SOLUTIONS Chapter 15 If you are not part of the SOLUTION, you’re part of the PRECIPITATE!

SOLUTIONSChapter 15

If you are not part of the SOLUTION,

you’re part of the

PRECIPITATE!

SOLUTION OBJECTIVES Differentiate between colloids, solutions, and suspensions Define solvent and solute and be able to identify them in a

solution. Describe the process of solvation, dissociation, and

dissolving. Be sure to study polarity of compounds to be able to predict

solubility. Differentiate between saturated, unsaturated, and

supersaturated. State and discuss the factors affecting the rate of solubility. State and discuss the factors that affect solubility. Be able to read a solubility curve graph. Relate the enthalpy of solution to endothermic and

exothermic dissolving processes. Differentiate between molarity and molality. Solve problems involving molarity, molality, mole fraction

mass percent, volume percent, and making solutions.

THREE TYPES OF MIXTURES

SOLUTION COLLOID SUSPENSION

MIXTURES

DEMO

SOLUTION: Salt and waterCOLLOID: Shaving gelSUSPENSION: Cornstarch in water

TYNDALL EFFECT: particles dispersed in mixture are big enough to scatter light.

SOLUTIONS A solution is a

homogeneous mixture of two or more substances in a single physical state.

SOLVENT: the substance doing the dissolving and in the greater amount.

SOLUTE: the substance being dissolved and in less abundant.

DEMO If I combine 50mL beads and

50.0mL sand, what will the final volume be?

If I combine 50.0mL water (green) and 50.0mL alcohol, what will the final volume be?

NINE BASIC TYPES OF SOLUTIONS

Liquid Solutions:solid in liquid (salt water)liquid in liquid (vinegar)gas in liquid (carbonated drink)

Gaseous Solutions: All gaseous mixtures are solutionssolid in gas (soot in air)liquid in gas (humid air)gas in gas (air)

NINE BASIC TYPES OF SOLUTIONS

Solid SolutionsHomogeneous mixtures of solids are usually made from liquid solutions that have been mixed and then solidified (frozen) solid in solid (alloys – brass, bronze) liquid in solid (dental fillings) gas in solid (charcoal gas mask)

SOLUTION EQUILIBRIUM

- A solution is in dynamic equilibrium when the number of solute particles returning to the crystal surface is equal to the number of solute particles leaving the crystal surface.

SOLUTION EQUILIBRIUM

A SATURATED solution is a solution with the maximum amount of solute dissolved at a given temperature (Any more added goes to the bottom). It has reached dynamic equilibrium.

An UNSATURATED solution has the ability to dissolve more solute at a given temperature

If a solution is SUPERSATURATED, then a hot solution is saturated and cooled. An unstable condition results because the solution holds more solute that it normally does at a given temperature.

LASERDISCSOLUTIONS – Chapter 201. What are ion-dipole attractions? During the process of

solvation, how do ion-dipole attractions compare to the ionic bonds inside an ionic crystal?

2. Describe the processes that are balanced when solubility equilibrium has been reached. If more solute is added to a solution in equilibrium, will this always result in more dissolved solute particles in the solution? Explain.

3. How does a solution become supersaturated? After crystals precipitate out of a supersaturated solution, would you still call it supersaturated? Explain.

THE DISSOLVING PRACTICE

Dissolving occurs when the solute is pulled apart by the solvent.

This takes place at the surface of the solute. The solvent surrounds the solute.

This process of surrounding the solute is called SOLVATION.

When the surrounding is done by water, this is called HYDRATION, a particular type of solvation.

When ionic compounds separate into their ions in a solvent, DISSOCIATION occurs.

DISSOCIATION OF NaCl

Solute-Solvent Combinations

1. Polar Solvent – Polar Solute:The polar solvent is attracted to the polar solute. The solvent gradually surrounds the solute. The particles attach themselves due to polar attraction. Solvation occurs. Like dissolves like. Ex. Salt and water

Solute-Solvent Combinations

2. Polar Solvent – Nonpolar Solute:Polar solvent particles are attracted to each other and not the solute. Solvation does not occur and a solution is unlikely. Ex. oil and water

DEMO: Marbles

Solute-Solvent Combinations

3. Nonpolar Solvent – Polar Solute:Nonpolar solvent particles have little attraction to the polar solute. Solvation does not occur and a solution is unlikely. Ex. Salt and oil

Solute-Solvent Combinations

4. Nonpolar solvent – nonpolar solute:Random motion of solute particles causes them to leave the surface of the solute and become evenly dispersed in the nonpolar solvent. Solvation occurs. Like dissolves like.

LASERDISCPolar and Nonpolar Solvents – Chapter 301. Why are some liquids immiscible? Explain in terms of

intermolecular forces?2. How would you predict whether carbon tetrachloride is

a polar or nonpolar solvent? What evidence have you observed that supports your prediction?

3. Based on chemical formula alone, can you tell whether iodine, I2, is a polar or nonpolar solute? What evidence have you observed that supports your prediction?

4. Based on the chemical formula alone, can you tell whether copper (II) chloride, CuCl2, is a polar or nonpolar solute? What evidence have you observed that supports your prediction?

DO NOWPick up handout – due TuesdayTurn in Superaturation labBook projects due today: William

B., Marwa J., Quinn R., Devon L., Andrew C., Madison B., Peter D., Seth T., Conner R., Kat M., Tatum N., Sam W., Dan N., Claire W., and Jamal R.

SOLUBILITY Solubility is how much solute can dissolve in a

given amount of solvent. It is measured in g/L or mol/L. It is usually the

grams of solute per 100g of solvent. A CONCENTRATED solution is said to have a

high ratio of solute to solvent. A DILUTE solution is the opposite of this.

For solids: solubility usually increases with increased temperature

For gases: solubility usually decreases with increased temperature

SOLUBILITY

How do you determine saturated, unsaturated, and supersaturated for a given substance?

FACTORS AFFECTING SOLUBILITY

1. Nature of the solvent and solute - “like dissolves like”. This means that polar solvent dissolves a polar solute and nonpolar solvent dissolves nonpolar solute. But polar does not dissolve nonpolar.

In other words – will it even dissolve?

FACTORS AFFECTING SOLUBILITY

2. Temperature - increase the temperature and solubility increases (except gases).

FACTORS AFFECTING SOLUBILITY

3. Pressure - increase the pressure and you increase solubility (only with gases).

Gases ONLY

4. How much is already dissolved – saturated versus unsaturated versus supersaturated…

Factors Affecting RATE of Solubility (How fast something will dissolve)

Agitation (shaking, stirring) Increased temperature (except gases)Smaller Particle size

Each allows more solvent to come in contact with the solute faster.

HEAT OF SOLUTION Energy is released and absorbed as substances

dissolve. ENDOTHERMIC: The solute particles must

separate and the solvent particles must separate to make room for the solute.

EXOTHERMIC: When the solute and the solvent mix, the particles are attracted to each other.

The overall change in energy is the heat of solution.

Most solutions are endothermic.

CONCENTRATIONConcentration is NOT dependent upon the

sample size and it can be measured.

Grams/100.0 grams measures solubility Parts per million (ppm) measures small

concentrations Parts per billion (ppb)measures pollutants Molarity used in lab chemistry Molality used for special calculations

We will learn to calculate molarity, molality, mass percent, volume percent,

and mole fraction

MOLARITY, MThis is the ratio between the moles of dissolved substance and the volume of the solution expressed in liters.

Molarity, M = moles of solute volume of solution in

liters A one-molar (1M) solution of HCl contains one mole of HCl in one liter of water. (Which means it contains 36.46g of HCl in 1 liter of water.)

SAMPLE PROBLEMSandy dissolves 45.0 g of NaCl in 2.5 liters of solution. What is the concentration in molarity of NaCl?

Mass = 45.0 g Molar Mass = 58.44 g/mol

V = 2.5 L Molarity = 45.0 g NaCl 1 mol NaCl

2.5 L 58.44 g NaCl = 0.31M

PRACTICEWhat is the molarity of 58.5g of

NaCl dissolved in 2.0L of solution?

MOLALITY, mThis is concentration expressed in terms

of moles of solute per kilogram of solvent. Volume is not a factor.

Molality, m = moles solute Kg solvent A 1.0 molal aqueous sugar soln:

1 mole sugar 1 kg water

SAMPLE PROBLEM

Calculate the molality of 98.0g RbBr in 0.824 Kg water.

 m = 98.0 g RbBr 1 mol RbBr

0.824Kg H2O 165.38g RbBr

   = 0.719m

PRACTICECalculate the molality of 85.2g

SnBr2 in 140.0g water.

MASS PERCENTScientists frequently express the

concentration of solutions in mass percent.

 Mass % = g of solute

x 100 g of solute + g of

solvent  A 5% solution of NaOH contains 5g NaOH

in each 100g of solution (95g solvent and 5g of solute).

VOLUME PERCENTConsumer products frequently express

their concentration of solutions in volume percent.

 Volume % = L of solute

x 100 L of solute + L of

solvent  A 5% solution of sodium hydroxide

contains .05L NaOH in each 1.0L of solution (0.95L solvent and 0.05L of solute).

MOLE FRACTIONThe concentration of solution can

also be expressed in mole fractions.

 Mole Fraction =

mole of solute

mole of solute + mole of solvent

PREPARING SOLUTIONS

A 3M solution of HCl is not bought but made from 12M stock solutions.

In addition, a 1.0M solution of NaOH is made from a calculated amount of solid NaOH added to water.

It is important to know how to make different concentration of solutions.

PREPARING SOLUTIONS

Prepare 1.0 liter of a 1.0M aqueous solution of NaOH

M = mol 1.0M = mol L 1.0L

mol = (1.0M)(1.0L) = 1.0mol1.0mol NaOH 40.00g NaOH = 40.00g

1 mol NaOH

So, you will put 40.00g NaOH in a flask, add 1.0L water, and mix well.

Preparing SolutionsPrepare 1.0L of a 6.0M aqueous

solution of KCl. The molar mass of KCl is 74.55g/mol.

DILUTING SOLUTIONS

To dilute a solution, you can form a ratio between molarity and volume.

 M1V1 = M2V2

DILUTING SOLUTIONSDiluting a 12.0M solution of HCl to 6.0M

HClWhat volume would you use to make

0.500L of 6.0M HCl solution?

V1 = M2V2 = (6.0M HCl)(0.500L HCl)

M1 (12.0M)

= 0.250L HClThus you would need 0.250L HCl and

0.250L water to make the solution.

PRACTICEPreparing a 0.1M solution of HCl What volume would you use of

12.0M HCl to make 1.0L of 0.1M HCl?

ANSWERPreparing a 0.1M solution of HCl

M1 = 12.0M HCl V1 = M2V2

V1 = ? M1

M2 = 0.1M HCl V1 = (0.1MHCl)(1.0L)

V2 = 1.0L (12.0M HCl)

= 0.0083L

M1V1 = M2V2 Add 8.3mL 12.0M HCl to

991.7mL water

COLLIGATIVE PROPERTIES

“colligative” – depending upon the collection

These are properties that depend on the concentration (the number of) of the solute particles.

Three Factors: Vapor pressure reduction Boiling Point Elevation Freezing Point Depression

LASERDISCColligative Properties (chapter 32)1. Support one of the following hypotheses about why

it helps to add salt to the water before cooking the pasta: a) The salt brings the water to a boil sooner; or b) The salt brings the water to a boil at a higher temperature.

2. Which antifreeze solution boils at the highest temperature? Which solution would you want in your car on a hot summer day? Explain your reasoning.

3. On cold winter days, unprotected radiator water can freeze and expand, which can ruin an engine by cracking the engine block. Which antifreeze mixture would you want in your car on a cold winter day?

Vapor Pressure: A Review

VAPOR PRESSURE REDUCTION

VAPOR PRESSURE REDUCTION

VAPOR PRESSURE REDUCTIONThe vapor pressure of a solvent

containing a non-volatile solute is LOWER than the vapor pressure of the pure solvent.

WHY: The solute takes up space at the surface of the liquid and prevents some solvent molecules from leaving the liquid.

RAOULT’S LAW

The vapor pressure of a solution of a non-volatile solute is equal to the vapor pressure of the pure solvent at that temperature multiplied by its mole fraction.

RAOULT’S LAWPo is the vapor pressure of the pure solvent at a particular temperature.xsolv is the mole fraction of the solvent.

So if the solution is 20% solute and 80% solvent, the mole fraction for the solvent is 0.8.So the Vpsoln = 0.8 vpsolvent

RAOULT’S LAWBecause changes in state depend upon vapor pressure, the presence of a solute also affects the freezing point and the boiling point of a solvent.

**It depends on the number and not the identity of the solute particles

BOILING POINT REVIEW

The boiling point of a substance is the temperature at which the vapor pressure of a liquid is equal to the external pressure on its surface (usually atmospheric pressure).

BOILING POINT ELEVATION An addition of solute lowers that vapor

pressure, therefore a higher temperature is necessary to get the vapor pressure up to the external (atmospheric) pressure so that the solution will boil.

Tb is the difference between the boiling point of the solution and the boiling point of the pure solvent. It is directly proportional to the number of solute particles per mole of solvent particles.

BOILING POINT ELEVATIONDissociation factor molality

Tb = (Kb)(df)(m)

constant (molal BP elevation constant) 

The dissociation factor, df, is how many particles the solute breaks up into. For a nonelectrolyte, the df is 1

Water’s “Kb” value is 0.515C/m

FREEZING POINT: A REVIEW

The freezing point of a substance is the temperature at which the vapor pressure of the solid and liquid phases are the same.

FREEZING POINT DEPRESSION

Club Soda demo

Animation for Freezing Point depression

FREEZING POINT DEPRESSION

When a dissolved solute lowers the freezing point of its solution, you have freezing point depression. Tf is directly proportional to the molality of the solution.

Tfp = (Kfp)(df) (m)

 constant (molal fp depression constant)Water’s “Kfp” value is 1.853C/m

CHEMICALS USED TO MELT ICE

Formula Lowest Temp

Pros Cons

(NH4)2SO4 -7.0°C fertilizer Damages concrete

CaCl2 -29.0°C Melts ice faster than NaCl

Attracts moisture; surfaces can be

slippery

MgCl2 -15.0°C Melts ice faster than NaCl

Attracts moisture

CH3COOK -9.0°C biodegradable corrosive

KCl -7.0°C fertilizer Damages concrete

NaCl -9.0°C Keeps sidewalks dry

Corrosive; damages concrete and vegetation

CMA -9.0°C Safest for concrete and vegetation

Works better at preventing re-icing

CALCULATIONS: NON-IONIC

Determine the freezing and boiling point for 85.0g of sugar (C12H22011) in 392g water. Need these things:1. Molality, m2. number of particles (should be one)3. Change in temperature,

CALCULATIONS: NON-IONIC

1. Molality392 g H2O 1 kg = 0.392Kg

1000 g  m = 85.0 g C12H22011 1 mol C12H22011 0.392 Kg H2O 342.34 g C12H22011

= 0.633m

2. df = 1 (non-ionic solute)

CALCULATIONS: NON-IONIC

Change in Freezing point temperature:Tfp = (Kfp)(df) (m) = 1.853C 1 0.633 m m

= 1.17C

New Freezing Point:0.00C – 1.17C = -1.17C

Freezing point of pure water

CALCULATIONS: NON-IONICChange in boiling point temperature:Tbp = (Kbp)(df) (m) = 0.515C 1 0.633 m m

= 0.326C

New Boiling Point:100.000C + 0.326C = 100.326C

Boiling point of pure water

PRACTICEWhat is the freezing point for a solution of 210.0g of glycerol dissolved in 350.0g of water? (The molecular mass of glycerol is 92.11 g/mol) FP = ? Solute = 210.0g glycerol Solvent = 350.0g H2O

PRACTICEWhat is the freezing point for a solution of 210.0g of glycerol dissolved in 350.0g of water? (The molecular mass of glycerol is 92.11 g/mol)1. m = 210.0g gly 1 mole gly 1000g

350.0g H2O 92.11 g gly 1 kg

= 6.51395 = 6.514m2. Df = 13. Tf = (Kf)(df)(m) = (1.853ºC/m)(1)(6.514m)

= 12.070 = 12.07ºC

New FP = 0.000ºC – 12.07ºC = -12.07ºC

CALCULATIONS: IONIC

Determine the freezing and boiling point for 21.6 g NiSO4 in 100.0 g water.

Need these things:1. molality2. number of ions 3. change in temperature

CALCULATIONS: IONIC1. Calculate molality:100.0g H2O 1 kg = 0.1000 Kg

1000 g  

m = 21.6 g NiSO4 1 mol NiSO4

0.1000 Kg H2O 154.76 g NiSO4

= 1.396 = 1.40 m

CALCULATIONS: IONIC

2. Determine the number of ions:

Substance is NiSO4

There are two ions, Ni+2 and SO4-2,

so the df is 2

CALCULATIONS: IONIC

3. Calculate the change in temperature and the new freezing point.

Tfp = 1.853C 2 1.40 m = 5.19C

m 0.00C – 5.19C = -5.19C

CALCULATIONS: IONIC

Tbp = 0.515C 2 1.40 m = 1.44C

m

100.00C + 1.44C = 101.44C

CALCULATIONS: IONICWhat is the boiling point of a solution containing 34.3 g of Mg(NO3)2 dissolved in 0.107 kg of water? (The formula mass of magnesium nitrate is 148.32 g/mol.)

Bp = ? Solute 34.3g Mg(NO3)2

Solvent 0.107kg

CALCULATIONS: IONIC1. m = 34.3g Mg(NO3)2 1 mole Mg(NO3)2

0.107kg 148.32 g Mg(NO3)2

= 2.16127796 = 2.16m

2. Df = 3 Mg+2 NO3-1 NO3

-1

3. Tb = (Kb)(df)(m) = (0.515ºC/m)(3)(2.16m)

= 3.3372 = 3.34ºC

New BP = 100.0000ºC + 3.34ºC = 103.34ºC

Do NowPick up handout – due Monday as a

homework check Ice Cream Supplies due NEXT TuesdayBook Projects due Thursday: Hayden

R., Matt H., Bob Wiley, Sean R., Sam K., Cameron H., Joe F., Marshall C., Josh M., Kainath M., Grace M., Sarah., Adiba K., Alexander G., and Stilian I.

RECAP A solution reduces the vapor pressure of

the pure solvent. Vapor pressure reduction results in

freezing point depression (lowering) and boiling point elevation.

To solve these problems, you need: Molality Dissociation factor (Number of ions) Molal BP/FP constant (given to you)

RECAP Calculating freezing point change:

Tfp = (Kfp)(df)(m)

Calculating boiling point change:Tb = (Kb)(df)(m)

Final step is to subtract from true freezing point or add to true boiling

point

DETERMINING MOLAR MASS Colligative properties provide a useful

means to experimentally determine the molar mass (molecular mass in one mole) of an unknown substance.

Steps:1. Solve for molality.2. Solve for moles of solute.3. Solve for molar mass

This is basically going backwards

DETERMINING MOLAR MASSA 10.0 g sample of an unknown organic compound is dissolved in 0.100 kg water. The boiling point of the solution is elevated to 0.433oC above the normal boiling point of water. What is the molar mass?

solute = 10.0g Solvent = 0.100kg waterTbp = 0.433oC

df = 1

DETERMINING MOLAR MASS

Step One: determine molalityTbp = Kbp df m

m = Tbp = 0.433oC = 0.841 m (mol/kg) Kbp 0.515oC/m

Assume df = 1 because it is an unknown ORGANIC compound (not ionic)

DETERMINING MOLAR MASS

Step Two: Detemine moles of solutem = mol solute

Kg solvent

mol solute = m x kg solvent= 0.841 mol/kg x 0.100 kg= 0.0841 mol

DETERMINING MOLAR MASS

Step Three: Determine Molar Massmol solute = mass solute

molar mass solute

molar mass = mass solute = 10.0 g mol solute 0.0841 mol

= 118.90606 = 119 g/mol

DETERMINING MOLAR MASSPRACTICE:A solution of 3.39 g of an unknown in 10.00 g of water has a freezing point of –7.31oC. The solution is a nonelectrolyte. What is the molar mass of the compound? (The molal freezing point constant of water is 1.853C/m.)

Mass solute = 3.39gMass solvent = 10.00gFP = –7.31oCΔTfp = 7.31oC

df = 1

DETERMINING MOLAR MASSPRACTICE:A solution of 3.39 g of an unknown in 10.00 g of water has a freezing point of –7.31oC. The solution is a nonelectrolyte. What is the molar mass of the compound? (The molal freezing point constant of water is 1.853C/m.)

Step One: determine molalityTfp = Kfp df m

m = Tfp = 7.31oC = 3.94m

Kbp df (1.853oC/m)(1)

DETERMINING MOLAR MASSPRACTICE:A solution of 3.39 g of an unknown in 10.00 g of water has a freezing point of –7.31oC. The solution is a nonelectrolyte. What is the molar mass of the compound? (The molal freezing point constant of water is 1.853C/m.)

Step Two: Determine moles of solutem = mol solute

Kg solventmol solute = m x kg solvent

= 3.44 m x 0.01000 kg= 0.0394 mol

DETERMINING MOLAR MASS

PRACTICE:A solution of 3.39 g of an unknown in 10.00 g of water has a freezing point of –7.31oC. The solution is a nonelectrolyte. What is the molar mass of the compound? (The molal freezing point constant of water is 1.853C/m.)

Step Three: Determine Molar Massmol solute = mass solute

molar mass solute

molar mass = mass solute = 3.39g mol solute

0.0394 mol= 86.040609 = 86.0g/mol

SEIZURE- Hiorns began by reinforcing the walls

and ceiling, and tanking the flat (apartment)with plastic sheeting.

- Then 70-80,000 liters of hot copper sulfate solution was pumped in through a hole in the ceiling from the flat above.

- Weeks went by, until the temperature of the solution dropped, and the crystals began to precipitate.

- Finally, any remaining liquid was pumped back out, to be recycled by the chemical industry.