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I. The Nature of Solutions

Solutions

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• DEFINITIONS – SOLUTION – SOLUTE – SOLVENT – HOMOGENEOUS MIXTURE

• CONCENTRATION – DILUTE VS CONCENTRATED

• THE NATURE OF SOLUTIONS

• SOLVATION

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Some DefinitionsSome DefinitionsA solution is a A solution is a

HOMOGENEOUS HOMOGENEOUS mixture of 2 or more mixture of 2 or more substances in a substances in a single phase. single phase.

One constituent is One constituent is usually regarded as usually regarded as the the SOLVENTSOLVENT and and the others as the others as SOLUTESSOLUTES..

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A. Definitions

• Solution - Solution - homogeneous mixture

Solvent Solvent - present in greater amount

Solute Solute - substance being dissolved

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Parts of a Solution• SOLUTE – the part of a solution that is

being dissolved (usually the lesser amount). Uniformly spread in the solvent

• SOLVENT – the part of a solution that dissolves the solute (usually the greater amount)

• Solute + Solvent = Solution

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What happens when a solute dissolves in a solvent?

• SolvationSolvation – – the process of dissolving

solute particles are separated and pulled into solution

solute particles are surrounded by solvent particles

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How Does a Solution Form?

As a solution forms, the solvent pulls solute particles apart and surrounds, or solvates, them.

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CHARACTERISTICS OF A LIQUID SOLUTION

• 1.- Homogeneous mixtures, particles are evenly spread.

• 2.- Dissolved particles are too small to be seen, therefore solutions are clear and do not disperse light.

• 3.- Can not be separated by filtration. Dissolved particles are too small and will pass trough any filter.

• 4.- Stable. Dissolved particles will not come out of the solution and will not settle.

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Solutions are not always liquids• Solutions are homogeneous mixtures of

two or more pure substances.

• In a solution, the solute (present in smaller amount) is dispersed uniformly throughout the solvent (present in largest amount).

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Solutions

Homogeneous mixtures.

Solvation is the process by which the solution forms.

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SOLUBILITY

• FACTORS THAT AFFECT SOLUBILITY

• NATURE OF SUBSTANCES

• TEMPERATURE

• PRESSURE

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Solubility

A measure of how much solute can be dissolved in an amount of solvent at a given temperature.

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A substance can be…

• Soluble in a solvent.

• Example: sugar is soluble in water.

• Miscible is the term used when the two components are liquids and they dissolve in one another.

• Example: alcohol and water are miscible

• Insoluble in a solvent

• Example: sand is insoluble in water.

• Immiscible is the term used when the two components are liquids and they do not mix.

• Example: oil and water are immiscible

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What affects Solubility?

1. Nature of Solute

2. TemperatureTemperature

3. Pressure

* graph

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Nature of Solute

• A polar solute molecule (alcohol) dissolves in a polar solvent (water).

• A nonpolar solute (oil paint) dissolves in a nonpolar solvent (turpentine)

“Like Dissolves Like”

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When temperature increases…

• Solubility of a gas decreases

• Solubility of a solid increases

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3 Pressure• Makes gas more soluble

ex. Soda canex. Soda can

• Has almost no effect on liquids and solids

-High pressure forces carbon dioxide into water to make soda.

- When you open the cap, there is less pressure on the soda b/c the soda fizzes and gas escapes.

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• Gases are more soluble athigh pressures

EX: nitrogen narcosis, the “bends,” soda

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April 20

• Objective: How do chemists represent solubility’s dependence on temperature?

• Vocabulary:

• Solubility

• Saturated Solution

• Unsaturated Solution

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Do now

• Read and annotate

• FOCUS QUESTION• According to your reading, what is

the difference between saturated and unsaturated solutions?

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Solubility

A measure of how much solute can be dissolved in an 100g of solvent at a given temperature.

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Review Concepts

• As temperature increases solubility

• Increases for solid solutes

• Decreases for gases solutes

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DefinitionsDefinitionsSolutions can be classified as Solutions can be classified as

saturatedsaturated or or ununsaturatedsaturated..A A saturatedsaturated solution contains solution contains

the maximum quantity of the maximum quantity of solute that dissolves at that solute that dissolves at that temperature. CAN NOT temperature. CAN NOT DISSOLVE MORE SOLUTEDISSOLVE MORE SOLUTE

An An unsaturatedunsaturated solution solution contains less than the contains less than the maximum amount of solute maximum amount of solute that can dissolve at a that can dissolve at a particular temperature.particular temperature.

CAN DISSOLVE MORE SOLUTECAN DISSOLVE MORE SOLUTE

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Solubility Curves

Represent the solubility of Represent the solubility of different solutes in 100 g of different solutes in 100 g of water at different temperatures. water at different temperatures.

Each point on the curve represents a saturated solution at a given temperature.

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C. Solubility

• Solubility CurveSolubility Curve– shows the dependence

of solubility on temperature

– In the Y axis (dependent variable)

– g solute/ 100g H2O

Remember

100g H2O = 100 ml H2O

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Example 1

• 1 -How much NaNO3 can be dissolved at 10 o C in

100 g of water

• The resulting solution is a saturated solution at 10 o C.

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What happens if the temperature of the solution in increased from 10 oC to 50 oC?• Would that solution still be saturated?

• What is the solubility of NaNO3 at 50 o C?

• How much more solute can be dissolved?

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Problem 6

• Since the solubility of a solid increases as the temperature increases

• Solubility at 50 o C ~115 g NaNO3/100 g H2O

Solubility at 10 o C ~ 80 g NaNO3/100 g H2O

115g-80g= 35 g

• Additional 35 g of NaNO3 can be dissolved in the saturated solution at 10C to make it saturated at 50C

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• When the problem gives 2 temperatures get the solubility at the 2 different temperatures and analyze what would happen.

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Example 2

What would happen if a 100 ml of a saturated solution of NH4Cl at 90 o C is cooled to 25 o C?

(Hint find the solubility at each temperature!)

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Example 3

• How much NH4Cl is found in 200 ml of a saturated solution at 90 o C?

• And in 50 mL at the same temperature?

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Group Activity

• Get in your groups and work with the handout problems.

• Each group is responsible to hand out one challenge problem (11 to 17) clearly showing the steps to solve it AND a sentence explaining the answer.

• Homework

1 - Each student is responsible for 2 other problems of the challenge section (3 in total counting the one done in class).

2.-Read and annotate your RB P 121-122

Do questions 13 to 18 page 123

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Exit Slip

• How much KNO3 can be dissolved in 200g of H2O at 60 oC?

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Solubility curves worksheet answers

• 1 KI

• 2 KClO3

• 3 SO2

• 4 134 g

• 5 SO2, NH3 and HCl

• 6 Sol at 50 o C ~115 g

at 10 o C ~ 80 g

difference 35 g

7 ~ 47 o C

• 8 KNO3 and NaNO3

• 9 NaCl

• 10 KNO3

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Answer to challenge problems

• 11 Approximately 58 g of NaNO3 has to be added to 50 mL of water to make the solution saturated.

• 12 Approximately 44 g

• 13 38 o C

• 14 116g will precipitate

• 15 60 g

• 16 12 g

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Exit slip

• Indicate what is the relationship between solubility and temperature for most of the substances in table G.

• Why HCl, NH3 and SO2 do not follow the same relationship?

• How much KNO3 can be dissolved in 100g of H2O at 60 oC?

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Supersaturated solutions

• Contain more solute that what is supposed to have a a given temperature.

• They are VERY unstable!!!!

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SupersaturatedSupersaturatedSodium AcetateSodium Acetate

• One application One application of a of a supersaturated supersaturated solution is the solution is the sodium acetate sodium acetate “heat pack.”“heat pack.”

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C. Solubility

SATURATED SOLUTION

no more solute dissolves

UNSATURATED SOLUTIONmore solute dissolves

SUPERSATURATED SOLUTION

becomes unstable, crystals form

concentration

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Solubility for ionic compoundsTable F

• This table is used to predict if a double replacement reaction will occur. If it the reaction produces an insoluble compound

it occurs. If the products of the reaction are filtered the insoluble compound will remain in the filter paper

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Table F

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• Pb(NO3)2 + 2KI PbI2 + 2KNO3

• NaCl + AgNO3 AgCl + NaNO3

• CuSO4 + Na2CO 3 Na2SO4 + CuCO3

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Questions• Is NaCl soluble?

Yes!

• Is AgBr soluble?

No!

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Set 1 Solubility curves

1. 2

2. 1

3. 1(turn page around)

4. 2

5. 4(turn page again)

6. 4

7. 2

8. 3

9. 1

10. 6 to 8 g

11.a- As P decreases, solubility decreases too.

b- As T increases, solubility decreases

12 a KNO3

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April 23• Ways of expressing concentration

• MOLARITY

• % BY MASS, BY VOLUME

• PPM

• Vocabulary

• Concentration – dilute – concentrated

• Molarity – volumetric flask

• Homework: RB page 128 q 24 to 37

49Do Now: check answers to HWP 123 RB

13.2

14.4

15.3

16.1

17.1

18.4

19.4

20.3

21.2

22.1

23.2

• Page 120

8)1

9)1

10)1

11)2

12)3

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CONCENTRATIONThe amount of solute in the solution.

Relative terms

• Diluted: Small amount of solute in relation to the amount of solvent

• Concentrated: Large amount of solute in relation with the solvent.

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Concentration of SoluteConcentration of SoluteConcentration of SoluteConcentration of Solute

The amount of solute in a solution The amount of solute in a solution is given by its is given by its concentrationconcentration.

Quantitative measurements are given by

Molarity (M) = moles soluteliters of solution

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Example 1• If you have 50 moles (mol) of solute (salt) in

25 liters (L) of solution. Find the molarity (concentration).

• Use Table T for the molarity formula.

Molarity = moles of solute / liters of solution

50 moles / 25 liters = 2M

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Example 2

• A solution contains 5 mol of NaCl in 1 Liter of solution. Find the molarity

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Example 3

• If a solution is 6 M , how many mol of solute are present in 1 L of solution?

• How many mol in 2 L of solution?

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Example 3

• How to prepare 1L of a 0.1 M solution of CuSO4. 5H2O?

• Step 1: Find the number of mol of the solute in the solution

• Step 2. Convert mol to mass

• Step 3. Place solute in volumetric flask

• Step 4. Add water to the mark

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Preparing SolutionsPreparing SolutionsPreparing SolutionsPreparing Solutions

• Weigh out a solid Weigh out a solid solute and dissolve in a solute and dissolve in a given quantity of given quantity of solvent.solvent.

• Dilute a concentrated Dilute a concentrated solution to give one solution to give one that is less that is less concentrated.concentrated.

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1.0 L of 1.0 L of water was water was

used to used to make 1.0 L make 1.0 L of solution. of solution. Notice the Notice the water left water left

over.over.

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PROBLEM: Dissolve 5.00 g of PROBLEM: Dissolve 5.00 g of NiClNiCl22•6 H•6 H22O in enough water to O in enough water to make 250 mL of solution. make 250 mL of solution. Calculate the Molarity.Calculate the Molarity.

PROBLEM: Dissolve 5.00 g of PROBLEM: Dissolve 5.00 g of NiClNiCl22•6 H•6 H22O in enough water to O in enough water to make 250 mL of solution. make 250 mL of solution. Calculate the Molarity.Calculate the Molarity.

Step 1: Step 1: Calculate moles Calculate moles of NiClof NiCl22•6H•6H22OO

5.00 g • 1 mol

237.7 g = 0.0210 mol

0.0210 mol0.250 L

= 0.0841 M

Step 2: Step 2: Calculate MolarityCalculate Molarity

[NiClNiCl22•6 H•6 H22OO ] = 0.0841 M

59Calculating Calculating ConcentrationsConcentrations

Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of Hof H22O. Calculate molality and % by mass of O. Calculate molality and % by mass of

ethylene glycol.ethylene glycol.

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Learning Check

How many grams of NaOH are required to prepare 400. mL of 3.0 M NaOH solution?

1) 12 g

2) 48 g

3) 300 g

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Step 1: Step 1: Change mL to L.Change mL to L.

250 mL * 1L/1000mL = 0.250 L250 mL * 1L/1000mL = 0.250 L

Step 2: Step 2: Calculate.Calculate.

Moles = (0.0500 mol/L) (0.250 L) = 0.0125 molesMoles = (0.0500 mol/L) (0.250 L) = 0.0125 moles

Step 3: Step 3: Convert moles to grams.Convert moles to grams.

(0.0125 mol)(90.00 g/mol) = (0.0125 mol)(90.00 g/mol) = 1.13 g1.13 g

USING MOLARITYUSING MOLARITYUSING MOLARITYUSING MOLARITY

moles = M•Vmoles = M•V

What mass of oxalic acid, What mass of oxalic acid, HH22CC22OO44, is, is

required to make 250. mL of a 0.0500 Mrequired to make 250. mL of a 0.0500 Msolution?solution?

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Try this molality problem

• 25.0 g of NaCl is dissolved in 5000. mL of water. Find the molality (m) of the resulting solution.

m = mol solute / kg solvent

25 g NaCl 1 mol NaCl

58.5 g NaCl= 0.427 mol NaCl

Since the density of water is 1 g/mL, 5000 mL = 5000 g, which is 5 kg

0.427 mol NaCl

5 kg water= 0.0854 m salt water

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• DO NOW – REVIEW MOLARITY

• % BY MASS, % BY VOLUME, ppm

• COLLIGATIVE PROPERTIES

• HW SOLUTIONS TAKE HOME TEST!!!!

• DUE MONDAY

APRIL 24

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Two Other Concentration Two Other Concentration UnitsUnits

grams solutegrams solutegrams solutiongrams solution

% by mass% by mass = =

% by mass% by mass

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Two Other Concentration Two Other Concentration UnitsUnits

grams solutegrams solutegrams solutiongrams solution

Ppm =Ppm =

Ppm = parts per millionPpm = parts per million

66Calculating Calculating ConcentrationsConcentrations

Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g

of Hof H22O. Calculate m & % of ethylene glycol (by mass).O. Calculate m & % of ethylene glycol (by mass).

Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g

of Hof H22O. Calculate m & % of ethylene glycol (by mass).O. Calculate m & % of ethylene glycol (by mass).

%glycol = 62.1 g

62.1 g + 250. g x 100% = 19.9%%glycol =

62.1 g62.1 g + 250. g

x 100% = 19.9%

Calculate weight %Calculate weight %

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Learning Check

A solution contains 15 g Na2CO3 and 235 g of

H2O? What is the mass % of the solution?

1) 15% Na2CO3

2) 6.4% Na2CO3

3) 6.0% Na2CO3

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Using mass %

How many grams of NaCl are needed to prepare 250 g of a 10.0% (by mass) NaCl solution?

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Determining Electrical Conductivity

• When a solution is soluble, it has ions that can conduct electricity (electrolytes)

Ex. NaClEx. NaCl

• When a solution is insoluble, it cannot When a solution is insoluble, it cannot conduct electricity (conduct electricity (non-electrolytesnon-electrolytes or or poor electrolytes)poor electrolytes)

Ex. AgBrEx. AgBr

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Colligative PropertiesColligative PropertiesOn adding a solute to a solvent, the properties On adding a solute to a solvent, the properties

of the solvent are modified.of the solvent are modified.

• Melting point Melting point decreasesdecreases

• Boiling point Boiling point increasesincreases

These changes are called These changes are called COLLIGATIVE COLLIGATIVE PROPERTIESPROPERTIES. .

They depend only on the They depend only on the NUMBERNUMBER of solute of solute particles relative to solvent particles, not on particles relative to solvent particles, not on the the KINDKIND of solute particles. of solute particles.

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Freezing point depression and boiling point elevation

• One mole of particles dissolved in a 1000g of water lowers the freezing point of the water by 1.86 C and increases the boiling point of water by 0.52 C.

• Note that electrolytes (ionic substances) produce a greater effect than non electrolytes because they produce more particles in solution.

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Change in Freezing Change in Freezing Point Point

The freezing point of a solution is The freezing point of a solution is LOWERLOWER than that of the pure solventthan that of the pure solvent

Pure waterPure waterEthylene glycol/water Ethylene glycol/water

solutionsolution

73Change in Freezing Change in Freezing Point Point

Common Applications Common Applications of Freezing Point of Freezing Point DepressionDepression

Propylene glycol

Ethylene glycol – deadly to small animals

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Common Applications Common Applications of Freezing Point of Freezing Point DepressionDepression

Which would you use for the streets of New York to lower the freezing point of ice and why? Would the temperature make any difference in your decision?

a) sand, SiO2

b) Rock salt, NaCl

c) Ice Melt, CaCl2

Change in Freezing Change in Freezing Point Point

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Change in Boiling Point Change in Boiling Point Common Applications Common Applications

of Boiling Point of Boiling Point ElevationElevation

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Calculate the Freezing Point of a solution Calculate the Freezing Point of a solution containing 4.00 mol of glycol in a 1000 g of containing 4.00 mol of glycol in a 1000 g of waterwater

KKff = 1.86 = 1.86 ooC/molC/mol

SolutionSolution

Change in FP= (1.86 Change in FP= (1.86 ooC/mol)(4.00 mol)C/mol)(4.00 mol)

∆∆TTFP FP = 7.44 = 7.44

FP = 0 – 7.44 = -7.44 FP = 0 – 7.44 = -7.44 ooCC(because water normally freezes at 0)(because water normally freezes at 0)

If NaCl is used instead the fp will be lower If NaCl is used instead the fp will be lower because the number of particles double.because the number of particles double.

Freezing Point Freezing Point DepressionDepression

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At what temperature will a solution with 4 At what temperature will a solution with 4 mol of NaCl in a 1000 g H2O freeze?mol of NaCl in a 1000 g H2O freeze?

SolutionSolution

NaCl Na + Cl

∆ ∆TTFPFP = (1.86 = (1.86 ooC/molal) • 8 mol C/molal) • 8 mol

∆ ∆TTFP FP = 14.88= 14.88 ooCC

FP = 0 – 14.88 = -14.88 FP = 0 – 14.88 = -14.88 ooCC

Freezing Point Freezing Point DepressionDepression

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RB Answers Colligative Prop

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41)2

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49)4

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