Unit #3 : Solutions and Solubility

Time: 22 hours

Time Title and Expectations

2.5 h A Backgrounder: Characteristics of Solutions

demonstrate an understanding of the properties of solutions, the concept of concentration and the importance of water as a solvent

explain solution formation that involves the dissolving of ionic or non-ionic substances in water (eg. oxygen in water, salt in water)and the dissolving of non-polar solutes in non-polar solvents (eg. grease in gasoline)

describe the dependence on temperature of solubility in water for solids, liquids and gases use appropriate scientific vocabulary to communicate ideas related to aqueous solutions (eg.

concentration, solubility, conjugate acid, precipitate) supply examples from everyday life of solutions involving all 3 states (eg. carbonated water,

seawater, alloys, air) describe examples of solutions for which the concentration must be known and exact (eg.

intravenous solutions, drinking water) explain the origins of pollutants in natural waters (eg. landfill, leachates, agricultural run-

off) and identify the allowable concentrations of metallic and organic pollutants in drinking water.

Explain hardness of water, its consequences (eg. pipe scaling) and water-softening methods (eg. ion exchange resins)

5.0 h Qualitative Analysis: Theory Put to Use

carry out experiments and other lab procedures involving solutions, and solve quantitative problems involving solutions.

describe common combinations of aqueous solutions that result in the formation of precipitates.

determine through experiments: qualitative and quantitative properties of solutions (eg. perform a qualitative analysis of ions in a solution, plot solubility curves for some common solutes in water) and solve problems based on such experiments

represent precipitation reactions by their net ionic equations write balanced chemical equations for reactions involving acids and bases (eg. dissociation,

displacement and neutralization reactions)

5.0h Concentration

solve problems involving concentration of solutions and express the results in various units (eg. moles per litre, grams per 100 mL, parts per million [billion], mass or volume percent)

prepare solutions of required concentration by dissolving a solid solute or diluting a concentrated solution.

Solve stoichiometry problems involving solutions

6.0 h Acids and Bases

demonstrate an understanding of the Arrhenius and Bronsted-Lowry theories of acids and bases

explain qualitatively, in terms of degree of dissociation, the difference between strong and weak acids and bases

demonstrate an understanding of the operational definition of pH (ie. pH = -log 10 [H+] )

determine through experimentation the effect of dilution on the pH of an acid or a base use a titration procedure to determine the concentration of an acid or base in solution (eg.

acetic acid in vinegar)

2.5 h Drinking Water in Ontario

demonstrate an understanding of the properties of solutions, the concept of concentration and the importance of water as a solvent

relate a scientific knowledge of solutions and solubility to everyday applications and explain how environmental water quality depends on the concentrations of a variety of dissolved substances

describe the technology and the major steps involved in the purification of drinking water and the treatment of waste water.

1.0 h End of Unit Task

Activities

1. The nature of the solute and solvent and to determine whether a solution will form.(eg. solubility of different substances in water alcohol and kerosene) “like dissolves like”

2. Temperature affects solubility of a substance “Solubility Curve”3. Qualitative Precipitation4. Titration5. pH dilution effects on a strong acid and a weak acid

Chapter 8

Types of Solutions

Sec 8.1

Solution: is a homogeneous mixture uniform throughout.

Solvent: is any substance that has other substances dissolved in it – usually present in larger quantity

Solute: present in smaller quantity and is dissolved by the solvent

Solutions have variable composition:

Dilute solutions: small amount of solute to solvent

Concentrated solutions: large amount of solute to solvent

Types of Solutions

A solution can be a gas, a liquid, or a solid.Examine fig 8.2 p.285

The most numerous and versatile solutions are those in which water is the solvent. Water can dissolve many substances, forming many unique solutions. All aqueous solutions have water as the solvent and are clear (transparent). They may be either coloured or colourless.

Miscible liquids: liquids that dissolve in each other eg. ethanol and water.

Immiscible liquids: liquids that do not readily dissolve in each other eg. oil and water.

Alloys: solid solutions of metals eg. bronze is an alloy of copper and (10%) tin.

Properties of Aqueous Solutions

Compounds can be classified as electrolytes or non-electrolytes using a conductivity tester. Electrolytes are mostly highly soluble, ionic compounds including bases and acids. Non-electrolytes are usually the molecular compounds.

Another way of classifying solutions uses litmus paper as a test to classify acids, bases or neutral substances (eg. most ionic and molecular compounds).

Solubility and Saturation

The ability of a solvent to dissolve a solute depends on the forces of attraction between the particles. There is always some attraction between solvent and solute particles, so some solute always dissolves.

The solubility of a solute is the amount of solute that dissolves in a given quantity of solvent, at a certain temperature. Usually has units of g/100 mL .

Saturated Solution: is formed when no more solute will dissolve in a solution, and excess solute is present.

Unsaturated Solution: is a solution that is not yet saturated. Therefore, it can dissolve more solute.

Supersaturated Solution: is a solution that has more dissolved solute than a saturated solution.

Demo: Solubility

Purpose: to prepare and observe the properties of unsaturated, saturated and supersaturated solutions of sodium acetate.

Materials: distilled water bunsen burnerTest tube sodium acetate

Procedure:1. Clean test tube2. Add 2 mL of water into t.t.

3. Add 1 crystal of solute. 4. Stir with glass rod and observe.5. Define the type of solution made.6. Add 1 more crystal to confirm theory.7. Add enough crystals to ½ fill t.t and

shake. Define type of solution.8. Heat mixture slowly wave through

bunsen burner until solid dissolves.9. Allow mixture to cool to room

temperature.10. “Seed” the mixture by adding a crystal.

Observe and note any temperature change. Why do you think that the crystal added to the solution in the final step of the lab is called a “seed” crystal?

Application question:1. Solutions are not restricted to solids

dissolved in water. Air is a solution of different gases dissolved in each other and it behaves in the same fashion as liquid solutions. For example: on a cool still morning the air is sometimes supersaturated with a certain gas. A passing vehicle may disturb a type of air

and cause either frost to form on the vehicle or produce a trail of fog directly behind the vehicle. A) name the substance that ir recrystallizing from the supersaturated air solution B) What is equivalent to a seed crystal in this example. C) Why do you think this phenomenum doesn’t occur at a warmer temperature?

Identifying Suitable Solvents

Read p.287Perform “Thought Lab” p.288

Section Review p.289 # 1-13

Factors That Affect Rate of Dissolving and Solubility

Sec 8.2

Temperature: at higher temperatures, most solid solutes dissolve at a greater rate, because the solvent molecules have greater kinetic energy. Thus, they collide with the undissolved solid molecules more frequently.

Agitation: agitating a mixture by stirring or shaking the container increases the rate of dissolving. Agitation brings fresh solvent into contact with undissolved solid.

Surface Area: decreasing the size of the particles increases the rate of dissolving.Breaking up a large mass into many smaller masses, you increase the surface area that is in contact with the solvent. This allows the solid to dissolve faster.

Water –“ The Universal Solvent”

75% of earth is covered in water a life supporting fluid for many forms of

life water dissolves so many more substances

than any other liquid that it is often referred to as the “universal solvent”

this success as a solvent is due to 3 features of water molecules

1. their small size2. their highly polar nature3. considerable capacity for

hydrogen bonding

Solubility and Particle Attraction

Reasons why a solute may or may not dissolve in a solvent are related to the forces of attraction between the solute and solvent particles.

Process of Dissolving At the Molecular Level

Read p. 291

Solubility and Intermolecular Forces

Recall: Intermolecular Forces1. Dipole-Dipole Attractions2. Hydrogen Bonding3. Van der Waal forces or London

forces

Read p.292 – 293

Investigation Exercise

Question: Which of the substances listed have high solubility in water? Which have low solubility?

Prediction:

1. Use the concepts of polarity and hydrogen bonding to predict the relative solubility in water for each substance listed under

Materials. You may use general terms such as low or high, or simply rank the solubilities. Provide your reasoning for you prediction.

Materials:

Acetic acid, HC2H3O2 (l)

Carbon dioxide CO2 (g)

Dimethyl ether, CH3OCH3 (g)

Methanol, CH3OH (l)

Oxygen, O2 (g)

Propane, C3H8 (g)

Tetrachloromethane, CCl4 (l)

Water

Evidence:

Table 1: Solubility of Molecular Compounds in Water

Chemical Solubility

Acetic acidCarbon dioxideDimethyl etherMethanol OxygenPropaneTetrachloromethane

Evaluation:1. What additional observations would be

useful to improve the quality of the evidence?

Ionic Compounds in Water

Examine fig 8.10 p.293

Ionic crystals dissolve due to the ionic-dipole attractions (attractive forces between an ion and a polar molecule).

When a compound such as NaCl dissolves, it dissociates into individual aqueous ions.

Eg. NaCl(s) Na+ (aq) + Cl- (aq)

Factors That Affect Solubility

Molecule Sizep.295Smaller molecules are often more soluble than larger molecules. Example: methanol and ethanol are soluble in water due to hydrogen bonding but pentanol (CH3CH2CH2CH2CH2OH) is less polar overall less soluble.

Temperature

When a solid dissolves in a liquid, energy is needed to break the strong bonds between particles in the solid. At higher temperatures, more energy is present. solubility of most solids increases with temperature.

Solubility of most liquids is not greatly affected by temperature because the bonds between particles in a liquid are not as strong as the bonds between particles in a solid.

Gas particles move quickly and have a great deal of kinetic energy. When a gas dissolves in a liquid, it loses some of this energy. At higher temp, the dissolved gas gains energy again coming out of solution and is less soluble. Thus, the solubility of gases decreases with higher temp.

Experiment p.296-297Plotting Solubility Curves

Express Lab p.298

Pressure and Solubility

Pressure changes have hardly any affect on solids and liquid solutions. The solubility of the gas is directly proportional to the pressure of the gas above the liquid.

Section Review: p.301 #1-8

Concentration of Solutions

Sec 8.3

Concentration: the amount of solute per quantity of solvent. It is expressed by the ratioconcentration = quantity of solute

quantity of solution

Measurements of Concentration

1. Percentage Concentration or concentration as a Mass/Volume Percent:

A mass/volume percent gives the mass of solute dissolved in a volume of solution, expressed as a percent. The mass/volume percent is also referred to as the percent (m/v).

Ex. Intravenous fluid: is usually a saline solution that contains 0.9 g NaCl dissolved in 100 mL of solution, or 0.9% (m/v).

Do Practice Problems p.305 # 1-4

1. a) (m/v) percent = 14.2 g x 100% 450 mL

(m/v) percent = 3.16%

b) (m/v) percent = 31.5 g x 100% 1800 mL

= 1.75%

c) percent (m/v) = 1.72 g x 100% 60 mL

= 2.9%2. percent (m/v) = 1.52 g x 100%

24.1 mL

= 6.31%

3.mass = percent (m/v) x volume (mL) = (0.00145 ) x (250 mL) = .362 g

4.mass NaCl = .0086 x 350 mL

= 3.0 g NaCl

mass KCl = .0003 x 350 mL = .1 g KCl

mass CaCl2 = .00033 x 350 mL = .12 g CaCl2

Concentration as a Mass/Mass Percent

referred to as percent (m/m)

Mass/mass = Mass of solute (g) x 100%Percent Mass of solution (g)Do Practice Problems p.308 #5 – 95. percent (m/m) = 17 g x 100%

65 g= 26%

b) percent (m/m) = 18.37 g x 100% 110.57 g

= 16.6%

c) percent (m/m) = 12.9 g x 100% 85.4 g= 15.1%

6. percent (m/m) = 55 g x 100% 155 g

= 35%

7.mass of C = percent (m/m) x mass of solution

= .017 x 5000 g= 85 g of Carbon

8.mass Cr = .105 x 60.5 g = 6.35 g

9.mass Au = .75 x 20 g= 15 g gold

Concentration as a Volume/Volume Percent

also referred to as percent (v/v)

Volume/Volume =Volume solute (mL)x 100%Percent Volume solution (mL)

Ex) Vinegar (acetic acid): the label is listed as 5% acetic acid (by volume). This means that there are 5 mL of pure acetic acid dissolved in 100 mL of the vinegar solution.

Do Practice Problems p.310 # 10 – 14

Concentration in Parts per Million and Parts per Billion

parts per million (ppm) parts per billion (ppb) usually mass/mass relationships

ppm = Mass of solute x 106

Mass of solution

ppb = Mass of solute x 109

Mass of solution

Practice Problems p.312 #15 – 18

15. a) ppm = .020 g x 106

60000 g

= .33 ppm

Molar Concentration

is the number of moles of solute that can dissolve in 1.0 L of solution

also known as molarity

Molar = Amount of solute (mol)Concentration Volume of solution (L)(mol/L)

C = n V

Practice Problems: p.316 # 19 – 24

Investigation 8-B p.317 Determining the Concentration of a Solution

Solution Preparation

Sec 8.4

solutions with precisely known concentrations are called standard solutions

to prepare a standard solution precision equipment such as a volumetric flask is required

read p.319 for technique

Preparation of Standard Solutions by Dilution

making a less concentrated solution of a known solution (stock solution) by adding a measured amount of additional solvent to it.

Ci Vi = Cf Vf

i = initialf – final

Practice Problems p.321 # 25 – 27

Investigation 8-CEstimating Concentration of an Unknown Solution p.322 –323

Section Review p.324 #1-4

Chapter 8 Review Questions p. 325-327

Chapter 9

Aqueous Solutions

Sec 9.1

Making Predictions About Solubility

The term soluble means that > 1 g solute will dissolve in 100 mL of water at room temperature.

Insoluble means that solubility is < 0.1 g per 100 mL.

Factors Affecting the Solubility of Ionic Substances

1. Effect of Ion Charge on Solubility: compounds with larger ions tends to be insoluble. WHY?

charge force that holds the ions together

2. The Effect of Ion Size on Solubility: bond between small ions is stronger than the bond between large ions with the same charge

solubility with smaller ions

ex) F- , Cl- , Br- , I- (solubility )

Predicting solubility will be done by using the solubility charts on p. 334 Table 9.1 or p.666 table F5.

Practice Problems p.335 #1 – 3

Section Review p.336 #1 - 6

Reactions in Aqueous Solutions

Sec 9.2

Review:

AB + CD CB + AD

Double Displacement Reaction

Possible products:1. precipitate2. gas3. water (neutralization)

Practice Problem p.339 # 4

Important!!!!!!

p.340

Formation of H2S and NH3 will always be a gas at room temperature.

Remember:

H2SO3 (aq) SO2 (g) + H2O (l)

H2CO3 (aq) CO2 (g) + H2O (l)

Net Ionic Equations

Terms:

Spectator Ions: ions in solution that are not important; these ions are like passive onlookers.

Total Ionic Equation: ionic compounds that dissociate into their respective ions in solution.

Net Ionic Equation: an ionic equation that is written without showing the spectator ions.

Guidelines for Writing a Net Ionic Equation:p.342 Table 9.2Practice Problems p.343 #5 – 6

Investigation 9-BQualitative Analysis p.345

Stoichiometry in Solution Chemistry

Sec 9.3

Practice Problems p.352 #7 – 10

7.Ca(CH3COO)2 Ca2+(aq) + 2CH3COO-(aq)

C 0.500 mol/L n 0.300 molV ?

moles of acetate = ½ mol calciumacetate

mol calcium = 0.150 molacetate

C = mol Vol

Vol = 0.150 mol0.500 mol/L

V = 0.300 LV = 300 mL

8.(NH4)3PO4 (s) 3 NH4 + (aq) + PO4 3- (aq)

mass 6.0 gvol 300 mLmol ?C ? ?

[NH4+] = 6.0 g x mol x 3 mol NH4 + x 1

149 g 1 mol (NH4)3PO4 .300 L

= 0.40 mol/L

[PO4 3-] = 6.0 g x mol x 1 mol PO4 3- x 1 149 g 1 mol (NH4)3PO4 .300 L

= 0.13 mol/L

Chapter 10

Acids and Bases

Properties of Acids & Bases p.372

Acid & Base Theories

The Arrhenius Theory of Acids & Bases:

Acid: an acid is a substance that dissociates in water to produce one or more hydrogen ions, H+

Eg) HCl(aq) H+ (aq) + Cl- (aq)

The hydronium ion (H+) is a hydrated proton.

The reaction occurs in water it looks like

HCl (aq) + H2O (l) H3O+ (aq) + Cl- (aq)HydroniumIon

Base: a base is a substance that dissociates in water to form one or more hydroxide ions. (OH-)

Eg) Ba(OH)2 (s) Ba 2+ (aq) + 2OH- (aq) Hydroxide Ion

Arrhenius theory is limited to acid-base reactions in water.

what if an acid-base reaction takes place in other solvents, other than water?

The Bronsted-Lowry Theory of Acids & Bases

An acid is a substance from which a proton (H+ ion) can be removed.

A base is a substance that can remove (H+ ion) from an acid.

According to the Bronsted-Lowry theory, there is only one requirement for an acid-base reaction. One substance must provide a proton, and another substance must receive the same proton.

an acid is a proton donor and a base is a proton acceptor.

Eg)

HCl(aq) + H2O (l) H30+ (aq) + Cl- (aq)

A1 B2 A2 B1

Acid Base conjugate conjugate Acid base

Two molecules or ions that are related by the transfer of a proton are called a conjugate acid-base pair.

Conjugate meaning “linked together”

The conjugate base of an acid is the particle that remains when a proton is removed from the acid.

The conjugate acid of a base is the particle that results when the base receives the proton from the acid.

Practice Problems p.378 #1-3

Section Wrap up p.379Table 10.3

Section Review p.379 #1-10

Strong and Weak Acids and Bases

Section 10.2

Strong Acid:an acid that dissociates completely (100%) into ions in water.Eg) HCl

HCl (aq) + H2O (l) H3O + (aq) + Cl- (aq)

1M 1M 1M

Weak Acid: is an acid that dissociates very slightly in a water solution.Ex) acetic acid

CH3COOH (aq) + H2O (l) CH3COO- (aq) + H3O + (aq)

<5% dissociation

Most acids are weak acids. Some contain only a single hydrogen ion that can dissociate. These acids are called monoprotic acids.

Acids that contain 2 or more hydrogen ions that can dissociate are called diprotic or polyprotic acids.Ex) H2SO4

H2SO4 (aq) H+ (aq) + HSO4 – (aq)

HSO4 – (aq) H+ (aq) + SO4 2- (aq)

**Whenever you see a reversible () chemical equation involving an acid, you can safely assume that the acid is weak.!!

Strong base: a base that dissociates into hydroxide ions (OH-) in water.

Ex) all oxides and hydroxides of the alkaline earth metals –Group 2 (IIA)- below beryllium are strong bases.

Weak base: is a base that dissociates very slightly in a water solution.

Ex) ammonia NH3

NH3 (aq) + H2O (l) NH4 + (aq) + OH- (aq)

Practice Problems p.384 #4,5

pH of a Solution

H2O (l) + H2O (l) H3O + (aq) + OH – (aq)

Self dissociation of water @ 25 C

Scientists determined that:

[H3O +] = 1.0 x 10 –7 mol/L

[OH-] = 1.0 x 10 –7 mol/L

The pH scale: Measuring by Powers of Ten

pH : the “p” stands for the word “power” – which is referred to as exponential power; the power of 10. the “H” stands for the concentration of

hydrogen ions (or H3O + ions) measured in mol/L.

in 1909, a Danish biochemist, Soren Sorensen, defined pH = -log[H+]

his definition based on Arrhenius’ hydrogen ion.

[H3O+] ranges from 10 mol/L for a concentrated strong acid to about 10-15 mol/L for a concentrated strong base.

Examine Table 10.7 p.387 “Understanding pH”

Examine Concept Organizer p.388

Practice Problems p.389 #6-9

6. a) pH = -log[H3O+]

given: [H3O+] = 0.0027 mol/L

pH = -log(0.0027)pH = 2.57

b) pH = 7.138c) pH = 4.01d) pH = 11.082

7. [H3O+] = 5.0 x 10 –3 mol/LpH = 2.30acidic

8. [H3O+] = 2.9 x 10 –4 mol/LpH = 3.54acidic

9. a) [H3O+] = 6.3 x 10 –3 mol/LpH = 2.20

b_ [H3O+] = 6.59 x 10 10 mol/LpH = 10.82

Section Review p.393 #1 – 10

Acid-Base Reactions

Section 10.3

Neutralization Reactions

Acid + Base salt + water

for most neutralization reactions, there are no visible signs that a reaction is occurring.

How can you determine that a neutralization rxn is taking place?

Using an acid-base indicator This is a substance that changes color in

acidic and basic solutions Most acid-base indicators are weak,

monoprotic acids The undissociated weak acid is one color

its conjugate base is a different color.

H (indicator) (aq) H+ + (indicator)- (aq)

Color 1 color 2

Ex) phenolphthalein is an indicator that is colorless between pH 0 and pH 8, but turns pink between pH 8 and pH 10.

Calculations Involving Neutralization Reactions

CAVA = CBVB

A = acidB = base

Practice Problems p.398 # 10 – 13

Acid-Base Titrations

Titration is a common laboratory technique used to determine the concentration of substances in solution. The concentration of the unknown solution is determined by quantitatively observing its reaction with a solution of known concentration.

Aim: to find the point at which the # moles of the standard solution is stoichiometrically equal to the original number of moles of the unknown solution.

This is referred to as the “equivalence point”.

At equivalence point;

# moles H+ ions in original solution have reacted with an equal # moles of OH- ions from the other solution.

Apparatus needed:

1. sample to be analyzed is transferred to an erlenmeyer flask.

2. Known concentration of solution (called the titrant) placed into a buret, is added frop by drop to the sample.

3. Indicator added to sample in erlenmeyer flask, indicates the endpoint of the reaction by changing color. The indicator is chosen so that it matches its equivalence point.

# moles H+ = # moles OH-

p.399-400 Titration Step by Step

p.402 – 403 The Concentration of Acetic Acid in Vinegar. (Manufacture claims on the label that the vinegar contains 5.0 % acetic acid which translates into 0.87 mol/L concentration of acetic acid.)

Section Review p.404 #1 – 7Chapter 10 Review p.405-407

Assignment #2Island at Risk: A Simulation

Answer Questions:1. Decide what you think the best solution is, for everyone involved. Why do you

consider it to be the best solution? (5 marks)

2. From what you learned in Unit 3 about solutions and water treatment, do you think there could be other chemical based solutions to this problem in the future? Give a chemical explanation of how the leachates might be contained, if such processing were possible. (5 marks)

.3. Try to suggest some safe, alternative methods of reducing weeds, insects and

harmful microorganisms. For example, could boiling water be used as an effective herbicide? Would this method of reducing weeds cause more problems than it would solve? Why or why not? (5 marks)

PEER-PEER EVALUATION:

Group Name: _____________________________Students Name: ____________________________

1. Did the group as a whole work cooperatively?

2. Did everyone contribute information to the project?

Written Report Evaluation:

Group Name:________________________

CONTENT 0 1 2 3 4 5 6 7 8 9 __________20

how much research went into the report? How thoroughly did you know your topic? Was anything of real importance left out which should reasonably have been

covered? scientific terms defined and explained using scientific terminology references included and properly used and listed points of view backed up scientifically

ORGANIZATION 0 1 2 3 4 5 ________8

was the topic put together well? Was the report on the topic? Was there a useful introduction and conclusion? Was the topic easy to follow

Total = / 28 marks

Total Evaluation:

Evaluation of Oral Presentation: / 35 marks

Questions: / 15 marks

Report Evaluation : / 28 marks

TOTAL OVERALL MARK: / 78 marks