1. SOLUTIONS A homogeneous mixture in which the components are uniformly intermingled

2. Vocabulary Terms to Know Colligative property Concentration Dilute solution Electrolyte Immiscible Miscible Molarity Saturated solution vs Unsaturated vs Supersaturated Solubility Solute Solvent

3. ELECTROLYTES Substances that break up in water to produce ions. These ions can conduct electric current Examples: Acids, Bases and Salts (ionic compounds)

4. ELECTROLYTES When substances break up into their ions in water, this process is called ionization or dissociation. For these compounds, we can write their dissociation equation: BaCl 2 (s)  Ba 2+ (aq) + 2Cl – (aq) Pb(NO 3 ) 2 (s)  Pb 2+ (aq) + 2NO 3 – (aq)

5. Terms Solvent – The substance present in the largest amount in a solution. The substance that does the dissolving. Solute – The other substance or substances in a solution. The substance that is dissolved.

6. SOLUBILITY Is the amount of a substance that dissolves in a given quantity of solvent (usually water) at a specific temperature to produce a saturated solution

7. SOLUBILITY of Polar vs NonPolar “Like dissolves Like”Like dissolves Like – Polar molecules dissolve polar molecules (Ionic compounds dissolve ionic compounds) – Nonpolar molecules dissolve nonpolar molecules (Molecular compounds dissolve molecular compounds) Lab/Demo for Electrolytes

8. SOLUBILITY RULES All common salts of Group I elements and ammonium are soluble All common acetates and nitrates are soluble All binary compounds of Group 7 (other than F) with metals are soluble except those of silver, mercury I and lead All sulfates are soluble except those of barium, strontium, calcium, silver, mercury I and lead Except for those in Rule 1, carbonates, hydroxides, oxides, sulfides and phosphates are insoluble

9. Terms Saturated – When a solution contains the maximum amount of solute Unsaturated When a solvent can dissolve more solute – Concentrated When a relatively large amount of solute is dissolved – Dilute When a relatively small amount of solute is dissolved Supersaturated – When the solution contains more solute than a saturated solution will hold at that temperature Super Cooled Water

10. Factors that affect solubility Temperature: – Solids Direct relationship; as temperature goes up solubility goes up – Gases Indirect relationship; as temperature goes up solubility goes down

11. Temperature vs Solubility

13. Gas Solubility (Temp)

14. Factors Affecting the Rate of Dissolution (dissolving) Stirring Temperature Surface Area

15. Gas Solubility (Pressure) S 1 S 2 P 1 P 2 = Visualization Solubility and pressure are directly related (you will NOT need to solve this equation) Henry’s Law (for gases!)

16. MOLARITY Molarity is the number of moles of solute per liters of solution – A way to quantify concentration (a way to put a number to concentration) M = molarity = moles of solute liter of solution

17. Practice Problem #1 Calculate the molarity of a solution prepared by dissolving 11.5 g of NaOH in enough water to make a 1.50 L solution. 11.5g NaOH 40.0g NaOH 1 mol NaOH =.288 mol moles liter M =.288 mol 1.50 L =.192M

18. Practice Problem #2 Calculate the molarity of a solution prepared by dissolving 1.56 g of HCl into enough water to make 26.8 ml of solution..0427 moles of HCl 26.8 mL =.0268L M = 1.59M solution

19. Practice Problem #3 Calculate the amount of liters needed to make a 3.00 molar solution of NaNO 3 with 44.0 grams of NaNO 3 ?.518 moles of NaNO 3 3.00M = moles liter M =.518 mol ? Liters ? Liters =.173 liters

20. DILUTIONS (uses molarity) M 1 x V 1 = M 2 x V 2 What volume of 16 M sulfuric acid must be used to prepare 1.5 L of a 0.10 M H 2 SO 4 ? 16mL of a 12M acid is diluted to 750 mL, what is the new molarity? 0.0094L or 9.4mL 0.25 M

21. DILUTIONS (Real solutions) To how much water should 25.0mL of 6.00M HCl be added to produce a 4.00M solution? M 1 x V 1 = M 2 x V 2 V 2 = 37.5 mL V 2 – V 1 = amount of water 37.5 mL – 25.0 mL = 12.5 mL 12.5 mL of water

22. Freezing Point Depression in Solutions The particles hinder the solvent from freezing by slowing the formation of solid crystals – This is used in winter with icy roads & sidewalks! – Making Popsicles Making Popsicles – Cold Ice Cold Ice

23. Colligative Properties A property that depends on only the number of solute particles, NOT their identity There are 3 important colligative properties: 1. Boiling Point Elevation 2. Freezing Point Depression 3. Vapor-pressure Lowering

24. Molality (NOT molarity) Similar to molarity in that it shows concentration… – Molality (m) = moles of solute Kilogram of solvent WE WILL NOT BE CALCULATING MOLALITY OR COLLIGATIVE PROPERTIES

25. Molality Practice (mol/kg) 2.4 moles of sucrose are dissolved in 320 mL of water. What is the molality? = 7.5m 2.4 moles sucrose.320Kg water WE WILL NOT BE CALCULATING MOLALITY OR COLLIGATIVE PROPERTIES

26. Molality Practice (mol/kg) 56.8g of NaCl is dissolve in 560 mL of water. What is the molality?.97mol/.560kg = 1.73m WE WILL NOT BE CALCULATING MOLALITY OR COLLIGATIVE PROPERTIES

27. Colligative Properties Freezing Point Depression/ Boiling Point Elevation  T f = m x k f x i OR  T b = m x k b x i m = molality i = number of ions dissolved per molecule k f = freezing point constant (given) k b = boiling point constant (given) k f = 1.86  C  Kg/mol (for water) k b = 0.51  C  Kg/mol (for water) WE WILL NOT BE CALCULATING MOLALITY OR COLLIGATIVE PROPERTIES

28. Colligative Practice Problem 56.8g of NaCl is dissolved in 560mL of distilled water. What is the freezing point of this solution?  T f = m x k f x i Step 1 -> determine m (molality)=moles/Kg Step 2 -> Plug into equation with k f (given) and i (number of ions, in this case i = 2) Step 3 -> solve for  T f Step 4 -> adjust original freezing point accordingly WE WILL NOT BE CALCULATING MOLALITY OR COLLIGATIVE PROPERTIES

29. Colligative Practice Problem m = 1.73  T f = m x k f x i  T f = (1.73m) x 1.86  C  Kg/mol x 2  T f = 6.4 What is new freezing point? -6.4  C WE WILL NOT BE CALCULATING MOLALITY OR COLLIGATIVE PROPERTIES

30. Colligative Problem: finding molar mass A solution of a nonelectrolyte (does not dissociate in water) contains 30.0 g of solute dissolved in 250.0 g of water. The boiling point of the water is observed to be 101.04°C. What is the molar mass of this substance? WE WILL NOT BE CALCULATING MOLALITY OR COLLIGATIVE PROPERTIES

31. A solution of a nonelectrolyte (does not dissociate in water) contains 30.00 g of solute dissolved in 250.0 g of water. The boiling point of the water is observed to be 101.04°C. What is the molar mass of this substance? Game plan: – Molar mass (of solute) units grams/mole We are given grams of solute; must solve for moles. – We are given BP of water; we must be using  T b = m x k b x i – Are there moles anywhere in the equation? m = moles of solute kg of solvent

32. A solution of a nonelectrolyte (does not dissociate in water) contains 30.00 g of solute dissolved in 250.0 g of water. The boiling point of the water is observed to be 101.04°C. What is the molar mass of this substance?  T b = m x k b x i  T b = 101.04°C - 100°C = 1.04 °C m = ? moles of solute/.2500Kg of water k b (always given)= 0.51  C  Kg/mol i = 1 (for all nonelectrolytes 1.04 °C = ?moles x.51  C  Kg/mol x 1.2500 kg WE WILL NOT BE CALCULATING MOLALITY OR COLLIGATIVE PROPERTIES

33. A solution of a nonelectrolyte (does not dissociate in water) contains 30.00 g of solute dissolved in 250.0 g of water. The boiling point of the water is observed to be 101.04°C. What is the molar mass of this substance? ? Moles = 0.5098 moles Molar mass = grams/moles 30.00g of solute 0.5098 moles = 58.8 g/mol WE WILL NOT BE CALCULATING MOLALITY OR COLLIGATIVE PROPERTIES

34. Solution Stoichiometry 1. How many grams of silver chromate will precipitate when 150. mL of 0.500 M silver nitrate are added to 100. mL of 0.400 M potassium chromate? 2 AgNO 3(aq) + K 2 CrO 4(aq)  Ag 2 CrO 4(s) + 2 KNO 3(aq)

35. How many grams of silver chromate will precipitate when 150. mL of 0.500 M silver nitrate are added to 100. mL of 0.400 M potassium chromate? 2 AgNO 3(aq) + K 2 CrO 4(aq)  Ag 2 CrO 4(s) + 2 KNO 3(aq) Given: 150. mL of 0.500 M AgNO 3 100. mL of 0.400 M K 2 CrO 4 Want: Grams of Ag 2 CrO 4 Game-Plan: – Convert from mL to Liters – Use molarity to convert to moles of reactants – Use mole ratio to convert to moles of Ag 2 CrO 4 – Use molar mass to convert to grams of Ag 2 CrO 4

36. How many grams of silver chromate will precipitate when 150. mL of 0.500 M silver nitrate are added to 100. mL of 0.400 M potassium chromate? 2 AgNO 3(aq) + K 2 CrO 4(aq)  Ag 2 CrO 4(s) + 2 KNO 3(aq) = 12.4g Ag 2 CrO 4 0.150 L AgNO 3

37. How many grams of silver chromate will precipitate when 150. mL of 0.500 M silver nitrate are added to 100. mL of 0.400 M potassium chromate? 2 AgNO 3(aq) + K 2 CrO 4(aq)  Ag 2 CrO 4(s) + 2 KNO 3(aq) = 100. mL K 2 CrO 4 13.3g Ag 2 CrO 4

38. How many grams of silver chromate will precipitate when 150. mL of 0.500 M silver nitrate are added to 100. mL of 0.400 M potassium chromate? 2 AgNO 3(aq) + K 2 CrO 4(aq)  Ag 2 CrO 4(s) + 2 KNO 3(aq) VS