properties of solution1 - islamic university of...

Physical Properties of Solutions Dr. Ramy Y. Morjan

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Chapter 13

Physical Properties of Solutions

Overview Composition of Matter

All matter can be divided into two major classes which are pure substance and

mixture. The pure substance could be Element or could be compound. The mixture

can be either Homogeneous mixture or heterogeneous as in figure 1.

Fig. 1 Composition of Matter

Homogeneous Mixture: The prefixes "homo"- indicate sameness A homogeneous

mixture has the same uniform appearance and composition throughout. Many

homogeneous mixtures are commonly referred to as solutions as example dissolving

sugar in H2O.

Heterogeneous Mixture: The prefixes: "hetero"- indicate difference.

the individual components of a mixture remain physically separated and can be seen

as separate components, for example mixing sand with H2O

Matter

Pure Substance Mixture

Elements Compounds Homogeneous Heterogeneous

Metals Non-metals

solid, liquid gas

constant composition variable composition

Physical methodsPreparing mixture

Separating mixture

Semimetals (metalloids)

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What is a Solution? A solution is a homogeneous mixture of two or more substances dissolved in each

other. Solutions mainly consist of two components one is called solvent and the other

is called solute.

Solvent: the substance in which the substances (solute) are dissolves to produce a

homogeneous mixture (solution) and it is the substance that present in large amount.

Solute: the substances that dissolves in a solvent to produce a homogeneous mixture

Although we shall concentrate mainly on liquid solutions, solutions can be solids,

liquids or gases.

Classification of solution: Solutions can be classified on the basis of their state:

solids, liquids or gases. Examples of various ways of preparing a two component

solution in each of the three states are summarized in the following table.

Solution type solvent solute Example

Solid

Solid

Metalloids, copper dissolved in nickel

Solid

Gas

Hydrogen dissolved in palladium

Solid solution

Solid

Liquid

Mercury dissolved in gold

Gas

Gas

Oxygen dissolved in Nitrogen (air)

Gas

Solid

Dry ice dissolved in (sublimed) nitrogen

Gas solution

Gas

Liquid

Chloroform dissolved in (evaporated into) nitrogen

Liquid

Solid

Sugar dissolved in water

Liquid

Gas

Carbon dioxide dissolved in water (soft drinks)

Liquid solution

Liquid

Liquid

Ethanol dissolved in water





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In order to form a liquid solution in which the solute and solvent are both liquids, the

two liquids must be miscible.

Miscible liquids mean that the two liquids can dissolve completely in each other at

any proportions (بأي نسبھ) as example water and ethanol. In the other hand immiscible

means they can’t dissolve in each other as example oil and water.

Types of Solutions

Well, as we have seen in the above section that solution are formed by mixing solute

and solvent with each other. In order to such as this process to takes place the

molecules of the solute and the solvent must get in direct contact with each other

leading to solubility of the solute in the solvent and forming the solution. The ratio of

solute and its solubility in the solvent as well as the amount of the solvent used

determines the nature (type) of the produced solution.

Solubility: solubility is defined as the maximum amount of substance (solute) that

will dissolve in a solvent at that temperature. Usually solubility is measured by g/L.

Chemist usually classifying solution according to the capacity of solvent to dissolve a

solute at a specific temperature; in other words solution can be classified according to

the ratio between solute and solvent at a given temperature. This classification led to

the following three groups of solution.

1) Saturated solution

Contains the maximum amount of solute dissolved in solvent. In other words; when

the solvent has dissolved all the solute it can, and left some un-dissolved solute it’s a

saturated solution. For example; if you add 20 g of sugar to 100mL of water at room

temperature, then all of the sugar will dissolved. However if you add 300g to the same

amount of water, some of the sugar remains un-dissolved. We said this solution is

saturated. In saturated solution there is equilibrium between solution and solute that

mean if you for example keep string the solution some of the un-dissolved solute will

dissolve and in the same time some of dissolved solute will come out so they are in

equilibrium.

2) Unsaturated solution

An unsaturated solution contains less of solute than it could hold i.e. solutions can

dissolve more solute under normal condition. For example a saturated solution of





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NaCl/H2O contains 25g of NaCl in 100mL water; the unsaturated solution of

NaCl/H2O will contain less than 25g NaCl in 100mL water. In an unsaturated solution

there is no equilibrium exists between the solute and the solution because there is no

undissolved solute

3) Supersaturated solution: A solution that is temporarily holding more than it can,

a seed crystal will make it come out. A saturation solution can be prepared by adding

excess solute to a saturated solution. Unsaturated solutions are not stable solutions, for

example if just one crystal of solute is added, or the liquid is shaken, the excess solute

will crystallize immediately and saturated solution will form.

Crystallization: The process in which dissolved solute comes out of solution and

form crystals.

Crystals: Crystals are solids that form by a regular repeated pattern of atoms or

molecules connecting together. In crystal the arrangements of the building blocks

(atoms and molecules) lie in an orderly array. In crystals, however, a collection of

atoms called the Unit Cell is repeated in exactly the same arrangement over and over

throughout the entire material. Diamond the very expensive jewel which is 3D

covalently bonded network of carbon atoms. Diamond is a gigantic molecule ( جزیئ

.with crystalline structure which gives it its unique properties ضخم(





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Aqueous solutions

An aqueous solution is a solution in which the solvent is water. It is usually shown in

chemical equations as a subscript (aq). The word aqueous means pertaining to, related

to, similar to, or dissolved in water. As water is an excellent solvent as well as

naturally abundant, it logically has become a ubiquitous solvent in chemistry.

Substances that do not dissolve well in water are called hydrophobic ('water fearing')

whereas those that do are known as hydrophilic ('water-loving'). An example of a

hydrophilic substance would be the sodium chloride (ordinary table salt).

A Molecular View of the Solution Process

In the previous sections we have discussed the formation of solutions, and you learned

that solutions are formed by dissolving a solute in a solvent. Now we need to learn

how the dissolving process takes place. To understand this process we need to

remember what we have learned about the intermolecular forces and intramolecular

forces (Chapter 12), however these forces play very important rolls in interpretation

the solubility and solution formation. (Please go back to chapter 12)

To make a solution the solute and solvent are mixed together, and in order to obtain a

solution the solute molecules will dissolve in the solvent molecules that means the

solute molecules will take the positions of some solvent molecules.

What happens when you add a solid solute to a liquid solvent?

Immediately after the addition of a solid solute to a liquid solvent, the solid –state

structure begins to disintegrate (یتحلل) little by little. The solvent molecules chip away

at the surface of the crystal lattice, prying out solute particles and surrounding them,

and finally dispersing them throughout the body of the solution. This process

Cl-

Na+Cl-Na+

H2O

H2O

Na+

Cl-

solute

Cl-

Na+Cl-Na+

H2O

H2O

Na+

Cl-

solute





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(solubility) depends on the relative strength of three attractive forces (intermolecular

forces) which are:

1) Solvent-solvent interaction (before dissolution)

2) Solute –solute interaction (before dissolving)

3) Solute- solvent interaction (during the dissolving process)

When the solution is formed i.e. for the solute to dissolve in the solvent both the

solvent –solvent intermolecular force and solute –solute intermolecular forces will be

replaced (تستبدل) by the solute- solvent interaction (see figure 3). From this disscation,

we can conclude (نستنتج) that if the solvent –solvent intermolecular force and solute –

solute intermolecular forces are weak , and the solute- solvent interaction is strong the

formation of the solution will be easy and vies versus. (العكس صحیح)

Fig. 3 Mechanism of dissolving solid solute in liquid solvent

Formation of solution & energy changes

The solubility process of solid solute in a liquid solvent is companied with changes in

the energy of both the solute, solvent and the formed solution. As we have just

learned that for a solid solute to dissolve in a liquid solvent the intermolecular force

(interaction) between the solvent –solvent, and solute – solute must be destroyed. In

the same time a new interaction between solute- solvent start to take place. In both

cases the energy cotenants of the solute and solvent and solution is changed. To make

liquid solvent

solid solute

there is intermolecular forcebetween the solvent molecules

solvent -solvent interaction

there is intermolecular forcebetween the solvent molecules

solute-soluteinteraction

In solution, there is intermolecular interaction between solve and solute. The old solvent -solvent interaction and solute-soluteinteraction disappear

solution





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this clear and easy; let us assume that the formation of a solution to takes place in

three steps as the following.

a) Separation of the solvent –solvent interaction

b) Separation of the solute – solute interaction

c) Formation of new interaction between solute- solvent

In both steps a and b the process need energy in order to overcome the intermolecular

force (interaction) between the solvent molecules alone and between the solute

molecules alone. Steps a & b are endothermic steps. In any process or a chemical

reaction if energy (heat) is absorbed, the process is said to be endothermic process or

reaction. In the other hand, in step 3 the solute and solvent are mixed together, in

terms of energy; this process can be either endothermic process or can be exothermic

process. In any process or a chemical reaction if energy (heat) is released the process

is said to be exothermic process or chemical reaction. The total change of energy

during the dissolving process is the summation of each change in each step i.e. a and b

and c in our example. The total change of heat is known as the heat of solution ∆H or

enthalpy ∆H of solution (reaction).

For steps a, b and c the

∆Hsol = ∆Ha + ∆Hb + ∆Hc

If ∆H is positive (∆H > zero) that indicates that energy (heat) is absorbed i.e.

endothermic process or reaction.

If ∆H is negative (∆H < zero) that indicates that energy (heat) is released i.e.

exothermic process or reaction.

When the value of ∆Hsol is positive and when it is negative?

v If the interaction between solute and solvent is stronger than both the

interaction between solvent-solvent molecules and the interaction between

solute-solute molecules the value of ∆Hsol will be negative (∆Hsol < zero) and

the process is exothermic. In this the solubility process is favourable and the

solid solute will dissolve easily (easily here means that there is no great energy

deficit) in the liquid solvent to form a solution.

v If the interaction between solute and solvent is weaker than both the

interaction between solvent-solvent molecules and the interaction between





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solute-solute molecules the value of ∆Hsol will be positive (∆Hsol > zero) and

the process is endothermic. In this the solubility process is unfavourable and

the solid solute will not dissolve or slightly dissolved in the liquid solvent to

form a solution. Figure 5 is explaining the energy changing during the

dissolving process.

Fig. 5 The energy changes during the dissolving process

Input of energy is required to separate solute-solute (step a) and solvent-solvent (step

b) attractions. Energy is released due to solute-solvent attractions (step c). If the

amount of energy released in step c is greater than the energy absorbed in steps a and

b the process is exothermic (∆Hsol is negative) and favoured for dissolution, but if the

amount of energy in step c is less than the absorbed energy in steps a and b the

process is endothermic (∆Hsol is positive) and un-favoured as in step d in figure 5.

step a

solventsolute

solventexpanded

solute

expanded soluteexpanded

solvent

net energy change

solution

negative valueexothermic

p ote

nti a

l ene

rgy step b

step c

H

positive valueendothermic

H

solubility is unfavourable

step d





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In addition to the energy factor, there is another factor that effect on formation of

solution which is the tendency toward disorder. When solute and solvent molecules

are mixed together; to form a solution, there is an increase in randomness or disorder.

In the pure state, the solvent and solute have enough degree of order, this order is

destroyed when the solute dissolve in the solvent. Therefore, the solution process is

accompanied by an increase in disorder. This increasing in disorder of the system is

favouring the solubility of any substance. Figure 6 is explaining the above discussion.

Does any solute dissolve in any solvent? The direct and simple answer to this question is NO. There are so many solutes that

do not dissolve in particular solvent. However; solubility of any solute in a solvent

depends mainly on the type of intermolecular force of both solutes and solvents.

solvent molecules are ordered solute molecules are ordered

solution

solvent molecules have tendency to be disordered

solute molecules have tendency to be disordered





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To understand why some solutes do not dissolve in some solvents let us learn a

general saying which says ‘like dissolves like’. The meaning of this says is that if any

substances in this case (solute & solvent) have the same type of intermolecular forces

they will likely dissolve in each other. In other words if two or more substances have

the same polarity they most likely will dissolve in each other.

Solvents and solutes can be broadly classified into polar and non-polar. A molecule

is polar if there is some separation of charge in the chemical bonds, so that one part of

the molecule has a slight positive charge and the other a slight negative charge. The

slight negative and positive charges are arising as a result of the difference in the

electronegativities between the elements that form the bond. Polarity can be

measured as the dielectric constant or the dipole moment of a compound.

Water is a well-known example of a polar molecule, it consist of two H atoms

covalently bonded to oxygen O atoms (covalent bond). H2O is a polar because there is

difference in electronegativities between the H atoms and the O atom (figure 7)

Fig. 7 H2O is polar molecule

The dipole moment (µ) of water is 1.85 deby. The value of µ is calculated using the

following equation

Where µ is dipole moment, q is the charge and r is the distance between the two

charges. Table 1 show some compounds and dielectric constant and dipole moment

(µ) for the values.





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Compound Dielectric Constant

Dipole Moment (µ )

Formamide 110 3.37

Water 79 1.85

Methanol 32 1.66

Benzene 2 0.00

Table 1

Please note that benzene has µ equal zero i.e. it is non polar compound. There are no

polar bonds in benzene as it contains covalent bond between only H and C which

have nearly similar electronegativities (2.2 for H and 2.55 for C).

The polarity of a solvent determines what type of compounds it is able to dissolve

and with what other solvents or liquid compounds it is miscible. As a rule of thumb,

polar solvents dissolve polar compounds BEST and non-polar solvents dissolve non-

polar compounds BEST: "like dissolves like".

Strongly polar compounds (ionic compound) like inorganic salts (e.g. table salt NaCl)

dissolve only in very polar solvents like water, while strongly non-polar compounds

like oils or waxes dissolve only in very non-polar organic solvents like hexane.

Similarly, water and hexane are not miscible with each other and will quickly separate

into two layers even after being shaken well. Table 12 in the next page has a list of

common solvents both polar and non-polar solvent with the values of dielectric

constant and dipole moment. You should be able to tell a good thing about the

solubility of any solvent with another solvent or the solubility of any solute (you need

to know its structure) in any given solvent.

A compound, such as water, that is composed of polar molecules. Polar solvents

can dissolve ionic compounds or covalent compounds that ionize. Non-polar

solvents, such as benzene, will only dissolve non-polar covalent compounds.

Now you should be able to expect if a given solute will dissolve in a given solvent or

will not. Of course; no need to memorize the values of the dielectric constant and

dipole moment and the electronegativities or any other numbers. However all what

you need is to use your understanding to the concept of polarity and to understand the

classification of the periodic table as this will help you to obtain a good judgment

about the solubility of compounds in each other.





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Table 2 Common polar and non-polar Solvents

Name Structure bp, oC dipole moment

dielectric constant

water H-OH 100 1.85 80 methanol CH3-OH 68 1.70 33 ethanol CH3CH2-OH 78 1.69 24.3

1-propanol CH3CH2CH2-OH 97 1.68 20.1

1-butanol CH3CH2CH2CH2-OH 118 1.66 17.8

formic acid

100 1.41 58

acetic acid

118 1.74 6.15

formamide

210 3.73 109

acetone

56 2.88 20.7

tetrahydrofuran (THF)

66 1.63 7.52

methyl ethyl ketone

80 2.78 18.5

ethyl acetate

78 1.78 6.02

acetonitrile 81 3.92 36.6

N,N-dimethylformamide (DMF)

153 3.82 38.3

diemthyl sulfoxide (DMSO)

189 3.96 47.2

hexane CH3(CH2)4 CH3 69 ---- 2.02

benzene

80 0 2.28

diethyl ether CH3CH2OCH2CH3 35 1.15 4.34 methylene chloride CH2Cl2 40 1.60 9.08 carbon tetrachloride CCl4 76 0 2.24





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Now, let us see how ionic compounds like NaCl dissolve in polar solvent like H2O? Ionic compounds form ionic solutions. Sodium chloride when dissolve in

water forms an ionic solution. Such ionic solutions conduct electricity. The solutions

are called electrolytes. Table salt or sodium chloride (NaCl) is an ionic compound.

Sodium is in the first group of the periodic table and chlore is in the group 7. The

difference in their electronegativities indicates that it is an ionic compound. Ionic

compound do not form hydrogen bonding with water. Salt is a solid at room

temperature. The sodium and chloride atoms are arranged in a pattern that we called

crystal lattice. This

process is described on

steps as following;

1) When we put salt in

water, the sodium and

chloride atoms are

released from their

crystalline lattice and

form positive sodium ions

(Na+) & negative chloride ions (Cl-). The (Na+) and (Cl-) are free to move within the

water i.e. over come the Intramolecular forces.

Fig. Formation of Na+ and Cl- in water





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2) Water is a polar solvent, it consist of 2 H atoms and one O atoms. These three

atoms are covalently bonded to each other (Intramolecular forces). As a result of the

difference in electronegativities between the O and H, a partial negative charge will

arise on the O atoms and a partial positive

charge will arise on the two H atoms. The

partial charges on the O and H give the

H2O its unique property as polar solvent.

However; water molecules are joining to

each other via an intermolecular force

(hydrogen bonds), this hydrogen bonds are

stands behind the high boiling point of

water, as well as it makes the H2O a good

solvent for so many compounds. Fig H bonds (Intermolecular forces)

In order for the solute to dissolve in the solvent, the intermolecular forces i.e. the

hydrogen bond between the water molecules must be over come, this distortion

required energy to over come the interaction between the solvent (water) molecules.

As a result of the over comes of the interaction between the solvent molecules; a room

are created between the water molecules and enabling the solute to get into the

solvent.





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3) Now we have the positive ions Na+ and the negative ions Cl- and the H2O with its

partial positive on H atoms and the partial negative on the O atom. The positive ion

Na+ will be attracted to the partially negative O atom via ion-dipole interaction, in the

same way the partial H atoms on water will be attracted by the Cl- negative ion as well

via ion-dipole interaction.

Fig. Ion-Dipole interaction between Na+ and negative O, and between H and Cl-

In the actual situation; so many partial negative O on water will be surrounding the

Na+ positive ion, and so many H atoms in water surrounding the partial negative Cl-

ion via Ion-Dipole interaction and the NaCl – H2O solution is formed. The process in

which an ion or molecules are surrounded by solvent molecules in a specific manner

is called solvation.

Ion-Dipole interaction and the NaCl – H2O solution is formed





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Calculating Concentration Units & Dilutions

The concentration of a chemical solution refers to the amount of solute that is

dissolved in a solvent. We normally think of a solute as a solid that is added to a

solvent (e.g., adding table salt to water), but the solute could just as easily exist in

another phase. For example, if we add a small amount of ethanol to water, then the

ethanol is the solute and the water is the solvent. If we add a smaller amount of water

to a larger amount of ethanol, then the water could be the solute!

Units of Concentration

Once you have identified the solute and solvent in a solution, you are ready to

determine its concentration. Concentration may be expressed by several different

ways, using percent by mass, mole fraction, molarity, molality, or normality.

1) Percent by Mass (%)

Percent by Mass (%) also called the percent by weight or weight percent. This is the

mass of the solute divided by the mass of the solution (mass of solute plus mass of

solvent), multiplied by 100.

x 100%

Percent by Mass has no units because it is a ratio of two similar quantities.

Example

Determine the percent composition by mass of a 100 g salt solution which contains 20

g salt.

Solution:

20 g NaCl / 100 g solution x 100 = 20% NaCl solution

moles of solute Percent by Mass =

Mass of solution





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2) Mole Fraction (X)

This is the number of moles of a compound divided by the total number of moles of

all chemical species in the solution. Keep in mind, the sum of all mole fractions in a

solution always equals 1. For example if we have a solution contains 2 mol of solute

A dissolved in 10 mol of solvent B; the mole fraction of the solute A is calculated by

using the following formula

moles of solute A Mole Fraction A

moles of solute A + moles of solute B

2/12 + 10/12 = 1

Example:

What are the mole fractions of the components of the solution formed when 92 g

glycerol is mixed with 90 g water? (molecular weight water = 18; molecular weight of

glycerol = 92)

Solution:

90 g water = 90 g x 1 mol / 18 g = 5 mol water

92 g glycerol = 92 g x 1 mol / 92 g = 1 mol glycerol

total mol = 5 + 1 = 6 mol

A water = 5 mol / 6 mol = 0.833

B glycerol = 1 mol / 6 mol = 0.167

It's a good idea to check your math by making sure the mole fractions add up to 1:

A water + B glycerol = .833 + 0.167 = 1.000

2 Mole Fraction A

2 + 10

10 Mole Fraction B

2 + 10





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3) Molarity

Molarity is probably the most commonly used unit of concentration. It is the number

of moles of solute per liter of solution (not necessarily the same as the volume of

solvent!).

Example:

What is the molarity of a solution made when water is added to 11 g CaCl2 to make

100 mL of solution?

Solution:

11 g CaCl2 / (110 g CaCl2 / mol CaCl2) = 0.10 mol CaCl2

100 mL x 1 L / 1000 mL = 0.10 L

molarity = 0.10 mol / 0.10 L i.e. molarity = 1.0 M

Example 2:

.0678 g of NaCl is placed in a 25.0 ml flask full of water. When the NaCl dissolves,

what is the molarity of the solution?

.0464 M NaCl

4) Molality

Molality is the number of moles of solute per kilogram of solvent. Because the

density of water at 25°C is about 1 kilogram per liter, molality is approximately equal

to molarity for dilute aqueous solutions at this temperature. This is a useful

approximation, but remember that it is only an approximation and doesn't apply when

the solution is at a different temperature, isn't dilute, or uses a solvent other than

water.

moles of solute Molarity =

litter of solution

moles of solute molality =

kilogram of solvent





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Example 1:

.20 mol of ethylene glycol / 2.0 Kg of solvent = .10 m ethylene glycol

Example 2:

What is the molality of a solution of 10 g NaOH in 500 g water?

Solution:

10 g NaOH / (4 g NaOH / 1 mol NaOH) = 0.25 mol NaOH

500 g water x 1 kg / 1000 g = 0.50 kg water

molality = 0.25 mol / 0.50 kg

molality = 0.05 M / kg i.e. molality = 0.50 m

5) Normality (N)

Normality is equal to the gram equivalent weight of a solute per liter of solution. A

gram equivalent weight or equivalent is a measure of the reactive capcity of a given

molecule. Normality is the only concentration unit that is reaction dependent.

Example:

1 M sulfuric acid (H2SO4) is 2 N for acid-base reactions because each mole of surfuric

acid provides 2 moles of H+ ions. On the other hand, 1 M sulfuric acid is 1 N for

sulfate precipitation, since 1 mole of sulfuric acid provides 1 mole of sulfate ions.

Dilutions

You dilute a solution whenever you add solvent to a solution. Adding solvent results

in a solution of lower concentration. You can calculate the concentration of a solution

following a dilution by applying this equation:

M1V1 = M2V2

where M is molarity, V is volume, and the subscripts 1 and 2 refer to the initial and

final values.

Example:

How many millilitres of 5.5 M NaOH are needed to prepare 300 mL of 1.2 M NaOH?

Solution:

5.5 M x V1 = 1.2 M x 0.3 L





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V1 = 1.2 M x 0.3 L / 5.5 M

V1 = 0.065 L

V1 = 65 mL

So, to prepare the 1.2 M NaOH solution, you pour 65 mL of 5.5 M NaOH into your

container and add water to get 300 mL final volume.

Factors affecting solubility

1) Effect of temperature

a) Solid solute

b) Gas solute

2) Effect of pressure

1) Effect of Temperature on Solubility (solid solute)

Solubility refers to the ability for a given substance, the solute, to dissolve in a solvent

at a specific temperature. Generally, an increase in the temperature of the solution

increases the solubility of a solid solute. A few solid solutes, however, are less

soluble in warmer solutions. The chart shows solubility curves for some typical

inorganic salts (all solids). Many salts behave like barium nitrate and disodium

hydrogen arsenate, and show a large increase in solubility with temperature.

The solubility increases with temperature





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Some solutes (e.g. NaCl in water) are fairly independent of temperature. A few, such

as cerium(III) sulphate, become less soluble in hot water.

There is no direct relationship between the value of ∆Hsol (if it is positive or negative)

and the variation of solubility with temperature. For example, the solution process of

CaCl2 is exothermic and that for NH3NO3 is endothermic, but the solubility of both of

them has increased by increasing the temperature. However, experimental work is the

best way to determine the effect of temperature on solubility that because the

solubility is depends on the natural of the solute and solvent, for example; only 1

gram of lead (II) chloride can be dissolved in 100 grams of water at room

temperature, while 200 grams of zinc chloride can be dissolved at the same

temperature.

b) Effect of Temperature on Solubility (Gas solute)

As the temperature increases, the solubility of a gas decreases as shown by the

downward trend in the graph.

More gas is present in a solution

with a lower temperature compared

to a solution with a higher

temperature. The reason for this gas

solubility relationship with

temperature is because the increase

in temperature causes an increase in

kinetic energy. The higher kinetic

energy causes more motion in

molecules which break

intermolecular bonds and escape from solution.

This gas solubility relationship can be remembered if you think about what happens to

a bottle of Pepsi as it stands around for awhile at room temperature. The taste of the

Pepsi becomes bad since more of the "tangy" carbon dioxide bubbles have escaped.

Boiled water also tastes "flat" because all of the oxygen gas has been removed by

heating.





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2) Effect of Pressure on Solubility

Liquids and solids solutes exhibit practically no change of solubility with changes in

pressure. Gases as might be expected, increase in solubility with an increase in

pressure. Henry's Law states that: The solubility of a gas in a liquid is directly

proportional to the pressure of that gas above the surface of the solution.

C α P

C = ĸP

C = molar concentration of a gas in a liquid

P = pressure of the gas over the solution in atmospheres

K = Constant that only depends on temperature. Its unit is mol/L.atm

If the pressure is increased, the gas molecules are "forced" into the solution since this

will best relieve the pressure that has been applied. The number of gas molecules is

decreased. The number of gas molecules dissolved in solution has increased as shown

in the graphic on the left. Carbonated beverages provide the best example of this

phenomena. All carbonated beverages are bottled under pressure to increase the

carbon dioxide dissolved in solution. When the bottle is opened, the pressure above

the solution decreases. As a result, the solution effervesces and some of the carbon

dioxide bubbles off.





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What is Colligative Properties of Solutions?

Solutions have different properties than either the solutes or the solvent used to make

the solution. Those properties can be divided into two main groups’ colligative and

non-colligative properties. Colligative properties depend only on the number of

dissolved particles in solution and not on their identity. Non-colligative properties

depend on the identity of the dissolved species and the solvent.

A property that depends only on the amount of solute in a solution and not the identity

of the solute is called Colligative properties

Colligative properties are

1. Boiling point elevation,

2. Freezing point lowering, and

3. Osmotic pressure.

4. Vapour pressure lowering,

Kf and Kb Kf and Kb are the freezing point depression constant and boiling point elevation

constant respectively. When a solute is added to a solvent, the boiling point of the

solution is always greater than the boiling point of the pure solvent. Adding a solute

also lowers the freezing point.

Solvent Formula Melting Point (oC) Boiling Point(oC) Kf(oC/m) Kb(oC/m)

Benzene C6H6 5.455 80.2 5.065 2.61

Ethanol C2H5OH -- 78.3 -- 1.07

Water H20 0.000 100.000 1.858 0.521

To determine the amount of change in boiling point you will need this equation:

Tb = Kb * m

• Tb change in boiling point





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• Kb boiling point elevation constant

• m molality of solution

Then Tb is added to the normal boiling point of the pure solvent. Note that the

identity of the solute is not important, just its concentration (expressed in molality).

Therefore, boiling point elevation is a colligative property.

To determine the freezing point of a solution, you need to calculate the decrease in

freezing point caused by the addition of a solute to the solvent. Use the equation:

Tf = Kf * m

• Tf change in freezing point

• Kf freezing point depression constant

• m molality of solution

Then the Tf is subtracted from the normal freezing point of the pure solvent.

Freezing point depression is also a colligative property.

Example:

Calculate the boiling point and freezing point of a solution of .30 g of glycerol

(C3H8O3) in 20.0 g of water.

moles glycerol = (.30 g) (1 mole / 92 g) = .0033 moles

molality of solution = .0033 moles / .020 kg = .16 m

Tb = (.521 oC/m)(.16 m) = .083 oC

Boiling point = 100.00 + .083 = 100.083 oC

Tf = (1.858 oC/m)(.16m) = .30 oC

Freezing point = 0.00 oC - .30 oC = -.30 oC

In an ionic solution, the total concentration of ions is important. Therefore, another

factor (i) is included in the equations.





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Tb = Kb * m * i

Tf = Kf * m * i

"i" is the number of ions from each formula unit. In the previous example, if NaCl had

been the solute your change in boiling point and freezing point would have needed to

be multiplied by 2. (i=2 because NaCl consists of a Na+ ion and a Cl- ion

To explain the difference between the two sets of solution properties, we will compare

the properties of a 1.0 M aqueous sugar solution to a 0.5 M solution of table salt

(NaCl) in water. Despite the concentration of sodium chloride being half of the

sucrose concentration, both solutions have precisely the same number of dissolved

particles because each sodium chloride unit creates two particles upon dissolution a

sodium ion, Na+, and a chloride ion, Cl-. Therefore, any difference in the properties of

those two solutions is due to a non-colligative property. Both solutions have the same

freezing point, boiling point, vapour pressure, and osmotic pressure because those

colligative properties of a solution only depend on the number of dissolved particles.

The taste of the two solutions, however, is markedly different. The sugar solution is

sweet and the salt solution tastes salty. Therefore, the taste of the solution is not a

colligative property. Another non-colligative property is the colour of a solution. A

0.5 M solution of CuSO4 is bright blue in contrast to the colourless salt and sugar

solutions. Other non-colligative properties include viscosity, surface tension, and

solubility.





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Making solutions

1) Pour in a small amount of solvent

2) Then add the solute and dissolve it

3) Then fill to final volume.

M x L = moles

Example1: How many moles of NaCl are needed to make 6.0 L of a 0.75 M NaCl

solution?

Example 2: How many grams of CaCl2 are needed to make 625 mL of a 2.0 M

solution?

Example 3: 10.3 g of NaCl are dissolved in a small amount of water then diluted to

250 mL. What is the concentration?

Example 4: How many grams of sugar are needed to make 125 mL of a 0.50 M

C6H12O6 solution?





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Dilution

Dilution is the addition of water to a solution or (solvents) that depends on the type of

solution you are making.

The number of moles of solute doesn’t change if you add more solvent.

The moles before = the moles after

M1 x V1 = M2 x V2

M1 and V1 are the starting concentration and volume.

M2 and V2 are the starting concentration and volume.

Stock solutions are pre-made to known M

Practice

2.0 L of a 0.88 M solution are diluted to 3.8 L. What is the new molarity?

You have 150 mL of 6.0 M HCl. What volume of 1.3 M HCl can you make?

Need 450 mL of 0.15 M NaOH. All you have available is a 2.0 M stock solution of

NaOH. How do you make the required solution?




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