1. TOPIC 8: SOLUTIONS AND SOLUBILITY - HONORS CHEMISTRY - Ms. Argenzio How can we describe solutions? How can we explain molecule-ion attraction in an aqueous solution? How can we use Table G to understand solubility curves? How can we use Table F to predict solubility and the products of double replacement reactions? What are the factors that affect solubility? What are the factors that affect the rate of dissolving? What are colligative properties and how do electrolytes affect them? How can we determine the concentration of solutions?

2. AIM: How can we describe solutions? ~~~~Homogenous Mixtures aka – SOLUTIONS ~~~~  Solutions are homogeneous  Dissolved particles will not come out of the solution no matter how long its stands unless temperature changes or evaporation occurs  Liquid and gas solutions are clear and transparent (dissolved particles cannot be seen)  Cannot be filtered  Single phased even though the components that make it up can be of different phases

3. AIM: How can we describe solutions? PARTS OF A SOLUTION  Solutions consist of solute and solvent (solution = solute + solvent )  Solute – dissolved substance ( smaller portion of solution)  Solvent – substance in which the solute is dissolved (the larger portion of the solution)

4. AIM: How can we describe solutions? TYPES OF SOLUTIONS  Gas solutions – (air) gases have completely dissolved in one another  Liquid Solutions – Consist of a solute which can be a solid, liquid, or gas dissolved in a solvent Ex:  Carbonated water – CO 2 gas in water  Salt water – solid in water  Antifreeze – ethylene glycol  Solid Solution – mostly solid dissolved in solid. It can also be a liquid or gas dissolved in a solid Ex:  Alloys Brass – Cu/Zn  Bronze Cu/Sn

5. AIM: How can we describe solutions? Homogeneous aka solutions/miscible (capable of being mixed)  A solution is produced when one substance dissolves in another  Salts are ionic Heterogeneous mixture – nonuniform/immiscible  Heterogeneous mixtures are immiscible. They do not mix and do not dissolve

6. Saturated Solutions are in Equilibrium

7. AIM: How can we explain molecule-ion forces of attraction in an aqueous solution ? Molecule ion attraction : Mixing ionic compounds with water form aqueous solutions of dissolved ions Polar water molecules attract to the ions tearing them apart from the other ions Crystal salt dropped in H 2 O

8. AIM: How can we explain molecule-ion forces of attraction in an aqueous solution ? http://www.youtube.com/watch?v=xd edxfhcpWo&feature=related http://www.youtube.com/watch?v=7P HhBBg-6X0&feature=related

9. AIM: How can we describe solutions?  Saturated: has the maximum amount of solute dissolved in them. No more solute can be made to dissolve (100 seat restaurant with 100 people in it)  Unsaturated: has less than the maximum amount of solute dissolved in them. This means that more solute can be added to the solution and the additional solute would still dissolve ( 100 seat restaurant with 40 people in it)  Supersaturated: Rare solution where the solution holds more solute than is theoretically possible, unstable where the excess will precipitate if the solution is agitated (100 seat restaurant with 120 people in it, then the manager comes in, he throws the extra 20 people out)

10. AIM: How can we describe solutions?  Using reference Table G!!!!!  Up curves represent salts (direct relationship: increase T, increase solubility)  Down curves represent gases (indirect relationship: increase T, decrease solubility )

11. AIM: How can we understand solubility curves?  Each line represents a saturated solution of a solute at different temperatures  Higher a line is at a given temperature the more soluble that substance is  Data tells us how many grams of solute can be dissolved in 100g of water a particular temperature

12. AIM: How can we understand solubility curves? Any point on line represents a saturated solution. Saturated solutions are at equilibrium  Rate of dissolution is equal to the rate of crystallization

13. AIM: How can we use Table F to predict solubility?  View video on Solutions and Solubility: http://www.youtube.com/watch?v=VTmfQ UNLlMY

14. AIM: How can we use Table F to predict solubility?  Table F gives us information on the solubility of various compounds in solutions  You can predict the solubility of a compound or the two products in a double replacement reaction

15. AIM: How can we use Table F to predict solubility?  DIRECTIONS: 1. Cross out the symbol of the first ion or element 2. Underline the symbol of the second ion or element 3. Use TABLE F to predict solubility

16. AIM: How can we use Table F to predict solubility?  EXAMPLE:

17. AIM: How can we understand and recognize double replacement reaction?  Double reactions contains 2 ionic compounds  Every double replacement reaction will not necessarily occur. There are 3 situations that ensure a DR occurs:  If one of the products is a solid  One of the products is a gas  A molecular substance such as water is formed

18. AIM: How can we predict the products and solubility of the products of DR reactions? KI + Pb(NO 3 ) 2  _________ + ______________ 1. Predict products – the reactants swap places 2. Determine solubility of products EXAMPLE: For the reaction between LiBr and BaNO 3 predict the products and determine the solubility of the products

19. “If you are not part of the solution you are part of the precipitate”  Precipitate – insoluble solid (cant be dissolved)  Table F – SOLUBLE – dissolved INSOLUBLE – not dissolved *right side of the table all insoluble and they all have exceptions

20. AIM: How can we use Table F and G?  DO NOW: Answer the Following Questions: 1. Identify the solute and solvent in air 2. If 106g of KNO 3 are added to 100g H 2 O at 60  C, what kind of solution is formed? 3. If 110g of KNO 3 are added to 100g H 2 O at 50  C, what kind of solution is formed? 4. If 60g of KNO 3 are added to 100g of water at 50  C, what kind of solution is formed? How can you make this solution saturated? 5. If 110g of KNO 3 are dissolved in 100g of H 2 O at 62  C, what kind of solution is formed? How can you make this solution saturated? 2/11/13 – D day

21. AIM: What are some factors that affect solubility?  Solubility : how much solute is dissolved in a given amount of solvent  Solubility is measured in grams of solute per 100 grams of solvent (mostly water)  Solubility must always be accompanied by a temperature NaCl 36g/100g H 2 O @ 25  C

22. AIM: What are some factors that affect solubility? FACTORS THAT AFFECT SOLUBILTY:  Nature of the solute  Temperature  Pressure

23. Nature of the Solute  All solute don’t dissolve to the same extent in H 2 O Examples: PbCl 2 1g/100g H 2 O @ 25  C ZnCl 2 200g/100g H 2 O @ 25  C

24. TEMPERATURE  Solid Solutes – as T increases, Solubility increases  Liquid Solutes – as T increases, Solubility increases  Gas Solutes – as T increases, Solubility decreases

25. PRESSURE  Solid Solutes – as P increases, no effect  Liquid Solutes – as P increases, no effect  Gas Solutes – as P increases, Solubility increases Applying the principles of solubility and pressure, explain why soda goes flat when you open the can.

26. AIM: Factors affecting dissolving rate?  DO NOW: Answer the Following Questions: 1. Identify the solute and solvent in air 2. If 106g of KNO 3 are added to 100g H 2 O at 60  C, what kind of solution is formed? 3. If 110g of KNO 3 are added to 100g H 2 O at 50  C, what kind of solution is formed? 4. If 60g of KNO 3 are added to 100g of water at 50  C, what kind of solution is formed? How can you make this solution saturated? 5. If 110g of KNO 3 are dissolved in 100g of H 2 O at 62  C, what kind of solution is formed? How can you make this solution saturated? 3/25/14 – A day

27. AIM: What are some factors that affect the Rate of Dissolving?  Size of the particles – smaller particles dissolve faster than larger particles  Stirring – increase the stirring, increase the rate of dissolving  Temperature – increase the temperature, increase the rate of dissolving  Nature of the solute and solvent – “LIKE DISSOLVES LIKE RULE”

28. AIM: What are some factors that affect solubility? LIKE DISSOLVES LIKE RULE  Like polarities will dissolve in like polarities  Polars will attract polars (like), and repel non-polars (opposites). Non-polars will attract non-polars (like), and repel polars (opposites)  Oil, which is non-polar, does not mix with water, which is polar ( opposites repel )

29. AIM: What are some factors that affect solubility? Critical thinking: A student was working in a lab. Grease spilled all over his hands. He went to the sink to clean his hands. The grease, which was non polar, did not come off. Once the student used soap and water, all of the grease came off. EXPLAIN

30. AIM: What are colligative properties ? Colligative property  physical properties of solutions that depend on the concentration of solute in a given amount of solvent When any pure solvent forms a solution two things occur: 1. Boiling point elevation 2. Freezing point depression

31. NOTE: The addition of electrolytes and non electrolytes will raise the boiling point and lower the freezing point of a solution. However, the electrolyte will have a greater effect Summary – the measure of changes in boiling point and freezing point are dependent on: -Solute (electrolyte vs. nonelectrolyte) -Number of particles in solution -Concentration AIM: How do electrolytes and non electrolytes affect the boiling point and freezing point of solutions?

32. AIM: What are electrolytes and how do they behave? ELECTROLYTES  Ionic substances (salts)  Acids (HX)  Bases (MOH) NON-ELECTROLYTE  Sugars (C 6 H 12 O 6 )  Alcohols R-OH Summary Electrolytes conduct and turn a light bulb tester on. When placed in aqueous solutions electrolytes dissociate, ionize or break apart None electrolytes do not conduct and will not turn on a light bulb tester. When placed in water non-electrolytes do not dissociate, ionize or break apart

33. Exercise: Determine if the following are electrolytes or non-electrolytes and sub classify E or NEAcidBaseIonic/Salt HCl C 6 H 12 O 6 NaOH CH 3 OH ZnCl 2

34. EXCEPTIONS ACID  CH 3 COOH = HX Although acetic acid or vinegar does not look like HX, it is still an acid – exception! BASE  NH 4 OH = MOH Although ammonium hydroxide does not look like MOH, it is still a base – exception!

35. AIM: How can we determine the concentration of solutions? Concentration : the amount of the solute dissolved in the solution. For solutions there are several ways to express concentration : 1. Grams of solute/100mL of solvent (TABLE G) 2. Molarity 3. Parts per million 4. % by mass 5. % by volume

36. AIM: How can we determine the molarity of solutions?  Molarity (M) – the number of moles of solute (n) divided by the volume (V) of the solution in liters **important to note that molarity is not moles of solute per liter of solvent !!

37. Percent by Mass  Labels often list the concentration of the ingredients as a percent by mass. Fertilizers list the active ingredients as a percentage of the entire mass of fertilizer  Percent by mass – the mass of the part divided by the mass of the whole solution  EX: What is the percent by mass of sodium hydroxide if 5.00g of NaOH are added to 50.00g of H 2 O?

38. Percent by Volume  Percent by volume – used when two liquids are mixed to form a solution  A label on a bottle of alcohol shows a common example  % by volume is the ratio of the volume of an ingredient divided by the total volume and expressed as a percent  EX: What is the percent by volume of alcohol if 50mL of ethanol is diluted with water to form a total volume of 3000mL

39. How can we understand what parts per million means?  If you divide a pie equally into 10 pieces, each piece would be a part-per-ten ; i.e., one-tenth of the total pie  If you cut this pie into a million pieces, each piece would be very small and would represent a millionth, or part per million, of the original pie

40. How can we determine concentration of a solution in parts per million?  Parts per million is the ratio between the mass of the solute and the total mass of solution  This method is useful for extremely dilute solutions when molarity and percent mass would be too difficult to interpret

41. Questions:  Carbon dioxide has a solubility of 0.0972g/100g H2O @40dC. Expressed in parts per million this concentration is closest to what value?  A substance has a solubility of 350ppm. How many grams of the substance are present in 1.0L of a saturated solution?