SOLUTION CHEMISTRY

SOLUTION CHEMISTRY

1) mixture = a blend of two or more kinds of matter, each of which retains its own identity and properties

• can be physically separated (filtration, evaporation, decanting, magnetism, etc)

I. MIXTURES:

a.) homogeneous mixture = a mixture that is uniform in composition throughout

•Ex: Food coloring and waterb.) heterogeneous mixture = a mixture that is NOT uniform in composition throughout

•Ex: Oil and water

I. MIXTURES:

1) solution = a homogeneous mixture

A. Types of Mixtures:

2) suspension = a mixture in which the particles are so large that they settle out unless the mixture is constantly stirred or agitated

•Heterogeneous mixture•Ex: Sand and water

A. Types of Mixtures:

3) colloid = a mixture consisting of particles that are intermediate in size between those in solutions and those in suspensions

•Heterogeneous Mixture•Ex: Milk, mayonnaise, smog, butter, whipped cream

A. Types of Mixtures:

a) What property of a colloid helps to prevent colloid particles from settling out of a mixture?

• Brownian motion = the random continuous motions of colloidal particles

3) Colloids

b) Tyndall effect = visible pattern caused by the reflection of light from suspended particles in a colloid (or from suspended particles in a suspension if the particles have not settled out)

Ex: visibility of a headlight beam on a foggy night

3) Colloids

Solutions Colloids Suspensions

Homogeneous

Heterogeneous

Heterogeneous

Particle size: 0.01-1 nm;

can be atoms, ions, molecules

Particle size: 1-1000 nm,

dispersed; can be aggregates

or large molecules

Particle size: over 1000 nm,

suspended; can be large particles or aggregates

B. Comparison of solution, colloids, and suspensions

Solutions Colloids Suspensions

Do not separate on

standing

Do not separate on

standingParticles settle

outCannot be

separated by filtration

Cannot be separated by

filtration

Can be separated by

filtration

Do not scatter light

Scatter light (Tyndall effect)

May scatter light, but are not

transparent

B. Properties of solution, colloids, and suspensions

1.) If particles settle or can be filtered out = suspension2.) If particles DO NOT settle or filter out shine a beam of light (Tyndall effect) through the mixture

•If the Tyndall effect is observed = colloid•If the Tyndall effect is NOT observed = solution

C. Determining if a mixture is a true solution, a colloid, or a suspension:

II. Nature of

Solutions

1) Solvent = the substance that does the dissolving in a solution

a) Typically present in the greatest amount

b) Typically a liquidc) Water is the most common

or “universal” solvent

II. THE NATURE OF SOLUTIONS:

•B/c water molecules are polar• The hydrogen side of each water (H2O) molecule carries a slight positive electric charge, while the oxygen side carries a slight negative electric charge.

• water can dissociate ionic compounds into their positive and negative ions.

• The positive part of an ionic compound is attracted to the oxygen side of water while the negative portion of the compound is attracted to the hydrogen side of water.

• Water won't dissolve or won't dissolve well. If the attraction is high between the opposite-charged ions in a compound, then the solubility will be low.

•Ex: hydroxides exhibit low solubility in water.

•Ex: nonpolar molecules don't dissolve very well in water (fats and waxes)

2) Solute = substance being dissolved in a solution

a) Typically present in the least amount

b) Typically a solid

II. THE NATURE OF SOLUTIONS:

Solvent Solute Common ExampleGas Gas Diver’s tankGas Liquid HumidityGas Solid Moth ball

Liquid Gas CarbonationLiquid Liquid VinegarLiquid Solid SeawaterSolid Gas Gas stove lighter

Solid Liquid Dental fillingsSolid Solid Sterling Silver (Ag +

Cu)

A. 9 Possible Solutions Combinations:

oNOT all solutions are liquids/solids!

oSolutions are formed in ALL 3 states!

C. Johannesson

B. Solvation

1.) Solvation = the process of dissolving

b.) Then- solute particles are separated and pulled into solution

a.) First- solute particles are surrounded by solvent particles

B. Solvation2.) Dissociation = separation of an ionic solid into aqueous ions

•Ex: NaCl + H2O – the Na ion and Cl ion become hydrated and gradually move away from the crystal into solution.

•Each ion in the solution acts as though it were present alone: So there is only a solution containing Na+ and Cl- ions uniformly mixed with H2O particles

•NaCl(s) Na+(aq) + Cl–(aq)

Animation of Salt Dissolving: http://www.northland.cc.mn.us/biology/Biology1111/animations/dissolve.html

B. Solvation3.) Ionization = breaking apart of some polar molecules into aqueous ions•Ex: HNO3(aq) + H2O(l) H3O+(aq) + NO3

–(aq)

B. Solvation4.) Molecular Solvation =molecules stay intact•Ex: C6H12O6(s) C6H12O6(aq)

1) Grinding: increases surface area exposed to solvent

2) Stirring: allows solvent continual contact with solute

3) Heating: increases kinetic energy; increases mixing

C. Factors Affecting the Rate of Dissolving (Increases Solution Rate):

C. Factors Affecting the Rate of Dissolving (Increase Solution Rate):

III.

Electrolytes &

Nonelectrolyt

es

Many people know me as containing “electrolytes”… that just means I contain dissolved, dissociated ions!

1) electrolyte = a substance that dissolves in water to give a solution that conducts electric current2) nonelectrolyte = a substance that dissolves in water to give a solution that does NOT conduct an electric current

A. Electrolytes and Nonelectrolytes

3) Solutions of electrolytes can conduct electric current:

a) The positive ions and the negative ions disassociate (separate) in solution. The mobile ions can move a charge from one point in the solution to another point

A. Electrolytes and Nonelectrolytes

4) Solutions of nonelectrolytes CANNOT conduct electric current:

a) When a nonelectrolyte dissolves in water there are NO charged particles in solution.

b) Ex: Solute exists as molecules

A. Electrolytes and Nonelectrolytes

5) Weak Electrolytesa) Only a portion of dissolved molecules ionize

A. Electrolytes and Nonelectrolytes

6) Solid ionic compounds CANNOT conduct electric current:

a) Ions are present but they are NOT mobile.

A. Electrolytes and Nonelectrolytes

StrongElectrolyte

Non-Electrolyte

solute exists asions only

- +

salt

- +

sugar

solute exists asmolecules only

- +

acetic acid

WeakElectrolyte

solute exists asions andmolecules

A. Electrolytes and Nonelectrolytes

IV.

Solubility

1) Solubility = quantity of solute that will dissolve in specific amount of solvent at a certain temperature. (pressure must also be specified for gases).

a)  Ex: 204 g of sugar will dissolve in 100 g of water at 20C

b)  soluble and insoluble are relative termsc)  solubility should NOT be confused with

the rate at which a substance dissolves

A. SOLUBILITY:

2) saturated solution = a stable solution in which the maximum amount of solute has been dissolved.

a) Visual evidence: a quantity of undissolved solute remains in contact with the solution

A. SOLUBILITY:

3) solubility equilibrium = state where the solute is dissolving at the same rate that the solute is coming out of solution (crystallizing).

a) Opposing processes of the dissolving and crystallizing of a solute occur at equal rates.

b) solute + solvent solution

A. SOLUBILITY:

4) unsaturated solution = a solution that contains less solute than a saturated solution under existing conditions

A. SOLUBILITY:

5) supersaturated solution = a solution that temporarily contains more than the saturation amount of solute than the solvent can hold

• Unstable – if disturbed, the excess solute will crystallize out of solution

A. SOLUBILITY:

A. Solubility

SATURATED SOLUTIONno more solute dissolves

UNSATURATED SOLUTIONmore solute dissolves

SUPERSATURATED SOLUTIONbecomes unstable, crystals form

concentration

1) Nature of solute and solvent a) “Like dissolves like” = rule of thumb for

predicting whether or not one substance dissolves in another “Alikeness” depends on:

•Intermolecular forces•Type of bonding•Polarity or nonpolarity of molecules:•ionic solutes tend to dissolve in polar solvents but not in nonpolar solvents

B. 3 FACTORS EFFECTING SOLUBILITY:

Solvent Type Solute Type Is Solution Likely?

Polar Polar

Polar Nonpolar

Nonpolar Polar

Nonpolar Nonpolar

B. 3 FACTORS EFFECTING SOLUBILITY:

b.) Solvent-Solute Combinations:

YES

YES

NO

NO

“Like Dissolves Like”

2) Pressure:a) Pressure has little effect on the

solubility of liquids or solids in liquid solvents.

b)  The solubility of a gas in a liquid solvent INCREASES when pressure increases. It is a direct relationship.

B. 3 FACTORS EFFECTING SOLUBILITY:

3) Temperature:a) The solubility of a gas in a liquid solvent

DECREASES with an increase in temperature.

b)  The solubility of a solid in a liquid solvent MOST OFTEN increases with an increase in temperature. However, solubility changes vary widely with temperature changes sometimes decreasing with temperature increases.

B. 3 FACTORS EFFECTING SOLUBILITY:

V. Solubility

Graph

1) Saturated = any point on the line or ABOVE the line

2) Unsaturated = any point BELOW the line

A. SOLUBILITY CURVE:

1. What is the solubility of the following solutes in water?a) NaCl at 60ºC=

b) KCl at 40ºC=

c) KNO3 at 20ºC=

38g

39g

31g

2. Are the following solutions saturated or unsaturated? Each solution contains 100 g of H20.

a) 31.2 g of KCl at 30ºC = b) 106g KNO3 at 60ºC=

Unsaturated

Saturated (on the line)

2. Are the following solutions saturated or unsaturated? Each solution contains 100 g of H20.

c)40 g NaCl at 10ºC=

d)150g KNO3 at 90ºC = Unsaturated

Saturated

3. For each of the following solutions, explain how much of the solute will dissolve and how much will remain undissolved at the bottom of the test tube?

a) 180 g of KNO3 in 100 g of water at 80ºC

  169g dissolved11g

undissolved

3. For each of the following solutions, explain how much of the solute will dissolve and how much will remain undissolved at the bottom of the test tube?

b) 180 g of KNO3 in 100 g of water at 20ºC

31g dissolved149g

undissolved

3. For each of the following solutions, explain how much of the solute will dissolve and how much will remain undissolved at the bottom of the test tube?

 C) 60 g of NaCl in 100 g of water at 60ºC39g dissolved 

21g undissolved

4. A saturated solution of KNO3 is formed from one hundred grams of water. If the saturated solution is cooled from 90°C to 30°C, how many grams of precipitate are formed?

 (200g-45g) =

155g

5. A saturated solution of KCl is formed from one hundred grams of water. If the saturated solution is cooled from 90°C to 40°C, how many grams of precipitate are formed?

(53g-39g) = 14g

VI.

Concentration:

Molarity &

Molality

A. Concentration1.) Concentration = The amount of solute in a solution.a.) Describing Concentration:

• % by mass - medicated creams• % by volume - rubbing alcohol• ppm, ppb - water contaminants• molarity - used by chemists• molality - used by chemists

A. Molarity1.) Molarity = unit of concentration of a solution.•Formula:

solution of literssolute of moles(M)Molarity

total combined volume

substance being dissolved

A. Molarity

2M HCl

LmolM

L 1HCl mol 2HCl 2M

What does this mean?

1) To solve for Molarity when given grams of solute, you will use gram/mole conversion and the equation for Molarity. Use the following steps:

a) Change grams of solute to moles. To do this you need to use the following conversion:

Grams of solute 1 mole of solute =moles of solute Molar mass of solute*

*make sure formulas are correct; Use ion list!   

B. Molarity Calculations

b) Change the volume of solution to liters (L) by moving the decimal 3 places to the LEFT. • Remember that 1000 mL = 1 L. Water will most often be the solvent in a solution.

c) Substitute the information into the molarity equation and solve.

B. Molarity Calculations

solution of literssolute of moles(M)Molarity

2) Ex: What is the molarity of a solution composed of 22.4g of sodium chloride dissolved in enough water to make 500 mL of solution?

22.4 g of NaCl 1 mole NaCl =0.383 moles NaCl

58.443 g NaCl

0.383 mole NaCl = 0.766 M NaCl 0.500 L soln Note: You need to

change mL of solution to L. So move decimal 3 places to the left!

Use the Molarity equation to solve for grams of solute. You will need to solve the Molarity equation for moles of solute. Then convert moles of solute to grams of solute.

B. Molarity Calculations

LmolM

a) Substitute the information and solve for moles by cross multiplying or multiplying by the reciprocal:

“X” moles LiBr___ = 0.4M LiBr

0.300 L solution 

X = 0.12 moles of LiBr are needed

Ex: How many grams of lithium bromide are present in 300mL of a 0.4M lithium bromide solution?

LmolM

b) Convert moles to grams using the following conversion:Moles of solute molar mass of solute = grams of solute

1 mole of solute • The substitution would look like:

0.12 moles LiBr 86.845 g LiBr = 10. g LiBr 1 mol LiBr

Ex: How many grams of lithium bromide are present in 300mL of a 0.4M lithium bromide solution?

C. Dilution Calculations1.) Sometimes solutions of lower concentrations are made from existing solutions. 2.) Moles of solute remain the same.3.) Formula: M1V1 = M2V2

• M1 is the original molarity concentration• V1 is the volume of the original solution (in L)

• M2 is the new concentration• V2 is the amount of the new solution needed (in L)

C. Dilution CalculationsEx 1: What volume of 15.8M HNO3 is required to make 250 mL of a 6.0M solution?

GIVEN:M1 = 15.8MV1 = ?M2 = 6.0MV2 = 250 mL

WORK:M1 V1 = M2 V2

(15.8M) V1 = (6.0M)(250mL)

V1 = 95 mL of 15.8M HNO3

2211 VMVM

D. Molality

1.) Molality (m) = a unit of concentration of a solution expressed in moles of solute per kilogram of solvent.a) Ex: A solution that contains 1 mol of solute, ammonia (NH3), dissolved in exactly 1 kg solvent, is a “one molal” solution.

D. Molality

solvent ofkg solute of moles(m)molality

mass of solvent only

c) Note: 1 kg water = 1 L water

b) Formula:

E. Calculations: Solving for Molality

1) If needed convert: • grams of solute moles of solute• grams of solvent kg of solvent (move

decimal 3 places to the left!)

2) Solve for molality m = moles of solute kg of solvent 

3) Ex: Find the molality of a solution containing 75 g of MgCl2 in 250 mL of water. 75 g MgCl2 1 mol MgCl2

95.211 g MgCl2

= 3.2m MgCl2 0.250 kg water

kgmolm

= 0.79 mol MgCl2

0.79 mol MgCl2

1) Use Molality equation and solve for moles2) Then convert moles of solute to grams of solute

F. Calculations: Using the Molality equation to solve for grams of solute:

kgmolm

“X” moles NaCl___ = 1.54m NaCl0.500 kg H2O  X = 0.770 moles NaCl

0.770 mol NaCl 58.443 g NaCl = 45.0 g NaCl

1 mol NaCl

Ex: How many grams of NaCl are required to make a 1.54m solution using 0.500 kg of water?

VII. Colligative Properties

C. Johannesson

A. Colligative Property

1.) Colligative property = A property that depends on the NUMBER of solute particles rather than the type of particle.

1.) Vapor Pressure LOWERING of a solution

a) Ex: We have two beakers. One contains H2O, the other contains a salt solution (NaCl) in H2O. Both beakers are placed into a sealed chamber:

B. Examples of colligative properties:

DAY 1

DAY 3: If we leave it for a couple of days and then come back and take a look, this is what it might look like.

b) WHAT HAPPENED?:•Na+ and Cl- ions are non-volatile ions. In other words, they will not leave and go into the vapor phase. •Those H2O molecules with enough kinetic energy will leave the surface of the solutions and enter into the vapor phase. •The fact that Na+ and Cl- ions are dissolved in H2O indicates that there is an attractive interaction between the solutes and the H2O.

B. Examples of colligative properties:

DAY 1

DAY 3: If we leave it for a couple of days and then come back and take a look, this is what it might look like.

•The interaction of the Na+ and Cl- ions for H2O will act to hinder the ability of the solvated H2O molecules to leave and go into the vapor phase (Na+ and Cl- ions are non-volatile). •This hindrance of the H2O molecules to enter the vapor phase will reduce the vapor pressure of the H2O in the salt-containing solution.

B. Examples of colligative properties:

DAY 1

DAY 3: If we leave it for a couple of days and then come back and take a look, this is what it might look like.

•The two beakers are in the same sealed container, thus, the vapor pressure above the solutions is identical. •The rate of H2O entering the solutions by collisions from the vapor state will be identical

•The rate of H2O leaving the liquid phase and entering the vapor phase is slower for the NaCl-containing solution

B. Examples of colligative properties:

DAY 1

DAY 3: If we leave it for a couple of days and then come back and take a look, this is what it might look like.

2.) Osmotic Pressure = If two solutions, with different solute concentrations, are separated by a semi-permeable membrane, there can be a net flow of solvent across the membrane.

B. Examples of colligative properties:

3.) Freezing Point depression= The solution will begin to freeze at a temperature BELOW that of the pure solvent.

B. Examples of colligative properties:

4.) Boiling Point elevation = the solution will begin to boil at a temperature ABOVE that of the pure solvent.

B. Examples of colligative properties:

1.) One of the ways in which they are different, is that when you add a solute to a liquid both the freezing point and boiling point of the solution CHANGE.

C. How are solutions different than pure liquids?

a) Freezing point – In a solution, the solute particles interfere with the attractive forces among the solvent particles.

•This prevents the solvent from entering the solid state at its normal freezing point

C. How are solutions different than pure liquids?

b) Boiling point- When the temperature of a solution containing a nonvolatile solute is raised to the boiling point of the pure solvent, the resulting vapor pressure is still less than the atmospheric pressure and the solution will not boil.

•Thus, the solution must be heated to a higher temperature to supply the additional KE needed to raise the vapor pressure to atmospheric pressure.

C. How are solutions different than pure liquids?

2) Water is the liquid we will be dealing with most often

a) The freezing point of pure water is 0°C. b) The normal boiling point of water is

100°C. c) But if you make a solution using water

as the solvent, the freezing point of that solution will NOT be 0°C nor will the boiling point be 100°C!

C. How are solutions different than pure liquids?

1) Freezing point depression (TFP) – solutions will freeze at LOWER temperature than the pure solvent

a) The more solute dissolved, the greater the effect.

D. Freezing Point Depression & Boiling Point Elevation:

b) Ex: ethylene glycol (antifreeze) protects against freezing of the water in the cooling system by lowering the freezing point to about -40°C

D. Freezing Point Depression & Boiling Point Elevation:

c) Ex: making of homemade ice cream- The ice cream mix is put into a metal container which is surrounded by crushed ice. Then salt is put on the ice to lower its melting point. This gives a temperature gradient across the metal container into the saltwater-ice solution which is lower than 0°C. The heat transfer out of the ice cream mix allows it to freeze.

D. Freezing Point Depression & Boiling Point Elevation:

2.) Boiling point elevation (TBP) – solutions will boil at HIGHER temperatures than the pure solvent  a) The boiling point of pure water is

100°C, but that boiling point can be elevated by the addition of a solute such as a salt.

b) The more solute dissolved, the greater the effect.

D. Freezing Point Depression & Boiling Point Elevation:

1) Solution concentrations are given in molality (m) 2) Colligative properties are directly proportional to the molal concentration of a solute 

E. Calculating Freezing Point Depression & Boiling Point

Elevation:

 3)Account for particle molality for ELECTROLYTES

•Ex: NaCl = 2 ions (Na+ & Cl-)•Ex: MgCl2 = 3 ions (Mg+2 and 2 Cl-)

4) A change in the concentration (m), changes the freezing point and boiling point of a solution

E. Calculating Freezing Point Depression & Boiling Point

Elevation:

a) You will need to figure out the change in temperature for the normal freezing point (FP) or boiling point (BP) and adjust them with the following steps: ∆T = mK

∆T = change in temperaturem = moles solute/kg solvent (MOLALITY) K = constant

•You will mostly use the constants for water: Kfp = 1.853 ˚C/m

Kbp = 0.515˚C/m

E. Calculating Freezing Point Depression & Boiling Point

Elevation:

• Other values can be found below:

E. Calculating Freezing Point Depression & Boiling Point

Elevation: FREEZING Point Depression Constants

CompoundFreezing Point (oC)

Kfp (oC/m)

water 0 1.853acetic acid 16.66 3.90benzene 5.53 5.12p-xylene 13.26 4.3

naphthalene 80.29 6.94cyclohexane 6.54 20.0

BOILING Point Elevation Constants

CompoundBoiling

Point (oC)Kbp(oC/m)

water 100 0.515ethyl ether 34.55 1.824

carbon disulfide 46.23 2.35

benzene 80.10 2.53carbon

tetrachloride 76.75 4.48

camphor 207.42 5.611

4.) Steps solving for ∆T:a) Solve for molality (see steps

from previous problems) m = moles of solute kg of solvent

E. Calculating Freezing Point Depression & Boiling Point

Elevation:

kgmolm

b) Find your K constant from the chart listed in your notes.

• Be sure to find the correct K constant for what you are solving for-- either FP or BP!!

c) Solve for ∆T ∆T = mK

E. Calculating Freezing Point Depression & Boiling Point

Elevation:

d) If solving for:• Freezing point: Take the normal freezing point of the solvent and SUBTRACT the ∆T value

•Round final answer to 3 significant figures!

• Boiling point: Take the normal boiling point of the solvent and ADD the ∆T value

•Round final answer to 6 significant figures!

E. Calculating Freezing Point Depression & Boiling Point

Elevation:

5.) At what temperature will a solution that is composed of 32.8 g of glucose (C6H12O6) in 225 g of water boil?32.8g C6H12O6 1 mol C6H12O6

180.155g C6H12O6

= 0.809 m C6H12O60.225 kg water

= 0.182 mol C6H12O6

0.182 mol C6H12O6

E. Calculating Boiling Point Elevation:

m =

TBP = mKbp TBP = (0.809 m) (0.515 C/m)

= 0.417CBP = 100C + 0.417 C = 100.417C

6.) Find the freezing point of a saturated solution of NaCl containing 28 g NaCl in 115 g water.

28.0g NaCl 1 mol NaCl

58.443g NaCl

= 4.17m NaCl 0.115 kg water

= 0.479 mol NaCl

0.479 mol NaCl

E. Calculating Freezing Point Depression:

m =

TFP = mKfp TFP = (8.34 m) (1.853 C/m)

= 15.5C

FP = 0C – 15.5 C = -15.5C

Particle molality = 4.17m X 2 = 8.34m