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SolutionsSolutionsWhy does a raw egg swell or shrink when Why does a raw egg swell or shrink when

placed in different solutions?placed in different solutions?

Chemistry I – Chapters 15 & 16

Chemistry I HD – Chapter 15

ICP – Chapter 22

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Some DefinitionsSome DefinitionsA solution is a A solution is a

______________________________ 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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Parts of a Solution• SOLUTE – the part

of a solution that is being dissolved (usually the lesser amount)

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

• Solute + Solvent = Solution

Solute Solvent Example

solid solid

solid liquid

gas solid

liquid liquid

gas liquid

gas gas

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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.temperature.

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 temperatureparticular temperature

5Example: Saturated and Unsaturated Fats

Unsaturated fats have at least one double bond between carbon atoms; monounsaturated means there is one double bond, polysaturated means there are more than one double bond. Thus, there are some bonds that can be broken, chemically changed, and used for a variety of purposes. These are REQUIRED to carry out many functions in the body. Fish oils (fats) are usually unsaturated. Game animals (chicken, deer) are usually less saturated, but not as much as fish. Olive and canola oil are monounsaturated.

Saturated fats are called saturated because all of the bonds between the carbon atoms in a fat are single bonds. Thus, all the bonds on the carbon are occupied or “saturated” with hydrogen. These are stable and hard to decompose. The body can only use these for energy, and so the excess is stored. Thus, these should be avoided in diets. These are usually obtained from sheep and cattle fats. Butter and coconut oil are mostly saturated fats.

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DefinitionsDefinitionsSUPERSATURATED SOLUTIONSSUPERSATURATED SOLUTIONS

contain more solute than is contain more solute than is possible to be dissolvedpossible to be dissolved

Supersaturated solutions are unstable. Supersaturated solutions are unstable. The supersaturation is only The supersaturation is only temporary, and usually temporary, and usually accomplished in one of two ways:accomplished in one of two ways:

1.1. Warm the solvent so that it will Warm the solvent so that it will dissolve more, then cool the dissolve more, then cool the solution solution

2. Evaporate some of the solvent carefully so that the solute does not solidify and come out of solution.

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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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IONIC COMPOUNDSIONIC COMPOUNDSCompounds in Aqueous SolutionCompounds in Aqueous Solution

Many reactions involve ionic Many reactions involve ionic compounds, especially reactions in compounds, especially reactions in water — water — aqueous solutions.aqueous solutions.

KMnOKMnO44 in water in water KK++(aq) + MnO(aq) + MnO44--(aq)(aq)

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How do we know ions are How do we know ions are present in aqueous present in aqueous solutions?solutions?

The solutions The solutions __________________________________________________

They are called They are called ELECTROLYTESELECTROLYTES

HCl, MgClHCl, MgCl22, and NaCl are , and NaCl are strong electrolytesstrong electrolytes. . They dissociate completely They dissociate completely (or nearly so) into ions.(or nearly so) into ions.

Aqueous Aqueous SolutionsSolutions

10Aqueous Aqueous SolutionsSolutions

Some compounds Some compounds dissolve in water but do dissolve in water but do not conduct electricity. not conduct electricity. They are called They are called nonelectrolytes.nonelectrolytes.

Examples include:Examples include:sugarsugarethanolethanolethylene glycolethylene glycol

11It’s Time to Play Everyone’s Favorite Game Show… Electrolyte

or Nonelectrolyte!

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Electrolytes in the BodyElectrolytes in the Body

Carry messages to Carry messages to

and from the brain and from the brain

as electrical signalsas electrical signals

Maintain cellular Maintain cellular

function with the function with the

correct correct

concentrations concentrations

electrolyteselectrolytes

Make your ownMake your own50-70 g sugar50-70 g sugarOne liter of warm waterOne liter of warm waterPinch of saltPinch of salt200ml of sugar free fruit 200ml of sugar free fruit

squashsquashMix, cool and drinkMix, cool and drink

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Concentration 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.

Molarity (M) = moles soluteliters of solution

141.0 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 NiClPROBLEM: Dissolve 5.00 g of NiCl22 in enough water to make 250 mL in enough water to make 250 mL of solution. Calculate the of solution. Calculate the Molarity.Molarity.

Step 1: Step 1: Calculate moles Calculate moles of NiClof NiCl22

mol 0.0385 = g 129.6

mol 1 • g .005

M 0.154 = L 0.250mol 0.0385

Step 2: Step 2: Calculate MolarityCalculate Molarity

[NiCl2] = 0.154 M

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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 LStep 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 molesStep 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 MOLARITY

moles = M•Vmoles = M•V

What mass of oxalic acid, What mass of oxalic acid, HH22CC22OO44, is, isrequired to make 250. mL of a 0.0500 Mrequired to make 250. mL of a 0.0500 Msolution?solution?

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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 g2) 48 g3) 300 g

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An An IDEAL SOLUTIONIDEAL SOLUTION is is one where the properties one where the properties depend only on the depend only on the concentration of solute.concentration of solute.

Need conc. units to tell us the Need conc. units to tell us the number of solute particles number of solute particles per solvent particle.per solvent particle.

The unit “molarity” does not The unit “molarity” does not do this!do this!

Concentration UnitsConcentration Units

19Two Other Concentration Two Other Concentration UnitsUnits

grams solutegrams solutegrams solutiongrams solution

MOLALITY, mMOLALITY, m

% by mass% by mass = =

% by mass% by mass

m of solution = mol solutekilograms solvent

20Calculating Calculating ConcentrationsConcentrations

Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of HH22O. Calculate molality and % by mass of ethylene O. Calculate molality and % by mass of ethylene glycol.glycol.

21Calculating Calculating ConcentrationsConcentrations

Calculate molalityCalculate molality

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).

conc (molality) = 1.00 mol glycol0.250 kg H2O

4.00 molal

%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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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

25Colligative 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.• Vapor pressure Vapor pressure decreasesdecreases• Melting point Melting point decreasesdecreases• Boiling point Boiling point increasesincreases• Osmosis is possible (osmotic pressure)Osmosis is possible (osmotic pressure)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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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

27Change 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 Bloomington 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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Boiling Point Elevation Boiling Point Elevation and Freezing Point and Freezing Point

DepressionDepression ∆∆T = K•m•iT = K•m•ii = van’t Hoff factor = number of particles i = van’t Hoff factor = number of particles

produced per molecule/formula unit. For produced per molecule/formula unit. For covalent compounds, i = 1. For ionic covalent compounds, i = 1. For ionic compounds, i = the number of ions compounds, i = the number of ions present (both + and -)present (both + and -)

CompoundCompound Theoretical Value of iTheoretical Value of iglycolglycol 11NaClNaCl 22CaClCaCl22 33CaCa33(PO(PO44))22 55

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Boiling Point Elevation Boiling Point Elevation and Freezing Point and Freezing Point

DepressionDepression ∆∆T = K•m•iT = K•m•i

Substance Kb

benzene 2.53 camphor 5.95 carbon tetrachloride 5.03 ethyl ether 2.02 water 0.52

m = molalitym = molalityK = molal freezing K = molal freezing point/boiling point constant point/boiling point constant

Substance Kf

benzene 5.12 camphor 40. carbon tetrachloride 30. ethyl ether 1.79 water 1.86

32Change in Boiling Point Change in Boiling Point Dissolve 62.1 g of glycol (1.00 mol) in 250. g of Dissolve 62.1 g of glycol (1.00 mol) in 250. g of

water. What is the boiling point of the water. What is the boiling point of the solution?solution?

KKbb = 0.52 = 0.52 ooC/molal for water (see KC/molal for water (see Kbb table). table).SolutionSolution ∆T∆TBPBP = K = Kbb • m • i • m • i

1.1. Calculate solution molality = 4.00 mCalculate solution molality = 4.00 m2.2. ∆T∆TBPBP = K = Kbb • m • i • m • i ∆∆TTBPBP = 0.52 = 0.52 ooC/molal (4.00 molal) (1)C/molal (4.00 molal) (1) ∆∆TTBPBP = 2.08 = 2.08 ooCC BP = 100 + 2.08 = 102.08 BP = 100 + 2.08 = 102.08 ooC C

(water normally boils at 100)(water normally boils at 100)

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Calculate the Freezing Point of a 4.00 molal Calculate the Freezing Point of a 4.00 molal glycol/water solution.glycol/water solution.

KKff = 1.86 = 1.86 ooC/molal (See KC/molal (See Kff table) table)SolutionSolution∆∆TTFPFP = K = Kff • m • i • m • i = (1.86 = (1.86 ooC/molal)(4.00 m)(1)C/molal)(4.00 m)(1)

∆∆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)

Freezing Point Freezing Point DepressionDepression

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At what temperature will a 5.4 molal solution At what temperature will a 5.4 molal solution of NaCl freeze?of NaCl freeze?

SolutionSolution

∆∆TTFPFP = K = Kff • m • i • m • i

∆ ∆TTFPFP = (1.86 = (1.86 ooC/molal) • 5.4 m • 2C/molal) • 5.4 m • 2

∆ ∆TTFP FP = 20.1= 20.1 ooCC

FP = 0 – 20.1 = -20.1 FP = 0 – 20.1 = -20.1 ooCC

Freezing Point Freezing Point DepressionDepression

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Preparing 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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ACID-BASE REACTIONSACID-BASE REACTIONSTitrationsTitrations

HH22CC22OO44(aq) + 2 NaOH(aq) --->(aq) + 2 NaOH(aq) ---> acidacid basebase

NaNa22CC22OO44(aq) + 2 H(aq) + 2 H22O(liq)O(liq)Carry out this reaction using a Carry out this reaction using a TITRATIONTITRATION..

Oxalic acid,Oxalic acid,

HH22CC22OO44

37Setup for titrating an acid with a baseSetup for titrating an acid with a base

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TitrationTitration1. Add solution from the buret.1. Add solution from the buret.2. Reagent (base) reacts with 2. Reagent (base) reacts with

compound (acid) in solution compound (acid) in solution in the flask.in the flask.

3.3. Indicator shows when exact Indicator shows when exact stoichiometric reaction has stoichiometric reaction has occurred. (Acid = Base)occurred. (Acid = Base)

This is called This is called NEUTRALIZATION.NEUTRALIZATION.