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SolutionsSolutions

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Some DefinitionsSome Definitions

A solution is aA solution is a mixture mixture of 2 or more of 2 or more substances in a substances in a single phase. single phase.

One part is usually One part is usually called the called the SOLVENTSOLVENT and the other part is and the other part is the the SOLUTESOLUTE..

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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 Zinc, Copper

solid liquid Salt, Water

gas solid Hydrogen gas, Metal

liquid liquid Creamer, Coffee

gas liquid Carbon Dioxide, water

gas gas Nitrogen in Oxygen

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DefinitionsDefinitionsSolutions can be classified as Solutions can be classified as

saturatedsaturated or or ununsaturatedsaturated..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

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.

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DefinitionsDefinitionsSUPERSATURATED SOLUTIONS:SUPERSATURATED SOLUTIONS:

contains more solute than should be contains more solute than should be possible to be dissolvedpossible to be dissolved

• Supersaturated solutions are unstableSupersaturated solutions are unstable– only temporaryonly temporary

2 Ways to make a supersaturated solution:2 Ways to make a supersaturated solution:

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

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

https://www.youtube.com/watch?v=XSGvy2FPfCw

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SupersaturatedSupersaturatedSodium AcetateSodium Acetate

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

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Two types of aqueous Two types of aqueous SolutionsSolutions

• Ionic compounds will disassociate in waterIonic compounds will disassociate in water

• Covalent Compounds do not disassociate they Covalent Compounds do not disassociate they only dissolveonly dissolve

• DissasociationDissasociation

http://www.youtube.com/watch?v=EBfGcTAJF4o&safe=active

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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 Conduct The solutions Conduct ElectricityElectricity

They are called They are called ELECTROLYTESELECTROLYTES

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

Aqueous Aqueous SolutionsSolutions

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Aqueous 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

Examples include:Examples include:sugarsugarethanolethanolethylene glycolethylene glycol

10Rate of Dissolution

Factors that affect the rate of dissolution (make solute dissolve faster):

1. Increase the surface area of the solute (crush solute into small pieces)

2. Agitate the solution (brings solvent into increased contact with solute)

3. Heating the solvent (increases KE of solute particles on surface)

11 Factors that affect a substance’s solubility: Amount of solute (in grams) Amount of solvent (in grams) Specified temperature (in °C)

All 3 of these factors are shown on a solubility curve.

A solubility curve shows the trend in solubility of a substance at a given temperature range

12 Three types of solutions, in terms of solubility (see solubility curve): On the line

Below the line

Above the line

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1. How many grams of NaCl are required to make a saturated solution at 50ºC?

2. A supersaturated solution at 70ºC contains 132 g of solute in 100 g of water. Which compound does this solution contain?

3. Which of these is an unsaturated solution?

F- 60 g of KNO3 dissolved in 200 g of H2O at 10°CG -90 g of NaNO3 dissolved in 100 g of H2O at

20°CH -35 g of KCl dissolved in 100 g of H2O at 60°CJ -40 g of NaCl dissolved in 75 g of H2O at 90°C

4. Which compound is least soluble in 100g of water at 25ºC?

5. Is 45g of KCl at 80ºC: Saturated, unsaturated, or supersaturated?

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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.

Molarity (M) = moles soluteliters of solution

Find the equation on your Reference Sheet

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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.

16PROBLEM: Dissolve 5.00 g of NiClPROBLEM: Dissolve 5.00 g of NiCl22•6 H•6 H22O O in enough water to make 250 mL of in enough water to make 250 mL of solution. Calculate the Molarity.solution. Calculate the Molarity.

PROBLEM: Dissolve 5.00 g of NiClPROBLEM: Dissolve 5.00 g of NiCl22•6 H•6 H22O O in enough water to make 250 mL of in enough water to make 250 mL of solution. Calculate the Molarity.solution. Calculate the Molarity.

Step 2: Step 2: Calculate moles of the Calculate moles of the SoluteSolute

Step 3: Step 3: Calculate Volume in LCalculate Volume in L

Step 1Step 1 List out all your variables List out all your variablesM (mol/L)= ?M (mol/L)= ?moles=? moles=? Volume(L)=?Volume(L)=?

Step 4: Step 4: Calculate Molarity using Calculate Molarity using the appropriate equationthe appropriate equation

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USING MOLARITYUSING MOLARITYUSING MOLARITYUSING MOLARITY

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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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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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

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Two 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 solute

kilograms solvent

21Calculating 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.

22Calculating Calculating ConcentrationsConcentrations

Calculate molalityCalculate molality

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).

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

4.00 molal

%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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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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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.

• 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

28Change 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 22

CaClCaCl22 33

CaCa33(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

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Change 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 m

2.2. ∆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 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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ACID-BASE REACTIONSACID-BASE REACTIONSTitrationsTitrations

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

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

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TitrationTitrationTitrationTitration

1. 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.