II III I I. The Nature of Solutions Solutions

A. Definitions  Solution -  Solution - homogeneous mixture Solvent Solvent - present in greater amount Solute Solute - substance being dissolved

A. Definitions Solute Solute - KMnO 4 Solvent Solvent - H 2 O

A. Definitions  Miscible – when two liquids are soluble in each other (alcohol & water)  Immiscible – when two liquids are not soluble in each other (oil & water)  Aqueous – dissolved in water

A. Definitions  unsaturated solution - If the amount of solute dissolved is less than the maximum that could be dissolved  saturated solution - solution which holds the maximum amount of solute per amount of the solution under the given conditions  supersaturated solution - solutions that contain more solute than the usual maximum amount and are unstable.

 Electrolyte – solution that conducts an electric current  Non electrolyte – solution that does not conduct an electric current A. Definitions

B. Solvation NONPOLAR POLAR “Like Dissolves Like”

B. Solvation  Soap/Detergent polar “head” with long nonpolar “tail” dissolves nonpolar grease in polar water

C. Solubility SATURATED SOLUTION no more solute dissolves UNSATURATED SOLUTION more solute dissolves SUPERSATURATED SOLUTION becomes unstable, crystals form concentration

C. Solubility  Solubility maximum grams of solute that will dissolve in 100 g of solvent at a given temperature varies with temp based on a saturated solution

C. Solubility  Solubility Curve shows the dependence of solubility on temperature

C. Solubility  Solids are more soluble at... high temperatures.  Gases are more soluble at... low temperatures & high pressures (Henry’s Law). EX: nitrogen narcosis, the “bends,” soda

D. Increasing the Rate of Solution 1. Agitation 2. Increasing Temperature 3. Increasing Surface Area

II III I II. Concentration Solutions

A. Concentration  The amount of solute in a solution.  Describing Concentration % by mass - medicated creams % by volume- rubbing alcohol molarity - used by chemists molality - used by chemists

B. % by Mass  Remember …  % = part x 100 whole  % by mass = mass solute x 100 mass solution

Example  What is the % by mass of a solution with 3.6 g of NaCl dissolved in g of water?  % = (3.6 / 103.6) x 100 = 3.5% NaCl

C. % by Volume  Remember …  % = part x 100 whole  % by volume = volume solute x 100 volume solution

Example  What is the % by volume of 75.0 ml of ethanol dissolved in ml of water?  % = (75.0 / 275.0) x 100 = 27.3%

D. Molarity  Molarity = moles of solute/liter of solution  Note: it’s liters of solution, not liters of solvent

Molarity Examples  Calculate the molarity of a solution made by dissolving 23.4 g of sodium sulfate in 125 ml of solution  23.4 g Na 2 SO 4  mol  125 ml  L  M = mol / L  M = 0.165mol / L  M = 1.32 M

E. Dilution  Preparation of a desired solution by adding water to a concentrate.  Moles of solute remain the same.

E. Dilution  M 1 V 1 = M 2 V 2  M 1 = initial molarity  V 1 = initial volume  M 2 = final molarity  V 2 = final volume  The units for V 1 & V 2 do not matter as long as they are the same  M 1 & M 2 MUST be in molarity

E. Dilution Problems  Suppose we want to make 250 ml of a 0.10 M solution of CuSO4 and we have a stock solution of 1.0 M CuSO4. How many mL of the stock solution do we need?  First do the math  M 1 V 1 = M 2 V 2  (0.10M)(250ml) = (1.0)(V 2 )  V 2 = 25 ml

E. Dilution  What volume of 15.8M HNO 3 is required to make 250 mL of a 6.0M solution? GIVEN: M 1 = 15.8M V 1 = ? M 2 = 6.0M V 2 = 250 mL WORK: M 1 V 1 = M 2 V 2 (15.8M) V 1 = (6.0M)(250mL) V 1 = 95 mL of 15.8M HNO 3

C. Dilution  What volume of 15.8M HNO 3 is required to make 250 mL of a 6.0M solution? GIVEN: M 1 = 15.8M V 1 = ? M 2 = 6.0M V 2 = 250 mL WORK: M 1 V 1 = M 2 V 2 (15.8M) V 1 = (6.0M)(250mL) V 1 = 95 mL of 15.8M HNO 3