H 2 O (Chap. 15)

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1 H 2 O (Chap. 15)

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H 2 O (Chap. 15) . H 2 O - Liquid. H 2 0 is covalently bonded results in a polar molecule hydrogen bonding occurs between water molecules unique properties because of all of this: high surface tension molecules are drawn inward because there is nothing pulling them outward. - PowerPoint PPT Presentation

Transcript of H 2 O (Chap. 15)

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H2O (Chap. 15)

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H2O - Liquid

• H20 is covalently bonded• results in a polar molecule• hydrogen bonding occurs between water

molecules• unique properties because of all of this:

1. high surface tension»molecules are drawn inward because

there is nothing pulling them outward

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• a surfactant can interfere with the hydrogen bonding

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2. low vapor pressure»hydrogen bonding helps hold the

water together and not let it escape and enter the vapor/gaseous phase (boil or evaporate)

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• ice is less dense than waterH2O - Solid

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• ice has an open framework of H20 molecules arranged in a hexagonal shape

• this expands the ice crystal-- increases its volume, mass stays the same, and therefore density decreases

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Solutions (Homogeneous Solutions (Homogeneous mixtures)mixtures)

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• a a solution solution is a is a solutesolute that is dissolved in a that is dissolved in a solventsolvent

• an an aqueous solution aqueous solution is a is a solutesolute that is that is dissolved in a dissolved in a waterwater

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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 metal alloy

solid liquid salt water

gas solid Hg in Ag?

liquid liquid alcohol and water

gas liquid carbonated water

gas gas air

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14• most ionic compounds and polar covalent

molecules dissolve readily in water (“like disolves like”)

– solvation: ions (+ and -) break away from the crystal and surrounded by the water molecules

• nonpolar covalent molecules (oil, grease, fuels, do not dissolve in water

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Electrolytes• because there are ions in a solution, they because there are ions in a solution, they

can conduct electricitycan conduct electricity• the electric current can travel on the the electric current can travel on the

mobile chargesmobile charges• HCl, MgClHCl, MgCl22, and NaCl are , and NaCl are strong

electrolytes.– they dissociate completely (or nearly they dissociate completely (or nearly so) into ions.so) into ions.

• a compound that does not conduct a compound that does not conduct electricity (because it is insoluble electricity (because it is insoluble molecular compounds) are molecular compounds) are nonelectrolytesnonelectrolytes

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17It’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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• suspensions–a mixture from which particle settle out

upon standing–particles are larger (1000 nm) that in a

solution and cannot stay suspended like they can in a solution (1nm)

–examples: flour in water, clay in water

Heterogeneous Aqueous Heterogeneous Aqueous SystemsSystems

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• colloids–particles (1nm-1000nm)do not settle out

and therefore have cloudy appearance if concentrated

–examples: glue, gelatin, paint, smoke–particles will scatter light which is called

the Tyndall effect

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• under a microscope, can see the movement of the particles (Brownian motion)–these collisions prevent the particles

from settling

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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

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24Example: 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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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

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281.0 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.

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

5.00 g • 1 mol237.7 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

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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

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33Two 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

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34Calculating 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.

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35Calculating 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

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39Colligative 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

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41Change 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

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46Change 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

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51Setup 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.