Pb(NO3)2(aq) + 2KI(aq) PbI2(s) + 2KNO3(aq)

In order to conduct this reaction both the lead (II) nitrateand the potassium iodide must be made into solutions

Na+Cl-

-

-+

+

+

+

-

-

Cl-

Na+

Na+

Na+

Na+

Cl-

Cl-

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

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 volumeof 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 SolutionPreparing a Solution

Understanding MolarityUnderstanding Molarity

Molarity = Molarity = Moles of soluteMoles of solute

Liters of solutionLiters of solution

What is the molarity of an ascorbic acid solution (C6H8O6) prepared by dissolving1.80 grams in enough water to make 125 mL of solution. How many milliliters of thissolution contain 0.0100 mol ascorbic acid.

1.80 g C6H8O6

1 mol C6H8O6

176 g C6H8O6

= 0.0102 mol C6H8O6

Molarity = 0.0102 mol C6H8O6

.125 L soln= 0.0818 M

A solution is prepared by mixing 1.00 g ethanol (CH3CH2OH) with 100 ml. of water. Calculate the molarity of this solution.

Understanding MolarityUnderstanding Molarity

Sample Problem

Preparing a SolutionPreparing 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 Let’s assume that the potassium iodide is going to be prepared from dried reagent and that we have decided prepared from dried reagent and that we have decided to prepare a 0.10 Molar solution (0.1 moles/liter)to prepare a 0.10 Molar solution (0.1 moles/liter)

Preparing a SolutionPreparing a Solution

When preparing a solution from dried reagent we simply When preparing a solution from dried reagent we simply use dimensional analysis and ask three questions.use dimensional analysis and ask three questions.

1. 1. What volume of solution do you want to use? (let’sWhat volume of solution do you want to use? (let’s assume that you want 50 mlassume that you want 50 ml

2. 2. What molarity do you want? ( We have already saidWhat 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)3. What is the molar mass of KI? (ans.: 167 g/mol)

Preparing a SolutionPreparing a Solution

50.0 ml 1 L

1000 mlWhat volume of What volume of solution do you solution do you

want to use?want to use?

0.100 mol

1 LWhat molarity What molarity do you want? do you want?

167 g KI

1 mol KIWhat is the molarWhat 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 aredissolved in 50 ml of distilled water

Let’s assume that a 0.10 M solution of lead (II) Let’s assume that a 0.10 M solution of lead (II) nitrate is going to be prepared from a pre-existing nitrate is going to be prepared from a pre-existing 0.75 M solution0.75 M solution

Preparing a SolutionPreparing a Solution

There are two approaches to preparing solutions. The second approach involvesThe second approach involves dilution from a pre-prepared reagentdilution from a pre-prepared reagent.

Preparing a SolutionPreparing a Solution

The second approach involves the use of the equation:

M1V1 = M2V2 where

MM11 = the molarity of the stock solution (0.750 M) = the molarity of the stock solution (0.750 M)

VV11 = the volume of Pb(NO = the volume of Pb(NO33))22 that will be removed from that will be removed from

the stock solutionthe stock solutionMM22 = the molarity of the solution you want (0.100 M) = the molarity of the solution you want (0.100 M)

VV22 = the volume of the solution you wish to use = the volume of the solution you wish to use

(50.0 ml)(50.0 ml)

Preparing a SolutionPreparing a Solution

M1V1 = M2V2

(0.750 M)( V1) = (0.100)(50.0)

V1 = 6.67 ml

In order to prepare this solution you simple take 6.67 mlof Pb(NO3)2 from the stock solution and bring the volume upto 50 ml with distilled water.

Preparing a SolutionPreparing a SolutionPractice ProblemsPractice Problems