Solution A homogeneous mixture of two or more substances.

Solution

A homogeneousA homogeneous

mixture of two ormixture of two or

more substances.more substances.

SolventSolvent - the substance present in the most amount.

SoluteSolute - the substance present in the least amount.

For Any Solution

SolventSolvent - there can be

only ONE

SoluteSolute - there can be

MORE than one

For Any Solution



What is the solventWhat is the solvent

in air?in air?



What are the solutesWhat are the solutes

in air?in air?



What is the solventWhat is the solvent

in stainless steel?in stainless steel?



What are the solutesWhat are the solutes

in stainless steel?in stainless steel?

SolventSolvent - Water

SoluteSolute - The substance dissolved in

water

Aqueous Solution

The amount of solute thatThe amount of solute that

can be dissolved in a givencan be dissolved in a given

amount of solvent, at aamount of solvent, at a

given temperature.given temperature.

Solubility

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Is NaClIs NaClsolublesolublein Hin H22O?O?

A solution containing the A solution containing the

maximum amount of a solutemaximum amount of a solute

that is possible to dissolve in athat is possible to dissolve in a

given volume of solvent at agiven volume of solvent at a

given temperature.given temperature.

Saturated Solution

NaCl has a NaCl has a

solubility ofsolubility of

357 grams 357 grams

per liter of per liter of

"cold" H"cold" H22O.O.

Saturated Solution

++

How manyHow many

moles ofmoles of

NaCl areNaCl are

in thein the

solution?solution?

Saturated Solution

NaClNaClsolutionsolution

357 g NaCl 1 mole NaCl357 g NaCl 1 mole NaCl

Saturated Solution


58 g NaCl58 g NaCl

6.2 mole NaCl6.2 mole NaCl

==

How manyHow many

"particles" "particles"

of NaCl areof NaCl are

in thein the

solution?solution?

Saturated Solution


357 g NaCl 357 g NaCl

Saturated Solution


58 g NaCl58 g NaCl

3.7 X 103.7 X 102424 particles NaClparticles NaCl

==

6.02 X 106.02 X 102323 particles NaClparticles NaCl

A comparison of A comparison of

the amounts of the amounts of

solute and solventsolute and solvent

in a solution.in a solution.

Concentration

"Strong" and "Weak""Strong" and "Weak"

give SOME comparison,give SOME comparison,

but only a general idea.but only a general idea.

Concentration

"Dilute" and "Concentrated""Dilute" and "Concentrated"

still don't provide enoughstill don't provide enough

for quantitative calculations.for quantitative calculations.

Concentration

To do calculations, we mustTo do calculations, we must

know "how much" soluteknow "how much" solute

and "how much" solvent and "how much" solvent

are present.are present.

Concentration

M = M =

Molarity

moles solutemoles solute

dmdm33 solution solution

1 mole = formula mass (g)1 mole = formula mass (g)

1 liter = 1 cubic decimeter1 liter = 1 cubic decimeter

dmdm33

1 liter = 1000 milliliters

1 ml = 1 cm3

1. What is the molarity

of a liter of

solution containing

100 grams of

copper (II) chloride?

Molarity = moles

dm3

1 0 0 g C u C l 2

1 l i t e r

Molarity =moles

dm3

1 0 0 g C u C l 2

1 l i t e r

Molarity =

grams

moles

dm3

1 0 0 g C u C l 2

1 l i t e r

Molarity =

grams

mole

moles

dm3

1 0 0 g C u C l 2

1 l i t e r

Molarity =

grams

1 mole

moles

dm3

1 0 0 g C u C l 2 1 m o l e

1 l i t e r 1 3 4 g

Molarity =

Cu = 1 X 64 = 64Cl = 2 X 35 = 70

134

moles

dm3

1 0 0 g C u C l 2 1 m o l e 1 l i t e r

1 l i t e r 1 3 4 g 1 d m 3

Molarity =moles

dm3

1 0 0 g C u C l 2 1 m o l e 1 l i t e r

1 l i t e r 1 3 4 g 1 d m 3

Molarity =moles

dm3

Have we workedthe problem?

1 0 0 g C u C l 2 1 m o l e 1 l i t e r

1 l i t e r 1 3 4 g 1 d m 3

. 7 5 M C u C l 2

Molarity =moles

dm3

2. How much NaCl is

needed to prepare 250ml of 0.5M salt water

How is this problemdifferent from the first?

1. What is the molarity of a liter of

solution containing 100 grams of

copper (II) chloride?

2. How much NaCl is needed to prepare 250ml of 0.5M salt water?

The second GIVES M

The first ASKS for M

2.How much NaCl is needed to prepare 250 ml of 0.5 M salt water

250 ml 0.5 mole NaCl

dm3

Preparation


dm3

Preparation

What units will we havewhen the problem is worked?


dm3

Preparation

need grams


dm3 1 mole NaCl

Preparation

need grams

250 ml 0.5 mole NaCl 58 g NaCl

dm3 1 mole NaCl

Preparation

need grams

250 ml 0.5 mole NaCl 58 g NaCl

dm3 1 mole NaCl

Preparation

Now What?need grams

250 ml 0.5 mole NaCl 58 g NaCl 1 dm3

dm3 1 mole NaCl 1000 ml

Preparation

need grams

250 ml 0.5 mole NaCl 58 g NaCl 1 dm3

dm3 1 mole NaCl 1000 ml

Preparation

7.3 grams NaCl

PracticeProblems

0.37 g CaCl2 1 mole 1000 ml

340 ml 110 g 1 dm3

0.01 M CaCl2

Practice Problem 1

50 cm3 0.2 mole Al(OH)3 78 g Al(OH)3 1 dm 3

dm 3 mole 1000 cm3

0.78 g Al(OH)3

Practice Problem 2

HomeworkProblems

50 g NaOH 1 mole 1000 cm3

200 cm3 40 g 1 dm3

6.25 M NaOH

Homework Problem 1

100 cm3 0.25 mole CaSO4 136 g 1 dm3

d m 3 mole 1000 ml

3.4 g CaSO4

Homework Problem 2

100 ml 0.5 mole HCl 36 g 1 dm3

dm3 mole 1000 ml

1.8 g HCl

Homework Problem 3

Making Dilutions

A solution can be made less concentrated by dilution with solvent

Making Dilutions

M1 V1 = M2 V2

original solution 1 = diluted solution 2

Volume units must be the same for both volumes in this equation.

Making Dilutions

How do you prepare

100ml of 0.40M MgSO4 from a stock solution

of 2.0M MgSO4?

Dilution Problem

M1 V1 = M2 V2

M1 = V1 =

M2 = V2 =

Dilution Problem

M1 V1 = M2 V2

M1 = 2.0M MgSO4 V1 = unknown

M2 = 0.40M MgSO4 V2 = 100ml

Dilution Problem

Step 1 - write the equation:

M1 V1 = M2 V2 Step 2 - manipulate the equation:

V1 = M2 V2 /M1

Step 3 - add the numbers:

V1 = (0.40M) (100ml) /2.0M

Dilution Problem

Step 4 - do the calculation:

V1 = 20ml

Step 5 - describe the preparation:

Add 80ml of distilled water to 20ml of the 2.0 M MgSO4 solution

Dilution Problem

HomeworkProblems

Molarity Calculations:

1. 0.975M (NH4)2C4H4O6 2. 0.257M CoSO4 3. 0.291M Fe(NO3)2

Preparations:

1. 390g NiCl2

2. 95.3g AgF 3. Use 50cm3 of the 1.0M

NaCl solution. Add 200cm3 of distilled water to make the total volume 250cm3.

Dilutions: 1. Add 19.2cm3 of 0.52M CoCl2 solution

to a graduate. Add distilled water to make the total volume 100cm3.

2. Add 72.5cm3 of 0.69M Ba(NO3)2 solution to a graduate. Add distilled water to make the total volume 200cm3.

3. Add 37.5ml of 2M NH4Br solution to a graduate. Add distilled water to make the total volume 500ml.

Solution Preparation

Solution A homogeneous mixture of two or more substances.

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