chapter 4 notes (addendeum, from chapter 11)

8/10/2019 Chapter 4 Notes (Addendeum, From Chapter 11)

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Chapter 11 - Properties of Solutions

I Solution Composition

b

. Because a mixture unlike a chemical compound has variable composition the relative amounts of

substances in

a

solutions must be specified.

B. Defining Solution Conten t

i. General terms to describe solution conten t

1

dilute

-

2. concentrated

-

ii. Precise definitions to describe solution conten t

1 Example:

A

solution is prepared by mixing

1 OO

g ethanol (C2HsOH) with 100.0 g water to give a final volume

Calculate the molarity mass percent mole fraction and molality of ethanol in this solution

2

Molarity M) moles per

h

CW

Maw

Percent mlm)

grams

of

solute

per

1

g

of

M

o f r o h t a s

solut~on Lita

Volume Percent vhr)

m

of solute

per

1 mL of

solut~on

MassNolurne Percent (I )

grams of solute

per

1 mL of

solut~on

I arts

per hundred (pph)

loo

---, ,,,

Parts

per m illion (pprn)

x

1

laC

--:

Pa rb per billion (ppb)

A

)m

XI^

~ - l P . r n

I

4.

Mole Fraction X) atio of the number of moles of a given com ponent to the

total number of moles of a solution.

x


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A solution is prepared by mixing

1

OO g ethanol C2HsOH) with 100.0 g water to give a final volume

of 101

mL.

Calculate the molarity, mass percent, mole fraction, and molality o f ethano l in this solution

5. Molality m) Moles of solute per kilogram solvent

Why use molality instead of molarity? Volum es change with temperature so molarity varies w ith

temperature). Molality is independen t of temperature

6

Normality

N)

equiva lents per liter of solution

a. the number of equivalents depends on the re ction

taking

place

i. acid base reactions- he equivalentmass of acid or bass that

can furnish or accept exactly 1 mole of protons m.

Normality

Acid or Base Molar Mass Equivalent Mass Molarity vs. Normality

HCI

H2S 4

NaOH

ii redox reactions - he equivalent is defined s the quantity

of

6 f ~ 8 ~ i / ~ i

oxidizing or reducing agent that can accept 1 mole o f electrons.

I 1.

Example:

AC~/.

/

Since

nOd

on present in

1

mole of KMn04

I ~ L-l 0 0/604

s y

consu mes 5 m oles of electrons, the equivalent mass is the

FP

molar m ass divided

by

5.

Y l

a looy C L ~ ~ . I


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,

.u

F/ ---.

m -

dm

f 1 7 s 4 h s y 8ow . y g g 1 23~40/1

~ ~ 2 3

*

/mol

H t q

l h mL &;*ln

7. Example:

The

electrolyte in automobile leid storage batteries is a

3.75

M sulfuric acid solution that has a density

Consider a solution ed y mixing 10.0g of methanol

CH30H,

density o f the solution is 0.8?0

dmL ith 50

5c

f

water_ alculate the composition

in

terms of:

a. mass %

l o~o f iCU3 K

50 0

7 4 i d t lo 0~H 6 d

5

c.)m ole fraction

&

I % c o i b u s

I 0 . 0 ~3ortx

2 ~ 4 ~

w

beg

h u 1 e u 3 7

+

~ O ~ O J H Z ~

d.

lLiolality

(rn) a 0%cu os

e. Molarity M) a


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A. The extent to which a solute dissolves in a particular solvent depends upon:

i. the nature of solvent-solute particles and the interactions between them.

ii. Temperature of the solution

iii. Pressure (for gaseous solutes)

B. Factors Affecting Solubility

i. like dissolves like

1. Intermolecular Forces wo substances with similar intermolecular forces

(hydrogen bonding, dipole forces, London Dispersion) will be soluble in on

another.

2. Example: Pentane

CSH12)

and hexane (C6H14)

a.

What is the strongest intermolecular force between pentane and

hexane molecules?

Pentane: H-Bonding

I

Dipole Forces

Hexane: H-Bonding Dipole Force

C

--

C

6.

Are hexane and pentane soluble in each

t

( . I f (

c. Which typeof@pmces w p e n p ; : d z a n e be soluble?

Polar Solvents onpolar Solvents

ii. Organic compounds which contain

OH

(hydroxyl) functional

roups

end to be

soluble in water.

1.

Examples: Methanol, Ethanol and Ethylene Glycol

H H H

H H

I

I

I

H C OH H c C OH H-C C H

Methyl Alcohol Ethyl Alcohol Ethylene Glycol

3

Hydrogen bonding is the principle intermolecular force holding each of the

above molecules together when

in

the liquid or solid state.

4. Each molecule is similar in structure to water, which explains the solubility

compatibility between the substances.

8

1

b

H

P .

solubility decreases.

Hydrophobic


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~

, sorbs k t -

C.

Effect of Temperature on Solubility

Increasing the temperature of a solution increases the

solubility of the solute. a ~ p ~ ~

mo US grn

Dissolving a solid in a liquid is usually

an

endothermic

process; heat must be absorbed to break down the crystal

lattice of the solid.

iii.

Saturated solution

saturation is the point at which a solution of a substance can

dissolve no more of that substance and additional amounts of that substance will

appear

as a precipitate.

1 This point of maximum concentration, the saturation point, depends on the

temperature of the liquid as well as the chemical nature of the substances

involved.

2 Supersaturated solution -

Cooling a saturated solution will result in that the

concentration is actually higher than the saturation point, the solution has

become supersaturated.

D Effect of Pressure on Solubility

i. Pressure has a major effect on solubility only for gas-liquid systems.

ii. Henry s Law - At a constant temperature, the amount of a given gas dissolved in a

given type and volume of liquid is directly proportional to the partial pressure of that

gas in equilibrium with that liquid.

Where:

f

-

s

-

g

kPg

C,

=

concentration of a gas

p, fi P,

=

partial pressure of the gas

k

=

constant for a given gas-liquid system.

iv. Example

The solubility of pure nitrogen in blood at body temperature,37OC, and one atmosphere is

6.2

x M

If a diver breathes

-

ir

X,?

=

0.78)

at a depth where the total pressure is

2 5

atm, calculate the

concentration of nitrogen in her blood.

k = c ]

6 , ~/dq fi/

~ . ' ~ y l ~ d h

s4 ,-

=

P / O G ~ YI

\+- x ~ k = d 2 /O q m , d &

*-/I ,%)L/~'P C ~ ~ / - L /A d ~ p l U

7 =

x

?T

-

@t7&>ca Q/.)

chapter 4 notes (addendeum, from chapter 11)

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