Chapter 17 and 18 Notes.ppt

Water Molecule Warm-up

 Pick up one water molecule model from the front desk.  Draw your water molecule in the space provided on your

notes.

 Label the hydrogen and oxygen on your water molecule.

 Describe what happens when your water molecule is brought near the water molecule of someone near you.

Water Molecule

 Water has polar bonds due to a difference in electronegativity of 1.4.

Question!

 Draw a water

Water Molecule

 Molecules are attracted to each other by

hydrogen bonds

H

H

O

O

H

H

H

H

O

O

H

H

∂ ∂- -∂ ∂++ Hydrogen bond Hydrogen bond

Polar Covalent bond

Polar Covalent bond

∂

∂++

∂

-Surface Tension

that tends to minimize the surface area of a liquid

 Tends to hold a drop of

liquid in a spherical shape

 A sphere has the smallest

 Draw arrows showing

the pull of surface

tension on the drop of water below.

Surface Tension

 Water’s surface tension

is stronger than most liquids

 Surface tension

Surface Tension

 Surfactant – substance that interferes with

the hydrogen bonds between water

Soaps & Detergents

Soaps and detergents are chains that have one end that is

like oil and has no charge and the other end is charged.

O O O O Oil Oil droplet droplet O O O O O O SO

SO₄₄-

-SO

SO₄₄-

-SO

SO₄₄-

-SO

SO₄₄-

-SO

SO44-- SOSO 4

4-

-O

Thermochemistry Review

 Specific heat capacity – the amount of energy

required to raise the temperature of 1.0 g of water by 1C.

 q = m  C  T

 C = specific heat capacity

 1.00 cal/g°C or 4.18 J/g°C

Question!

 Why would a burn from steam at 100C be

Impact of Thermochemical

Properties

Date Galveston High/Low (°F)

El Paso

High/Low (°F) Feb 8 68°/ 51° 68°/ 39°

Feb 9 63°/ 51° 69°/ 41°

Feb 10 64°/ 57° 70°/ 41°

Average Difference

Water Vapor

 Evaporation – the change of state from liquid to

Water

Vapor

 Condensation – the change of state from a gas to

a liquid

 Water has a high heat of vaporization (2260 J/g)  2260 J of energy are needed to change 1.0 g of water

from a liquid to a gas or a gas to a liquid

 Water’s high boiling and melting points are caused by hydrogen bonding. When compared to ammonia (NH₃) the hydrogen bonds are stronger. It takes more energy to disrupt these bonds, giving water a higher boiling and

Ice

 Freezing – the change of state from a liquid to a

solid

 Water has a high heat of fusion (334 J/g)

 334 J of energy is needed to change 1.0 g of water from

Ice

 Density – the ratio of the mass of an object

to its volume

D =

D =

M

M

V

Question!

Ice

Bellwork

Wednesday, February 13

 Today we are learning about solutions.

 What does it mean to dissolve? Be ready to

describe it.

Review Problems

1. Calculate the heat released when a cup of tea cools

from 61C to 25C. The mass of the tea is 225g. (Assume the specific heat of the tea is equivalent to that of water)

= (225g)(4.18 J/gºC)(36ºC)

= (225g)(4.18 J/gºC)(36ºC)

= 33900 J

= 33900 J

Givens: q = mCq = mCTT C = 4.18 J/g°C

C = 4.18 J/g°C m = 225 gm = 225 g ∆

Review problems

2. Calculate the amount of heat needed to boil

(vaporize) a kettle filled with water (250 g).

Given: phase change, so DA only

m = 250g H_vap= 2260 J/g q = ?

250g 2260 J = 1 g

Review Problems

3. How much heat energy is needed to melt 30.0 g of ice

at 0C to water at 80C?

Givens: Heat change and phase change

C = 4.18 J/g°C H_fus= 334 J/g m = 30.0 g ∆T = 80°C - 0°C = 80°C q = ?

To melt, use DA

30.0 g 334J = 1.00 x 104 J

1 g

To heat, use q = mCΔT

q = (30.0g)(4.18J/gºC)(80ºC) = 1x104 J

Review

4. Which atom in the water molecule is most

electronegative?

A. Hydrogen B. Oxygen C. Neither

5. How are water molecules held together?

A. magic

Review

6. Water has a ________________ vapor pressure

due to _________________.

A. low; low surface tension

B. high; london dispersion forces C. low; hydrogen bonds

D. high; hydrogen bonds

7. Water has a ____________ heat capacity and a

______________ heat of vaporization.

Review

8. Due to the high heat capacity and vaporization

energy of H₂O, coastal areas have ___ temperatures.

A. moderate

B. extreme

C. low

D. high

9. Which of the following are characteristics of surface

tension?

A. reduces vapor pressure

B. holds liquids in spherical shapes

C. is caused by hydrogen bonding

D. is very strong in water

Review

10. A surfactant is a substance that interferes

with ________________ and decreases the _________________.

A. free time; productivity

B. hydrogen bonding; surface tension

Solutions

 Aqueous solution – a water solution containing

dissolved substances

 Solvent – the dissolving medium

 Solute – particles dissolved in the solution

 Would the salt or the water be the solute in a

salt-water solution?

Question!

 Would the salt or the water be the solute in a

Solutions

 Solvation – the process that occurs when a

solute dissolves in a solvent

Solutions

 “Like Dissolves Like”

 Polar substances dissolve

in polar solvents.

 H₂O: would dissolve things

that are polar and ionic

 Nonpolar substances

dissolve in nonpolar

solvents

 Gasoline: would dissolve

Write “soluble” next to the

compounds listed below that

would dissolve in water.

 Gasoline ______________  Salt ______________

Bellwork

Ionic Ionic Covalent Covalent

Polar

Solutions

 Electrolytes –

compounds that conduct an electric current in

aqueous solutions or in a molten state

 Includes all ionic

compounds

 Some very polar molecules

Solutions

 Nonelectrolytes – compounds that do not

conduct an electric current in either aqueous solutions or the molten state

 Molecular compounds

 Organic compounds (carbon containing

compounds)

 Nonpolar compounds

Solutions

 Strong electrolytes:

 In a strong electrolyte, nearly all the

ions dissolve

in aqueous solution

 NaOH, HCl and NaCl

 Weak electrolytes:

 In a weak electrolyte, only very small

amounts of ions dissolve in aqueous solution

 Distilled H

2O, ammonia and

Solutions

 How electrolytes form:



NH

3

+ H

O

→



HCl + H

2

O

→

NH

NH

+

+

OH

OH

-H

-Warm-up

Friday, February 11

Please get out your notes and lab

paper. Read the lab Background

Hydration

 Water of hydration – water

in a crystal

 When copper (II) sulfate is

dissolved in water and then the water allowed to evaporate,

copper (II) sulfate pentahydrate is formed

 CuSO₄ + 5H₂O ↔ CuSO₄5H₂O

 Hydrate - compound that

contains water of hydration

The picture above shows

The picture above shows

copper (II) sulfate

copper (II) sulfate

pentahydrate. Notice the

pentahydrate. Notice the

water molecules surrounding

water molecules surrounding

the copper molecules.

Hydration

 When cobalt (II) chloride is dissolved in

water and then the excess water is

allowed to evaporate, cobalt (II) chloride hexahydrate is formed

↔

↔

CoCl

CoCl





6H

6H

O

O

CoCl

In the hydrate CuSO

.

5H

0,

how many water molecules

would be present in 3

Naming Hydrates

 To name a hydrate, use the following:

 They are named by giving the name of the salt

followed by a prefix that stands for the number of water molecules associated with a formula unit of the salt.

 Ex: CuSO₄ • 5H₂O

copper (II) sulfate

Calculating the Percent

Hydration

4•5H2O.

Cu 1 S 1 O 9 H 10

Total mass of hydrate: 249.6 g Mass H₂O:

H 2 O 1

Total mass of H₂O: 18.0 g X 5 = 90.0g

+

+

+

+

x

x 63.563.5 == 63.563.5 x

x 32.132.1 == 32.132.1 x

x 16.016.0 = 144.0= 144.0 x

x 1.0 1.0 == 10.010.0

x

x 1.01.0 == 2.02.0 x

Hydration

Mass H₂O Mass Hydrate

90g

90g

249.6g

249.6g

X 100

X 100

X 100

X 100

=

=

% H

% H

O

O

=

Question!

 A student performs an experiment in

which she heats a hydrated compound

Bellwork

Monday, February

 A pot of boiling water contains 200.0ml of water.

If the pot was heated from 25.0°C to 100.°C, how much heat was added to make the water boil and vaporize? (Hint: the density of water is 1g/ml)

Please turn in your lab to

Please turn in your lab to

the red box as you come in.

Hydration

 Efflorescence – process in which hydrated

compounds lose water to its surroundings.

 Occurs when a substance has a vapor pressure

higher than that of water

 Some hydrated compounds can effloresce and will

form anhydrous compounds

 In this process the compound loses water

Compound(high pressure)

Compound(high pressure)

H

Hydration

 Deliquescence – process in which anhydrous or

somewhat hydrated compounds gain water from the surroundings.

 Occurs when the solution formed has a vapor pressure

lower than that of the water in the air

 Some anhydrous or somewhat hydrated compounds will

deliquesce and form hydrated, or more hydrated, compounds

 In this process water is gained

Compound(low pressure)

Compound(low pressure)

H

Hydration

 Desiccants – hygroscopic (water absorbing)

substances that are used as drying agents.

 Small silica gel packets used in electronic

merchandise shipping

 Sodium hydroxide is a dessicant and becomes a

Suspensions

 Suspensions – mixtures from

which particles settle out upon standing.

 Suspensions have particle sizes

larger than solutions

 Generally the sizes are larger

than 100 nm

 The particle size in a solution is

about 1 nm

Colloids

 Colloids – heterogeneous mixtures containing

particles that are intermediate in size between solutions and suspensions

 Generally the sizes are

between 1 nm and 100 nm

 Examples include glues,

Colloids

 Tyndall effect - the

Colloids

 Brownian motion – the chaotic movement of

Emulsions

 Emulsion – a colloidal dispersion of liquids in

liquids.

 Emulsifying agents are used to keep the emulsion

stable.

 Soap and detergents are good emulsifying agents  Mayonnaise is an emulsion. Do you know the main

Describe the following as

solutions, suspensions,

colloids or emulsions.

 Italian dressing  Jello

 Milk  Tea

Warm-up: Smarties!

 Table 1: Just put smartie in your mouth  Table 2: Swirl smartie, but don’t chew  Table 3: Chew smartie, then let dissolve  Questions

 What are three things that can speed up the

dissolution of the smartie?

 Will temperature always affect how substances

Chemistry I

Factors affecting

rate of solution

:

 Agitation

 brings fresh solvent in contact w/ solute  does not affect amount the amount of

Factors affecting rate of

solution

:



Temperature



heat increases rate of

solution by increasing

 Particle size

 smaller particles dissolve faster because of

greater surface area

Question

 What would dissolve faster?



Solubility

- a measure of the amount of

solute that will dissolve in a given amount of

solvent.

 we usually use g solute/100 g solvent

0ºC 20ºC 100ºC Ba(OH)₂ 1.67 31.89

---NaCl 35.7 36.0 39.2

Factors affecting

solubility:

 The nature of the solute and the solvent

 Temperature  Most solids

become more soluble at a higher

temperature. A few become less soluble.

 solubility of gases decreases as temperature

increases

 Pressure (for

solutions of gases in

liquids)

 increasing

pressure increases solubility of gas

Question



Pressurized CO

2

is pumped into three

Bellwork

 Place the following into the categories shown.

 Sugar  CH₄  NaI  HCl

 KOH

 Saturated solution -contains the maximum

amount of solute that will dissolve in a given amount of solvent at a given temperature.

 Unsaturated solution- more solute could be

dissolved

-indicated by no solid present

 Supersaturated solution- more solute is

EXAMPLE:

 How could you determine if a solution

is saturated, unsaturated or supersaturated?

 If solid is present – it is Saturated

 If no solid is present – it is unsaturated

or supersaturated

 To determine which, add a crystal  Supersaturated will crystallize

 Unsaturated will dissolve the crystal

Bellwork

Thursda

y,

Februar

y 21

 Draw and

 Miscible - liquids that are soluble in each

other

 Ex. ethanol and water

 Immiscible- liquids that are not soluble in

each other

 Henry’s Law- At a

given temperature, the solubility of a gas in a liquid is directly

proportional to the pressure of the gas above the liquid.

EXAMPLE:

 At 25oC, the solubility of a certain gas in H₂O is

0.75 g/L of H₂O at a pressure of 1.0 atm. What is the solubility of the same gas at 10.0 atm

pressure? S₁ = 0.75 g/L P₁ = 1.0 atm S₂ = ?

P₂ = 10.0 atm

S

S₁₁ S S₂₂ P

P₁₁ P P== ₂₂

0.75 x

0.75 x

1.0 10.0

1.0 10.0==

x = 7.5 g/L

 Concentration -a measure of the amount of solute per given amount of solvent or solution.

 Measured in units of molarity

 Dilute solution- Contains a small amount of solute  Concentrated solution- Contains a large amount

Molarity(M) = moles of solute liters of solution

 A 1 M solution contains 1 mole of solute per 1

EXAMPLES:

 What is the concentration in molarity of a

solution made with 1.25 mol NaOH in 4.0 L of solution? _{# mol}

# L

#mol = 1.25 mol V = 4.0 L

M = ?

M = 1.25 mol

4.0 L = 0.3125 M = 0.31 M

M =

Question

 A student measures 58.5g of NaCl and

Dilutions

 Diluting a solution won’t change the number of

moles of solute. It only changes the concentration.

M₁V₁ = M₂V₂

This formula works only for dilution of a solution!!!!

 How would you make 500. mL of 6.0 M NaOH

from 12.0M NaOH?

How many grams of NaOH would be necessary?

M1 = 12.0M

M1 = 12.0M V1 = ?V1 = ?

M2 = 6.0M

M2 = 6.0M V2 = 500. mLV2 = 500. mL

(12.0M)(x) = (6.0M)(500. ml)

(12.0M)(x) = (6.0M)(500. ml)

X = 250. ml

X = 250. ml

To make the solution,

To make the solution,

you would measure out

you would measure out

250. ml of the

250. ml of the

concentrated base and

concentrated base and

add it to water until you

add it to water until you

have 500. ml of solution.

have 500. ml of solution.

250. ml NaOH

250. ml NaOH 1L NaOH1L NaOH 12.0 mol NaOH12.0 mol NaOH 40.0 g NaOH40.0 g NaOH = = 1000 ml

1000 ml 1 L NaOH1 L NaOH 1 mol NaOH1 mol NaOH

120 g

120 g

NaOH

Question

 A solution is made with 10.5g AgCl dissolved

EXAMPLE:

 For a lab we need 500. mL of 4.25 M HCl.

Concentrated HCl is 12.0 M. How do we make up the solution?

M₁V₁ = M₂V₂

M₁ = 4.25 M

V₁ = 500. mL

M₂ = 12.0 M

V₂ = ?

(4.25M)(500. mL) = (12.0 M)(V₂)

V₂ = 177 mL

Add 177 ml of concentrated hydrochloric acid to

Add 177 ml of concentrated hydrochloric acid to

enough water to make 500 ml of solution.

Solution Stoichiometry

 use Molarity as a conversion factor#

# mol

1 L

1 L

EXAMPLES:

 What volume of 0.246 M nitric acid is required

to react completely with 0.386 L of 0.0515 M

Ba(OH)₂?

2HNO₃ + Ba(OH)₂ → Ba(NO₃)₂ + 2H₂O

0.386 L Ba(OH)₂ 0.0515 mol Ba(OH)₂ 1 L Ba(OH)₂

2 mol HNO₃

1 mol Ba(OH)₂ 0.246 mol HNO₃ 1 L HNO₃

= 0.162 L HNO₃ molarity mol-mol

ratio

 Calculate the number of grams of carbon

dioxide that can react with 0.135 L of a 0.357 M solution of potassium hydroxide, according to the following reaction:

2KOH + CO2  K2CO3 + H2O

0.135 L KOH 0.357 mol KOH

1 L KOH 2 mol KOH 1 mol CO₂

1 mol CO₂ 44.0 g CO₂

= 1.06 g CO

ratio

% Solutions

 Percent by volume

% v/v = volume solute (mL) x 100 volume solution (mL)

 Percent mass/volume

EXAMPLES:

 Rubbing alcohol is 70.0% isopropyl alcohol by

volume. How many mL of isopropyl alcohol are in a 250 mL bottle of this solution?

% v/v = 70.0 %

v solution = 250 mL v solute = X

% v/v = volume solute (mL) x 100 volume solution (mL)

70.0 % = __X__ x 100 250 mL

 If the % mass/volume (m/v) for a solute is 6.80

%, and the volume of the solution is 42.6 mL, what mass of the salt is present?

% m/v = 6.80 %

v solution = 42.6 mL g solute = X

% m/v = mass solute (g) x 100 volume solution (mL)

6.80 % = __X__ x 100 42.6 mL

X = 2.8968 g