing temperature decreases solubility for gases When an endothermic solution is heated, then cooled – solubility decreases and a precipitate forms Precipitation: the process by which a solute leaves the solution and becomes solid again

14. Rule 3: Increased pressure increases solubility for gases Rule 4: When a solute dissolves in an exothermic manner, (sometimes) an increase in temperature decreases solubility

15. OYO’s 11.1 You must dissolve 50.0 g of KCrO 4 (a yellow- orange solid) into 250 mL of water. It’s not dissolving well. What could you try? 11.2 A lab has a bottle of a solution made by dissolving CH 4 gas in octane. The lid keeps popping off. What could you do to prevent the lid from popping off?

16. 11.3 You have to make a 500 mL solution of 5.5 M NaOH. It gets really hot – and it’s not dissolving well. What should you do?

17. V. Applying Stoichiometry to Solutions Remember, Molarity (M) = moles/liter Therefore, L x M (mol/L) = mol mol x 1/M (L/mol) = L (Just write M as mol/L and you can figure it out. You could write mol/L or L/mol and you don’t have to memorize the above)

18. Examples 1. Sodium carbonate reacts with calcium hydroxide to produce calcium carbonate and sodium hydroxide. How many grams of calcium carbonate will be made from 50.0 mL of 1.23 M sodium carbonate and excess calcium hydroxide? 2. Sodium chloride is reacted with excess aluminum nitrate to make aluminum chloride and sodium nitrate. How many L of a 0.245M solution of NaCl will be necessary to make 100.0 g of AlCl 3 ?

19. OYO’s 11.4 The following reaction is carried out in the laboratory: 5H 2 O 2 (aq) + 2KMnO 4 (aq) + 3H 2 SO 4 (aq) → 5O 2 (g) + 2MnSO 4 (aq) + K 2 SO 4 (aq) + 8H 2 O If 125 mL of 0.5 M H 2 O 2 is reacted with an excess of the other reactants, how many grams of oxygen will be produced?

20. 11.5 One way to purify a mixture of two metals is to add to the mixture something that reacts with one of the metals and turns it into an aqueous compound. The mixture can then be filtered, isolation the other metal. A mixture of magnesium and aluminum was treated in such a way, to isolate the magnesium. Sodium hydroxide was added, which only reacts with the aluminum: 2Al(s) + 2NaOH (aq) + 6H 2 O(l) → 2NaAl(OH) 4 (aq) + 3H 2 (g) The aluminum was converted to an aqueous salt, and the magnesium stayed as a solid. The solution was then filtered, isolating the pure magnesium. If a mixture of these two metals contained 1.2 kg of Al, how many liters of a 6.5 M NaOH solution would be needed to make sure that all the aluminum was taken out of the mixture?

21. VI. Molality (m) Molarity (M) = moles (solute) /liters (solution) - liter includes the solute + solvent Molality (m) = moles (solute)/ kg of solvent = moles (solute)/mass of solvent only

22. Examples A. A chemist mixes 3.4 moles NaCl with 4.5 kg H 2 O. What is the molality? B. A solution is made with 45.0 g ZnCl 2 and 150.0 g H 2 O. What is its molality?

23. OYO’s 11.6 Calculate the molality of the following solutions: a. 2.3 moles of NaOH dissolved in 1.2 kg of water b. 250.0 g KNO 3 dissolved in 2.43 kg of water c. 24.5 g NaBr dissolved in 578 g of water

24. VII. Freezing Point Depression A. Why things freeze - Things freeze when molecules slow down - For this to happen, NRG must be removed from the liquid

25. C. We can predict by how much the temperature will go down ∆T = -i K f m ∆T = change in the temperature -i = the number of molecules (or ions) the solute splits into K f = the freezing point depression constant (it is different for every solvent). Units = o C/m m = molality (moles of solute/kg of solvent) The negative sign before i means that you must subtract this amount from the original freezing point (use 0.00 o C for the freezing point of water.

26. K f H2O = 1.86 o C/m K f acetic acid = 3.59 o C/m Example 11.3 A. When 15.0 grams of NaCl are added to 100.0 g H 2 O, what is the change in the water’s freezing point? B. What is the freezing point of a solution that is made by mixing 2.5 g Al(NO 3 ) 3 in 150.0 g of water?

27. C. What is the freezing point of a solution that is made by dissolving 100.0 grams of table sugar (C 12 H 22 O 11 ) into 950.0 grams of water? K f H2O = 1.86 o C/m

28. OYO’s 11.7 In order to prevent a car’s water from freezing, people all “antifreeze” to the water. The antifreeze is typically ethylene glycol (C 2 H 6 O 2 ). What is the freezing point of a radiator solution made from 4.00 kg water and 1.00 kg ethylene glycol? K f H2O = 1.86 o C/m 11.8 If a solution is made by dissolving 2.30 g Mg(OH) 2 in to 25.0 grams of acetic acid (C 2 H 4 O 2 ), what is the freezing point? (K f acetic acid = 3.59 o C/m, freezing point of pure acetic acid is 16.6 o C)

29. VIII. Boiling Point Elevation A. About boiling point when solutes are present - The boiling point is higher when a solute is dissolved in a solvent - The bond of attraction is even stronger than the H 2 O– H 2 O bond - You need to add even more NRG to break bonds - When something boils, the gas H 2 O molecules must escape. The more ions that are present, the more NRG is needed to break the attraction between them and the H 2 O molecules

30. B. The formula for boiling point elevation ∆T = i K b m Note: -There is not a negative sign in front of “i” – the boiling point goes up when solutes are added to solvents - K b is the “Boiling point elevation constant,” and is different from the K f

31. For Example: In cooking pasta, a chef added 20.0 grams of table salt (NaCl) to 1.45 kg of water and brought the water to a boil. What is the temperature of this boiling solution? K b H2O =0.521 o C/m (Use 100.0 o C as the boiling point for water)

32. OYO’s 11.9 Acetic acid has a K b of 3.08 o C/m and a normal boiling point of 118.5 o C. What would be the boiling point of a solution made by dissolving 50.0 g of CaCO 3 in 500.0g of acetic acid? 11.10 If you wanted to boil a solution made by dissolving 250.0 grams of table sugar (C 12 H 22 O 11 ) into 1.3 kg of water, to what temperature would you have to raise it? (K b H2O = 0.521 o C/m)

33. IX. Solutions, Enthalpy & Entropy Enthalpy (H) = used to describe heat changes that take place during a process or reaction ∆H rxn = ∆H products - ∆H reactants - ∆H is negative for exothermic - ∆H is positive for endothermic A reaction is favorable if exothermic

34. ∆H soln = ∆H 1 + ∆H 2 + ∆H 3

35. But, if most things dissolve endothermically, and exothermic processes are favorable – why do things dissolve at all? There are 2 factors that govern all chemical and physical processes (determine whether a reaction will occur naturally – spontaneously): 1. NRG – exothermic or endothermic 2. Disorder – everything tends to disorder (think of your bedroom!)

36. Entropy (S): a direct measure of disorder 2 nd Law of Thermodynamics: The universe naturally tends to disorder (entropy increases) The solute-solvent mixture is more disordered than the solute-solute or the solvent-solvent -- Entropy increases

37. OYO’s 11.11 Draw the diagram of the solution process. Label the steps (1, 2, & 3) and ∆H 1, ∆H 2, ∆H 3. Put a negative or plus sign under each ∆H to describe if the step was endo- or exo- thermic. 11.12 What 2 factors govern all chemical and physical reactions? 11.13 In terms of those 2 factors, what type of reaction will always be spontaneous?