2. Pb(NO 3 ) 2 (aq) + 2KI(aq)  PbI 2 (s) + 2KNO 3 (aq) In order to conduct this reaction both the lead (II) nitrate and the potassium iodide must be made into solutions

3. Na + Cl - - - + + + + - - Na + Cl - Na + Cl - Solvent Solute Solution When a solute dissolves in a solvent the solute is separated into individual components The intermolecular forces in the solvent are altered The solute and the solvent interact to form the solution

4. In order to prepare a solution we need to extend the concept of the mole to include the concept of molarity Molarity is defined as moles of solute per volume of solution in liters Solute-a substance that dissolves in a liquid Solvent-the dissolving medium in a solution Solution- a homogeneous mixture (moles/liter) Preparing a Solution

5. Understanding Molarity Molarity = Moles of solute Liters of solution What is the molarity of an ascorbic acid solution (C 6 H 8 O 6 ) prepared by dissolving 1.80 grams in enough water to make 125 mL of solution. How many milliliters of this solution contain 0.0100 mol ascorbic acid. 1.80 g C 6 H 8 O 6 1 mol C 6 H 8 O 6 176 g C 6 H 8 O 6 = 0.0102 mol C 6 H 8 O 6 Molarity = 0.0102 mol C 6 H 8 O 6.125 L soln = 0.0818 M

6. A solution is prepared by mixing 1.00 g ethanol (CH 3 CH2OH) with 100 ml. of water. Calculate the molarity of this solution. Understanding Molarity Sample Problem

7. Preparing a Solution There are two approaches to preparing solutions. The first involves preparing the solution from dried reagent. Let’s assume that the potassium iodide is going to be prepared from dried reagent and that we have decided to prepare a 0.10 Molar solution (0.1 moles/liter)

8. Preparing a Solution When preparing a solution from dried reagent we simply use dimensional analysis and ask three questions. 1. What volume of solution do you want to use? (let’s assume that you want 50 ml assume that you want 50 ml 2. What molarity do you want? ( We have already said that we want 0.10 M KI) that we want 0.10 M KI) 3. What is the molar mass of KI? (ans.: 167 g/mol)

9. Preparing a Solution 50.0 ml 1 L 1000 ml What volume of solution do you want to use? 0.100 mol 1 L What molarity do you want? 167 g KI 1 mol KI What is the molar mass of KI? mass of KI? = 0.835 g KI In order to prepare the above solution, 0.835 g of KI are dissolved in 50 ml of distilled water

10. Let’s assume that a 0.10 M solution of lead (II) nitrate is going to be prepared from a pre-existing 0.75 M solution Preparing a Solution The second approach involves There are two approaches to preparing solutions. The second approach involves dilution from a pre-prepared reagent dilution from a pre-prepared reagent.

11. Preparing a Solution The second approach involves the use of the equation: M 1 V 1 = M 2 V 2 where M 1 = the molarity of the stock solution (0.750 M) V 1 = the volume of Pb(NO 3 ) 2 that will be removed from the stock solution the stock solution M 2 = the molarity of the solution you want (0.100 M) V 2 = the volume of the solution you wish to use (50.0 ml) (50.0 ml)

12. Preparing a Solution M 1 V 1 = M 2 V 2 (0.750 M)( V 1 ) = (0.100)(50.0) V 1 = 6.67 ml In order to prepare this solution you simple take 6.67 ml of Pb(NO 3 ) 2 from the stock solution and bring the volume upto 50 ml with distilled water.

13. Preparing a Solution Practice Problems