Solutions

59
Solutions Mechanisms and Phenomena, Separations, and Concentration Problems

description

Solutions. Mechanisms and Phenomena, Separations, and Concentration Problems. Solutions. How they form, What they do, Separations, and Concentration Problems. Standards. 6. Solutions are homogeneous mixtures of two or more substances. As a basis for understanding this concept: - PowerPoint PPT Presentation

Transcript of Solutions

Page 1: Solutions

Solutions

Mechanisms and Phenomena, Separations,

andConcentration Problems

Page 2: Solutions

Solutions

How they form, What they do, Separations,

andConcentration Problems

Page 3: Solutions

Standards• 6. Solutions are homogeneous mixtures of two or more

substances. As a basis for understanding this concept: • a. Students know the definitions of solute and solvent.• 6. b. Students know how to describe the dissolving process at

the molecular level by using the concept of random molecular motion.

• 6. c. Students know temperature, pressure, and surface area affect the dissolving process.

• 6. d. Students know how to calculate the concentration of a solute in terms of grams per liter, molarity, parts per million, and percent composition.

• 6. e.* Students know the relationship between the molality of a solute in a solution and the solution’s depressed freezing point or elevated boiling point.

• 6. f.* Students know how molecules in a solution are separated or purified by the methods of chromatography and distillation.

Page 4: Solutions
Page 5: Solutions
Page 6: Solutions

Examples of Chemical Change vs. Physical Change

Burning BoilingRusting FreezingBread Rising MeltingPhotosynthesis Mixing

Salt dissolving in H2O Salt precipitating from H2O

Page 7: Solutions

Chemical ChangeChemical changes occur when the atoms in

a substance change their electron configurations.

Chemical changes are often noticeable due to the appearance of a new substance or the release or absorption of energy (light, heat, electricity).

Page 8: Solutions

Reaction SymbolsSymbol Meaning(s), (l), (g) Substance is a solid, liquid, or gas(aq) Aqueous, substance is dissolved in H2O

“Produces” or “makes”“Produces” through reversible reaction

heat or Δ Heat is added to the reactants

Pt A catalyst is used to speed up the reaction

Page 9: Solutions

Definitions• Heterogeneous – a mixture with large

particles (clumps of molecules or ions). • Homogeneous – a mixture with very small

particles (individual molecules or ions).

• Solution – a homogeneous mixture of two or more substances. (ex. salt in water)Solvent – the main ingredient of a solution. (ex.

water)Solute – the other ingredients dissolved into the

solvent. (ex. salt)

Page 10: Solutions

More Definitions• Solution – a homogeneous mixture of two or more

substances. (ex. salt in water)Solvent – the main ingredient of a solution. (ex. water)Solute – the other ingredients dissolved into the

solvent. (ex. salt)• Colloid – a heterogeneous mixture of two or more

substances with medium-sized particles. Usually stable. (ex. whipping cream, mayonnaise, Jell-O)

• Suspension – a heterogeneous mixture of two or more substances with larger-sized particles. Usually unstable. (ex. clay in water, smoke in air)

Page 11: Solutions

Diffusion of Gases

Page 12: Solutions

Ionic compounds

Page 13: Solutions

Ionic compounds

Page 14: Solutions

Ionic compounds

Page 15: Solutions

Polar molecules

OHH

+ +

OH

H+

+

O

HH+

+

Page 16: Solutions

Ionic compounds

Na+ Cl–

Na+ Cl–

Na+ Cl–

Na+ Cl–

Cl–

Na+

Na+Cl–

Page 17: Solutions

NaCl Dissolved in Water

OHH

+ +

OH

H +

+

–O HH+

+

Cl–

OH H

++

Page 18: Solutions

NaCl Dissolved in Water

OH H

++

OH

H++

OH

H ++

– Na+

OHH

+ +

Page 19: Solutions

NaCl Dissolved in Water

Page 20: Solutions

NaCl Dissolved in Water

Page 21: Solutions

Sucrose Dissolved in Water

Page 22: Solutions

PrecipitatePrecipitate – A solid that comes out of the

solution.

Iron (II) hydroxide benzoic acid

Page 23: Solutions

Precipitate

lead (II) nitrate

Page 24: Solutions

Concentrations• Concentration – how much solute is dissolved

within a certain amount of solution. • g solute/L solution• ppm – parts per million (mass)• ppb – parts per billion • Molarity (M) – mol solute/L solution• Molality – mol solute/kg solvent• % composition = mass solute/mass total

Page 25: Solutions

Concentrations Examples

ppm – parts per million (mass)

Molarity (M) =

g soluteL solution

= 3.7 g NaCl1 L solution

=

mol solute L solution =

3.0 mol MgCl2

1 L solution= 3.0 M MgCl2 solution

25 g LiF1,000,000 g solution = 25 ppm (mass)

Page 26: Solutions

Concentrations Examples

Molality – mol solute/kg solvent

% composition = mass solute/mass total

Page 27: Solutions

Concentration Example #1 • If 12 moles of KCl are dissolved in a 3 L

solution with water, what is the molarity of the KCl solution?

12 mol KCl3 L solution = 4 M KCl solution

Page 28: Solutions

Concentration Example #2 • How many moles of sodium chloride (NaCl)

are dissolved in 5 L of NaCl solution with a concentration of 3M?

5 L solution1

x 3 mol NaCl1 L solution = 15 mol NaCl

OR3 mol NaCl1 L solution = x mol NaCl

5 L solution

Page 29: Solutions

Solubility• soluble – something that can dissolve into a

solution. • insoluble – something that cannot dissolve

into a solution. • saturated – no more solute can dissolve into

the solution. • unsaturated – more solute can still dissolve

into the solution.

Page 30: Solutions

Solubility• supersaturated – more than the normal

amount of solute is dissolved in the solution. A slight change may force the excess solute to precipitate out of solution.

Page 31: Solutions

Solubility Graph

Page 32: Solutions

Solubility Graph

Page 33: Solutions

Solubility Graph

Page 34: Solutions

Electrical Conductivity• Distilled water does not conduct electricity• Electrolytes – dissolved salts or acids that

allow water to conduct electricity because of positive or negative ions.

Page 35: Solutions

Diluting Solutions• While the molarity (M) and the volume (V)

may change, the number of moles (M × V) will not change.

M1 × V1 = M2 × V2 before after

Page 36: Solutions

Diluting Example #1• 5.0 liters of a 3.0 M NaCl solution is diluted so

the volume of the solution is now 30 liters. What is the molarity of the diluted solution?

M1 × V1 = M2 × V2 (3.0 M) × (5.0 L) = M2 × (30 L)

15 = 30 M2

0.5 M = M2

30 30

Page 37: Solutions

Like Dissolves Like• Polar molecules dissolve polar molecules.

Water, alcohols, vinegar• Nonpolar molecules dissolve nonpolar

molecules. Fats, oils, gasoline

• Usefulness of soap in washing off greasy hands comes from soap molecules having both polar and nonpolar ends.

Page 38: Solutions

Factors that Affect Solubility • Surface Area – Smaller particles have more

surface area than larger particles. Surface area allows the solute to dissolve faster.

Page 39: Solutions

Factors that Affect Solubility • Surface Area – Smaller particles have more

surface area than larger particles. Surface area allows the solute to dissolve faster.

Page 40: Solutions

Factors that Affect Solubility • Stirring – increases solute’s contact with low

concentration solvent. The solute dissolves faster.

• Temperature – Higher temperatures usually allow more solids to dissolve in liquids but less gases.

• Pressure – Higher pressures usually allow more gases to dissolve in liquids.

Page 41: Solutions

Gas Dissolved in Water

Page 42: Solutions

Concentration Effects(aka Colligative Properties)

• Depressed Freezing Point (anti-freeze, salting the roads and sidewalks in winter) – freezes at a lower temperature because of the solute.

• Elevated Boiling Point (anti-boil radiator fluid) – boils at a higher temperature because of the solute.

Page 43: Solutions

Separations • Decanting – carefully pouring the solution from a

container while leaving the precipitate.• Centrifugation – spinning a suspension very fast so the

precipitate settles on the bottom. • Filtration – using a filter to trap the precipitate while the

solution passes through. • Evaporation – heating and evaporating out the solvent

so only the solute remains. • Chromatography – Utilizing different substances’ varying

attraction to certain materials so one substance travels faster than the other.

• Distillation – Separating 2 liquids heating them both and having them boil out at their different boiling points.

Page 44: Solutions

Decanting • Decanting – carefully pouring the solution

from a container while leaving the precipitate.

Page 45: Solutions

Centrifugation• Centrifugation – spinning a suspension very

fast so the precipitate settles on the bottom.

Page 46: Solutions

Filtration • Filtration – using a filter to trap the precipitate

while the solution passes through.

Page 47: Solutions

Evaporation • Evaporation – heating and evaporating out

the solvent so only the solute remains.

Page 48: Solutions

Chromatography • Chromatography – Utilizing different

substances’ varying attraction to certain materials so one substance travels faster than the other.

Page 49: Solutions

Distillation • Distillation – Separating 2 liquids heating

them both and having them boil out at their different boiling points.

Page 50: Solutions

Concentration Example #3• Pure gold nuggets are very rare. Instead most

mining companies settle for rocks that contain only small amount of gold. If a 1,000 g rock contains 200 mg of gold, then what is the concentration in parts per million of gold in the rock?

200 mg Au1,000 g rock

= 0.200 g Au1,000 g rock

x

= 200 ppm Au (mass)

200 g Au1,000,000 g rock

1,0001,000

=

Page 51: Solutions

Concentration Example #3• Pure gold nuggets are very rare. Instead most

mining companies settle for rocks that contain only small amount of gold. If a 1,000 g rock contains 200 mg of gold, then what is the concentration in parts per million of gold in the rock?

200 mg Au1,000 g rock

x 1 g1,000 mg

x = 106

1,000 106

1,000

= 200 ppm Au (mass)

200 g Au1,000,000 g rock

Page 52: Solutions

Concentration Example #4• 11.1 g of CaCl2 are dissolved within a 500 mL

solution with water. What is the molarity of the CaCl2 solution?

11.1 g CaCl2

500 mL solutionx 1 mol CaCl2

111 g CaCl2

x

= 0.2 M CaCl2 solution

1,000 mL 1 L

=

Page 53: Solutions
Page 54: Solutions

Carbonated Soda

Under High pressure Under Normal Pressure

Save for Equilibrium Shifts Unit since technically there is an unstable acid compound that CO2 forms in solution.

Page 55: Solutions

NeFONCBBe

He

Li

H

Kr

ArCl

Br

XeI

SPSiMg Al

Ca

Na

K

Page 56: Solutions

NeFONCBBe

He

Li

H

Kr

ArCl

Br

XeI

SPSiMg Al

Ca

Na

K

Page 57: Solutions

Bad Example #X • Carbon dioxide (CO2) is present in a mixture of

gases at STP with a concentration of 500 ppm. If there are a 44.8 L of the gas mixture , how many moles of carbon dioxide?

44.8 L mixture 1

x

= 0.001 mol CO2

500 mol CO2

106 mol mixture1 mol mixture22.4 L mixture

x

Page 58: Solutions

4 e– in valence shell

Page 59: Solutions

Solubility Graph