Solutions and their Solutions and their BehaviorBehavior

Chapter 18Chapter 18

1.1. Identify factors that determine the rate at Identify factors that determine the rate at which a solute dissolveswhich a solute dissolves

2.2. Identify factors that affect the solubility of a Identify factors that affect the solubility of a solute in solutionsolute in solution

3.3. Calculate the solubility of a gas in a liquid Calculate the solubility of a gas in a liquid under various pressure conditionsunder various pressure conditions

Objectives

Factors affecting the rate Factors affecting the rate of dissolvingof dissolving

How could you speed up the How could you speed up the dissolving of sugar in a glass of iced dissolving of sugar in a glass of iced tea?tea?

Which dissolves faster, table salt or Which dissolves faster, table salt or rock salt?rock salt?

Factors affecting the rate Factors affecting the rate of dissolvingof dissolving

TemperatureTemperature – increasing the – increasing the temperature speeds up the rate of temperature speeds up the rate of dissolvingdissolving

AgitationAgitation – stirring speeds up the rate – stirring speeds up the rate of dissolvingof dissolving

Particle sizeParticle size – smaller particles – smaller particles dissolve faster than large particles dissolve faster than large particles (surface area)(surface area)

SolubilitySolubility

The The solubilitysolubility of a substance is the of a substance is the amount that dissolves in a given amount that dissolves in a given quantity of solvent at a given quantity of solvent at a given temperature.temperature.

SolubilitySolubility

A A saturated solutionsaturated solution contains the contains the maximum amount of solute at a maximum amount of solute at a constant temperature. constant temperature.

SolubilitySolubility

An An unsaturated solutionunsaturated solution contains less contains less solute than a saturated solution.solute than a saturated solution.

A A supersaturated solutionsupersaturated solution contains contains more solute than a saturated solution. more solute than a saturated solution. (This (This occurs when a solution is saturated occurs when a solution is saturated and then allowed to cool but all of the and then allowed to cool but all of the solid remains dissolved. solid remains dissolved. It is an unstable It is an unstable solution, adding a crystal causes solution, adding a crystal causes precipitation.)precipitation.)

What would happen…What would happen…

……if you added more sugar to a if you added more sugar to a saturated sugar solution and saturated sugar solution and stirred?stirred?

……if you added more sugar to an if you added more sugar to an unsaturated sugar solution and unsaturated sugar solution and stirred?stirred?

Factors affecting the Factors affecting the solubility of a substancesolubility of a substance

How could you increase the amount How could you increase the amount of sugar that would eventually of sugar that would eventually dissolve in a glass of tea?dissolve in a glass of tea?

Factors affecting the Factors affecting the solubility of a substancesolubility of a substance

Only Only twotwo factors affect the amount factors affect the amount of solute that can dissolve.of solute that can dissolve.

1.1. Temperature Temperature affects solubility of affects solubility of both solids and gases in liquid both solids and gases in liquid solvents.solvents.

2.2. PressurePressure affects solubility of gases affects solubility of gases in liquid solvents.in liquid solvents.

Factors affecting the Factors affecting the solubility of a substancesolubility of a substance

The solubility of most solid The solubility of most solid substances substances increasesincreases as the as the temperature of the solvent temperature of the solvent increasesincreases. .

For a few substances, the For a few substances, the reversereverse occurs.occurs.

The Effect of Temperature The Effect of Temperature on the Solubility of Solidson the Solubility of Solids

The Effect of Temperature The Effect of Temperature on Gas Solubilityon Gas Solubility

Increasing the temperature of a Increasing the temperature of a dissolved gas solution dissolved gas solution decreasesdecreases the the concentration of the gas. concentration of the gas. Have you ever tried a Have you ever tried a hothot Dr. Pepper? Heat it Dr. Pepper? Heat it in a pan on the stove, pour a cup, and it has in a pan on the stove, pour a cup, and it has no bubbles!no bubbles!

Thermal pollutionThermal pollution occurs when hot water occurs when hot water is added to a lake, the dissolved oxygen is added to a lake, the dissolved oxygen levels fall in the water and it kills the levels fall in the water and it kills the fish.fish.

The Effect of Temperature The Effect of Temperature on Gas Solubilityon Gas Solubility

The Effect of Pressure on The Effect of Pressure on Gas SolubilityGas Solubility

Increasing the pressure of a gas over Increasing the pressure of a gas over the surface of a solvent increases the the surface of a solvent increases the solubility of the gas in the solvent.solubility of the gas in the solvent.

In a bottled soda, the pressure of COIn a bottled soda, the pressure of CO22 over the liquid is high and when the over the liquid is high and when the cap is opened, the pressure is reduced cap is opened, the pressure is reduced and bubbles begin to come out of the and bubbles begin to come out of the solution.solution.

The Effect of Pressure on The Effect of Pressure on Gas SolubilityGas Solubility

Henry’s LawHenry’s Law: at a given : at a given temperature the solubility (S) of a temperature the solubility (S) of a gas in a liquid is directly gas in a liquid is directly proportional to the pressure (P) of proportional to the pressure (P) of the gas above the liquid.the gas above the liquid.

SS11/P/P11 = S = S22/P/P22

Question:Question:

If the solubility of a gas in water is If the solubility of a gas in water is 0.77 g/L at 3.5 atm pressure, what is 0.77 g/L at 3.5 atm pressure, what is the solubility (in g/L) at 1.0 atm? the solubility (in g/L) at 1.0 atm?

(The temperature is held constant at (The temperature is held constant at 2525ooC.)C.)

AnswerAnswer: S: S11/P/P11 = S = S22/P/P22

0.77 0.77 g/Lg/L / 3.5 atm = S / 3.5 atm = S22 / 1.0 atm / 1.0 atm

SS22 = 0.22 g/L = 0.22 g/L

ObjectivesObjectives

1.1. Solve problems involving the Solve problems involving the molarity of a solutionmolarity of a solution

2.2. Describe how to prepare dilute Describe how to prepare dilute solutions from concentrated solutions from concentrated solutions of known molarity.solutions of known molarity.

Solution Composition: Solution Composition: Molarity Molarity

Concentration of a solutionConcentration of a solution is the is the amount of solute in a given volume amount of solute in a given volume of solution. of solution.

A A concentrated solutionconcentrated solution has a high has a high concentration of solute.concentration of solute.

A A diluted solutiondiluted solution has only a low has only a low concentration of solute.concentration of solute.

A unit of solution A unit of solution concentrationconcentration

A common unit of concentration: A common unit of concentration:

Molarity (Molarity (MM))

(Notice that the ratio includes liters of total solution,

not liters of solvent.)

Solution Composition: Solution Composition: Molarity Molarity

Consider both the moles of solute Consider both the moles of solute and the volume of solution to find and the volume of solution to find concentration. concentration.

Solution Composition: Solution Composition: Molarity Molarity

To find the moles of solute in a given To find the moles of solute in a given volume of solution of known molarity volume of solution of known molarity use the definition of molarity. use the definition of molarity.

Solution Preparation: Solution Preparation: Molarity Molarity

• To make a molar solution: – Weigh out a sample of solute that contains the

necessary moles of solute. – Transfer to a volumetric flask. – Add only enough solvent to mark on flask to get total

solution volume.

Problem:Problem:

Calculate the molarity when 3.0 mol Calculate the molarity when 3.0 mol of solute is dissolved in enough of solute is dissolved in enough water to make 2.0 L of solution.water to make 2.0 L of solution.

Answer:Answer:

mol solute/L solution = molaritymol solute/L solution = molarity

3.0 mol / 2.0 L = 1.53.0 mol / 2.0 L = 1.5MM

ProblemProblem A solution contains 17.0 grams of A solution contains 17.0 grams of

sodium chloride in 200. mL of solution. sodium chloride in 200. mL of solution. What is the molarity of the solution?What is the molarity of the solution?

Answer:Answer: (convert grams to moles) (convert grams to moles)17.0 grams 17.0 grams .. (1 mol/58.5 g) = 0.2906 mol (1 mol/58.5 g) = 0.2906 mol

mol solute/L solution = molaritymol solute/L solution = molarity0.2906 mol / 0.200 L = 0.145 molar 0.2906 mol / 0.200 L = 0.145 molar NaClNaCl

ProblemProblem

How many grams of sodium chloride are How many grams of sodium chloride are needed to prepare 275 mL of a 0.200needed to prepare 275 mL of a 0.200MM solution of NaCl?solution of NaCl?

Answer:Answer:

molarity molarity .. L solution = mol solute L solution = mol solute

0.2000.200M M .. 0.275 L = 0.0550 moles of NaCl 0.275 L = 0.0550 moles of NaCl

0.0550 mol 0.0550 mol .. (58.5 g/1 mol) =3.22 grams (58.5 g/1 mol) =3.22 grams NaClNaCl

Dilution Dilution When diluting a solution to a lower When diluting a solution to a lower

concentration, only water is added concentration, only water is added in the dilution – the amount of in the dilution – the amount of solute remains the same. solute remains the same. ((remember: remember: mol solute = molarity mol solute = molarity .. L solution) L solution)

(mol solute)(mol solute)concconc = (mol solute) = (mol solute)diluteddiluted

MMconcconc .. L Lconcconc = = MMdiluteddiluted .. L Ldiluteddiluted

MMconcconc .. V Vconcconc = = MMdiluteddiluted .. V Vdiluteddiluted

MM11.. V V11 = = MM22

.. VV22

Dilution Dilution Diluting a solution Diluting a solution

Transfer a measured amount of original Transfer a measured amount of original solution to a flask containing some solution to a flask containing some water. water.

Add water to the flask to the mark (with Add water to the flask to the mark (with swirling) and mix by inverting the flask. swirling) and mix by inverting the flask.

MM1 1 .. V V11 = = MM2 2

.. VV22

ProblemProblem How many mL of a concentrated stock How many mL of a concentrated stock

solution of 5.0solution of 5.0MM HCl is needed to prepare HCl is needed to prepare 250. mL of 1.75250. mL of 1.75MM HCl? HCl?

Answer: Answer: MM1 1

.. V V11 = = MM2 2 .. VV22

5.05.0MM .. V V11 = 1.75 = 1.75MM .. 250 mL 250 mL

VV11 = 87.5 mL = 87.5 mL

(notice: you do not have to convert to liters in (notice: you do not have to convert to liters in dilution problems)dilution problems)

ObjectivesObjectives

Explain on a particle basis why a Explain on a particle basis why a solution has a lower vapor pressure solution has a lower vapor pressure than the pure solvent of that solutionthan the pure solvent of that solution

Explain on a particle basis why a Explain on a particle basis why a solution has an elevated boiling point solution has an elevated boiling point and a depressed freezing point and a depressed freezing point compared to the pure solventcompared to the pure solvent

Calculate the molality and mole fraction Calculate the molality and mole fraction of a solutionof a solution

Calculate the molar mass of a molecular Calculate the molar mass of a molecular compound from the freezing point compound from the freezing point depression or boiling point elevation of depression or boiling point elevation of a solution of the compounda solution of the compound

Colligative Properties Colligative Properties

Colligative propertyColligative property – a – a solution property that depends on solution property that depends on the number, but not the type, of the number, but not the type, of solute particles present solute particles present

There are three colligative There are three colligative properties of solutions:properties of solutions:

1.1. Vapor pressure loweringVapor pressure lowering

2.2. Boiling point elevationBoiling point elevation

3.3. Freezing point depressionFreezing point depression

Vapor Pressure LoweringVapor Pressure Lowering

When solute particles When solute particles are added to a are added to a solvent, the solute solvent, the solute forms associations forms associations with the solvent. with the solvent. Fewer solvent Fewer solvent particles are free to particles are free to vaporize, so the vaporize, so the vapor pressure is vapor pressure is lower over a solution lower over a solution than over the pure than over the pure solvent.solvent.

Colligative PropertiesColligative Properties Adding solute Adding solute

particles causes the particles causes the liquid range to liquid range to become wider. become wider. The The lowering of vapor lowering of vapor pressure causes the pressure causes the boiling point to rise boiling point to rise and the freezing and the freezing point to fall. point to fall. ( interferes with ( interferes with crystal formation)crystal formation)

Boiling point Boiling point increasesincreases

Freezing point Freezing point decreasesdecreases

Colligative PropertiesColligative Properties

The higher the number of particles (ions, The higher the number of particles (ions, molecules, etc.) the great the change in the molecules, etc.) the great the change in the colligative property.colligative property.

Which would have the greatest elevation Which would have the greatest elevation in boiling point?in boiling point?

1.1. 1.01.0MM C C66HH1212OO66

2.2. 1.01.0MM NaCl NaCl

3.3. 1.01.0MM CaCl CaCl22

4.4. 1.01.0MM AlCl AlCl33

Answer:Answer: All 4 solutions have the same All 4 solutions have the same

concentration, but the number of concentration, but the number of particles that form is not the same:particles that form is not the same:

CC66HH1212OO6(s) 6(s) CC66HH1212OO6(aq) 6(aq)

(1 particle)(1 particle) NaClNaCl(s)(s) Na Na++

(aq)(aq) + Cl + Cl--(aq)(aq)

(2 particles)(2 particles) CaClCaCl2(s)2(s) Ca Ca2+2+

(aq)(aq) + 2Cl + 2Cl--(aq)(aq)

(3 particles)(3 particles) AlClAlCl3(s) 3(s) Al Al3+ 3+

(aq)(aq)+ 3Cl+ 3Cl--(aq)(aq)

(4 particles, greatest (4 particles, greatest effect)effect)

(Remember: ionic compounds dissociate.)(Remember: ionic compounds dissociate.)

Another unit of solution Another unit of solution concentrationconcentration

Another common unit of Another common unit of concentration: concentration:

Molality (Molality (mm))

(Notice that the ratio includes liters of solvent,

not liters of total solution.)

MolalityMolality

Calculate the molality of a solution Calculate the molality of a solution prepared by dissolving 45.0 grams of KCl prepared by dissolving 45.0 grams of KCl in 500. grams of water.in 500. grams of water.

Answer:Answer:

45.0 grams KCl 45.0 grams KCl .. (74.55 g/1 mol) =0.604 (74.55 g/1 mol) =0.604 molmol

mm = mol solute/kg solvent = mol solute/kg solvent

0.604 mol KCl/ .500 kg water = 1.210.604 mol KCl/ .500 kg water = 1.21 molalmolal

MolalityMolality How many grams of sodium chloride must How many grams of sodium chloride must

be dissolved in 250.0 grams of water to be dissolved in 250.0 grams of water to prepare a 0.75 molal NaCl solution?prepare a 0.75 molal NaCl solution?

Answer:Answer:mol solute = mol solute = mm .. kg solvent kg solvent

mol = 0.75mol = 0.75mm .. 0.250 kg 0.250 kgwater water = 0.19 mol = 0.19 mol NaClNaCl

0.19 mol NaCl 0.19 mol NaCl . . (58.5 g/1 mol) = 11 grams (58.5 g/1 mol) = 11 grams NaClNaCl

Calculating Solution Calculating Solution Boiling PointBoiling Point

The boiling point is raised by the addition The boiling point is raised by the addition of a nonvolatile solute. To calculate the of a nonvolatile solute. To calculate the change in boiling point, you must consider change in boiling point, you must consider the concentration of the particles in the concentration of the particles in solution.solution.

Remember: water normally boils at 100Remember: water normally boils at 100ooC.C.

T = KT = Kbb ii mmchange in boiling point = change in boiling point =

(boiling point constant) (number if ions) (boiling point constant) (number if ions) (molality)(molality)

ProblemProblem

What is the boiling point of 2.75What is the boiling point of 2.75mm NaCl NaCl(aq)(aq) solution? (Ksolution? (Kbb for water is 0.512 for water is 0.512ooC/C/mm))

Answer:Answer:

T = KT = Kbb ii mmT =T = (0.512(0.512ooC/C/mm) (2) (2.75) (2) (2.75mm))T =T = 2.822.82ooC C (this is the (this is the changechange in boiling in boiling point)point)Boiling point = 100Boiling point = 100ooC + 2.82C + 2.82ooC = C = 102.82102.82ooCC

Calculating Solution Calculating Solution Freezing PointFreezing Point

You calculate the freezing point of a You calculate the freezing point of a solution in a similar way.solution in a similar way.

Remember: water normally freezes at Remember: water normally freezes at 0.00.0ooC.C.

T = KT = Kff ii mm

change in freezing point = change in freezing point =

(freezing point constant) (number if (freezing point constant) (number if ions) (molality)ions) (molality)

ProblemProblem

What is the freezing point of a solution that What is the freezing point of a solution that contains 2.50 mol of CaClcontains 2.50 mol of CaCl22 in 1250 g of in 1250 g of water?water?

(K(Kf f for water is 1.86for water is 1.86ooC/C/mm))

Answer:Answer:

T = KT = Kff ii mm

T = (1.86T = (1.86ooC/C/m) m) (3) (2.50 mol/1.250 kg)(3) (2.50 mol/1.250 kg)

= 11.2= 11.2ooC C (this is the freezing point change)(this is the freezing point change)

Freezing point = 0.0Freezing point = 0.0ooC C -- 11.2 11.2ooC = C = --11.211.2ooCC

Determining Molar MassDetermining Molar Mass

You can use the change in boiling point or You can use the change in boiling point or freezing point to calculate the molar mass of freezing point to calculate the molar mass of an unknown solute.an unknown solute.

If you are given a mass of an unknown If you are given a mass of an unknown solute, you can use the boiling point solute, you can use the boiling point elevation to elevation to determine the molalitydetermine the molality of the of the solution, then use the volume of solvent to solution, then use the volume of solvent to determine the moles of solutedetermine the moles of solute. Use the given . Use the given mass and the moles of solute to mass and the moles of solute to determine the determine the molar mass of the solute.molar mass of the solute.

Problem:Problem: The boiling point of water is raised to The boiling point of water is raised to

102.5102.5ooC when 65.50 grams of a C when 65.50 grams of a nonvolatile nonvolatile molecularmolecular solute is dissolved solute is dissolved in 500. g of water. Calculate the molar in 500. g of water. Calculate the molar mass of the solute.mass of the solute.

Answer:Answer:

T = KT = Kff ii mm2.52.5ooC = (.512C = (.512ooC/C/mm) (1) (mol solute/.500 ) (1) (mol solute/.500 kg)kg)mol solute = 2.44 molmol solute = 2.44 mol65.50 grams/2.44 mol = 26.8 g/mol65.50 grams/2.44 mol = 26.8 g/mol