Unit 10: SolutionsChapter 15-16

This tutorial is designed to help students understand scientific measurements.

Objectives for this unit appear on the next slide.◦Each objective is linked to its description.◦Select the number at the front of the slide to

go directly to its description.Throughout the tutorial, key words will

be defined.◦Select the word to see its definition.

Objectives18 Define solubility including the terms soluble/insoluble and

miscible/immiscible; unsaturated, saturated, and supersaturated; solute, solvent, and aqueous

19 Define conductivity including how electrolytes and dissociation affect conductivity

20 Describe heterogeneous mixtures including suspensions, colloids, and emulsions

21 Explain how mixtures can be separated by six methods 22 Explain saturation points regarding liquids and gases and read a

solubility curve chart 23 State Henry’s Law and give examples of the law 24 Define concentration in terms of molarity, molality, and solve

problems 25 Solve dilution problems involving molarity 26 Identify and define colligative properties including calculations

for boiling point elevation and freezing point depression

18 Defining a Solution Two chemicals can often be mixed together

to form a solution. ◦In this case, the chemical that has the larger

quantity is known as a solvent. If water is the solvent, the solution is known as

aqueous.◦The chemical that has the smaller quantity is

known as a solute. ◦ It is stated that a solute will dissolve in the

solvent This is not limited to a solid dissolved into a liquid. Gases can be dissolved in liquids (pop). Liquids can be dissolved in liquids (gasoline). Gases can be dissolved in gases (air in the room). Etc.

Defining a solutionWhen identifying solutes, they

are often labeled as:◦Soluble: will dissolve in a certain

solvent◦Insoluble: will not dissolve in a

certain solventIf it is a liquid in a liquid, the

solutes are labeled as:◦Miscible: will dissolve in a certain

liquid◦Immiscible: will not dissolve in a

certain liquid

Defining a solutionWhen a solution is made, the

solute dissolves in the solvent.The amount of solute can vary.

◦Solvents can only hold so much solute.

◦Once too much is added, the remaining solute sinks to the bottom.

The terms saturated, unsaturated, and supersaturated explain this situation.

UnsaturatedA solution in which more solute

can be added.

SaturatedA solution in which the solution is

holding as much solute as possible with any extra settling to the bottom.

SupersaturatedA unique case where a solution

has more solute dissolved than in should be allowed to.

19 Conductivity The conductivity of a solution is

dependent on the presence of electrolytes.◦An electrolyte is an ion produced

when a solid dissociates into a solvent.

◦If electrolytes are present, the solution will conduct electricity.

DissociationIons are found in solution when a

compound dissociates during the dissolving process.

Each compound will break into a distinct number of parts.◦Covalent molecules remain as one.◦Ionic molecules break into their ions.

DissociationWhen an ionic compound is

placed in water, the partially positive hydrogens of the water molecule will attract the anion.

The partially negative oxygen of the water molecule will attract the cation.

These attractions will pull apart the compound thus separating it into ions.

Dissociation

Dissociation FactorsEach compound will break into

certain numbers of parts when they dissociate.◦This is the dissociation factor.

For covalent molecules, the molecule will not break apart so the dissociation factor is always 1.

For ionic compounds, the compound will break into its ions so it is determined by adding the ions together.

Dissociation FactorsConsider NaCl

◦There is one Na+ and one Cl-.◦Therefore, the dissociation factor is

2.Here are some more examples:

Molecule/Compound

Dissociation Factor

Reason

CH4 1 CovalentCaCl2 3 Ionic

Ca(NO3)2 3 Ionic (polyatomic ions remain together)

20 Heterogeneous MixturesNot all mixtures will make

solutions.Some are not uniform and are

called heterogeneous mixtures.Two common varieties of

heterogeneous mixtures are:◦Suspensions: will separate if not

agitated Surfactants and micelles will make up

suspensions These describe soaps and detergents.

◦Colloids: will not separate

21 SeparationsWith the different homogenous

(solutions) and heterogeneous mixtures that exist, it is important to be able to separate the parts.

There are six main ways to separate mixtures:◦Decanting◦Filtration◦Evaporation◦Centrifuge◦Distillation◦Chromatography

DecantingDecanting is a

separation technique where the solid is allowed to settle to the bottom.

The liquid is then poured off leaving two parts.

Image from: http://www.sciencequiz.net/jcscience/jcchemistry/practicals/decanting.htm

FiltrationFiltration occurs by

pouring a mixture through a porous paper.

The larger particles are caught in/above the paper.

The smaller particles pass through to the collection container.

Image from: http://en.wikipedia.org/wiki/File:Vacuum-filtration-diagram.png

EvaporationEvaporation is the

removal of the liquid from a solution.

The liquid is heated past its boiling point driving it away.

The solute then sinks to the bottom.Image from: http://www.school-for-champions.com/science/evaporation.htm

CentrifugeThe centrifuge is an

instrument in which the mixture is separated based on its densities.

The sample is spun rapidly forcing the more dense portion to the bottom of the tube.

After using a centrifuge, the sample is generally decanted.

Image from: http://www.daviddarling.info/encyclopedia/C/centrifuge.html

DistillationDistillation is used when two

or more liquids are present with distinct boiling points.

The mixture is heated to just above the boiling point of one liquid.

As it boils, the gas is funneled down a cooling tube causing the gas to condense.

It is then collected in an additional flask.

This allows both liquids to be kept.

Image from: http://glossary.periodni.com/glossary.php?en=distillation

Chromatography Chromatography is a widely

used technique to separate mixtures.

It is typically used for dyes and organic molecules.

The sample is added to a mobile phase which is passed over a stationary phase.

The stationary phase will interact with one part of the mixture thus slowing it down.

As this occurs, the mobile phase pulls the other part of the mixture further down the column.

Image from: http://www.waters.com/waters/nav.htm?cid=10048919&locale=en_US

ChromatographyThere are several types of

chromatography.Here are a few:

◦Gas chromatography◦High Pressure Liquid

Chromatography◦Ion Chromatography◦Size-Exclusion Chromatography◦Thin-Layer Chromatography

Separation RecapTechnique

Separates Heterogeneous Mixtures

Separates Homogeno

us Mixtures

Notes

Decanting Yes No

Filtration Yes NoUses porous paper; requires different sizes of particles

Evaporation Yes Yes Will lose the liquid portion of sample

Centrifuge Yes NoRequires the use of another technique

(typically decanting to finish the separation)

Distillation Yes YesRequires distinct

boiling points; keeps both liquids

Chromatography Not Preferred Yes Mostly used with

dyes and organics

22 SolubilitySolubility describes how much

solute a solution is allowed to hold during certain conditions.

The solubility of certain substances can be shown using a solubility curve.◦The curve shows how a solute

dissolves in a solvent given certain conditions.

Solubility CurvesThe line represents the

saturation point at each temperature.◦ For example: at 20°C, sugar is

saturated when 200 grams are added to 100 grams of water.

◦ Any point below the line is unsaturated.

◦ Any point above the line is saturated.

The curves often show more than one solute for each solvent.◦ In this case, water is the solvent.

Image taken from: http://www.btinternet.com/~chemistry.diagrams/solubility_curves.htm

23 Henry’s LawHenry’s Law explains a

phenomena that occurs when a gas is dissolved in a liquid.

The gas in the liquid is directly proportional to the partial pressure of the gas above that liquid.

If the gas above the liquid is removed, gas that is dissolved will escape to fill this space.

Henry’s Law Example - POPOne of the key components of pop is

the carbonation.The carbonation is caused by dissolving

carbon dioxide in the liquid.When the lid is removed from a bottle

of pop, the gas above the liquid is removed.◦This will cause some gas in solution to

escape and fill the space above the liquid meaning less carbon dioxide is dissolved.

◦After time, this will cause the pop to go flat.

Henry’s Law Example-POP

The pop starts with some gas dissolved and some in the space above the liquid.

Once the can is opened, gas escapes out the top of the can.

To return to the correct relationship, some gas will leave the solution to fill the space again.

24 ConcentrationSince solutions can have more or

less solute, it is important to know how much solute is dissolved.

This is measured with concentration.

Concentration is typically given in either molarity or molality.

MolarityMolarity provides a relationship

between the moles of solute and liters of solution.

Its mathematical relationship is:Molarity =

For example:Assume each red dot represents one mole.Assume there is 500 ml of water.The molarity of this solution would be:

MolalityMolarity provides a relationship

between the moles of solute and kilograms of solvent.

Its mathematical relationship is:Molality =

For example:Assume each red dot represents one mole.Assume there is 450 grams of water.The molality of this solution would be:

Solving concentration problemsAssume you 200 ml have a 5 molar

solution and you want to know how many moles this would be.◦5 molar can be written as ◦200 ml would have to be converted to

liters So 200 ml x = 0.200 liters

◦Dimensional analysis will allow moles to be calculated.

0.200 liters x = 1 mole

25 DilutionsOften, it is necessary to take a

concentrated solution and make it less concentrated.◦This is called a dilution.

To make a dilution, a portion of the concentrated amount is taken and more solvent is added.

DilutionsSuppose a 2 molar solution was desired but

only a 4 molar solution was present.◦Assume that water was the solvent.

To make the 2 molar solution, take a sample of the concentrated and double the water.

4 molar solution 2 molar solution

DilutionsThe previous problem can be calculated

mathematically as well.Assume a 500 ml sample of a 2 molar solution

was desired. If this was the case, we would need 1 mole of

solute. ◦ 0.500 liters x 1.0 moles

To get 1 mole of solute from the 4 molar solution, we would need 250 ml.◦ 1.0 moles x

Once we take the 250 ml, we would add an additional 250 ml of water to bring the solution to 500 ml.

DilutionsThe calculation used on the

previous slide can be condensed to the following:

M1V1 = M2V2

It is important to realize that the volume of the concentrated is only the volume needed. Enough solvent would have to be added to reach the desired concentration.

Desired concentration

Desired volume

Volume of concentrated required

Molarity of concentrated sample

26 Colligative PropertiesColligative properties are

physical properties that are affected by the amount of solute rather than the actual identity of that solute.

Two of the more common properties that are affected by the solute are:◦Freezing points◦Boiling points

Freezing Point DepressionThe freezing point of the solvent will decrease

based on the amount of solvent present.The change in the freezing point is calculated

using the equation:

∆T = kfdm

∆T = change in temperatureKf = freezing constant (find on the PT)d = dissociation factor m = molality

Boiling Point ElevationThe boiling point of the solvent will increase

based on the amount of solvent present.The change in the boiling point is calculated

using the equation:

∆T = kbdm

∆T = change in temperatureKb =boiling constant (look on back of PT)d = dissociation factor m =molality

Colligative PropertiesRecall that water boils at 100°C and

freezes at 0°C.If we dissolve NaCl into water (assume

2 m) the freezing a boiling point will change.

Freezing Boiling∆T = kfdm ∆T = kbdm∆T = 1.86°C/m x 2 x 2 m ∆T = 0.52°C/m x 2 x 2 m∆T = 7.44 °C ∆T = 2.08°C

Therefore, the new freezing point is -7.44 °C and the new boiling point is 102.08 °C.

This concludes the tutorial on solutions.

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