1 ionic compounds compounds in aqueous solution many reactions involve ionic compounds, especially...

1

IONIC COMPOUNDSIONIC COMPOUNDSCompounds in Aqueous SolutionCompounds in Aqueous Solution

Many reactions involve ionic compounds, Many reactions involve ionic compounds,

especially reactions in water — especially reactions in water — aqueous aqueous solutions.solutions.

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

2

An Ionic An Ionic Compound, Compound,

CuClCuCl22, in , in WaterWater

CCR, page 149CCR, page 149

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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 conduct conduct electricity!electricity!

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

4

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

Aqueous Aqueous SolutionsSolutions

5Aqueous Aqueous SolutionsSolutions

Acetic acid ionizes only to a small extent, so Acetic acid ionizes only to a small extent, so it is a it is a weak electrolyte.weak electrolyte.CHCH33COCO22H(aq)H(aq) ---> ---> CHCH33COCO22

--(aq) + H(aq) + H++(aq)(aq)

6AqueouAqueou

s s SolutionSolution

ssAcetic acid ionizes only to a small Acetic acid ionizes only to a small extent, so it is a extent, so it is a weak weak electrolyte.electrolyte.

CHCH33COCO22H(aq)H(aq) ---> ---> CHCH33COCO22--(aq) + (aq) +

HH++(aq)(aq)

7Aqueous 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

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If one ion from the If one ion from the “Soluble Compd.” list is “Soluble Compd.” list is present in a compound, present in a compound, the compound is water the compound is water

soluble.soluble.

Water Solubility of Ionic Water Solubility of Ionic CompoundsCompounds

Screen 5.4 & Figure 5.1

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Water Solubility of Ionic Water Solubility of Ionic CompoundsCompounds

Common minerals are often formed with Common minerals are often formed with anions that lead to insolubility:anions that lead to insolubility:

sulfidesulfide fluoridefluoride

carbonatecarbonate oxideoxide

Azurite, a copper carbonate

Iron pyrite, a sulfideOrpiment, arsenic sulfide

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An acid -------> HAn acid -------> H++ in water in waterAn acid -------> HAn acid -------> H++ in water in water

ACIDSACIDSACIDSACIDS

Some Some strongstrong acids areacids are

HClHCl hydrochlorichydrochloric

HH22SOSO44 sulfuricsulfuric

HClOHClO44 perchloricperchloric

HNOHNO33 nitricnitricHNOHNO33

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An acid -------> HAn acid -------> H++ in water in waterAn acid -------> HAn acid -------> H++ in water in water

ACIDSACIDSACIDSACIDS

HCl(aq) ---> HHCl(aq) ---> H++(aq) + Cl(aq) + Cl--(aq)(aq)

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The The Nature Nature of Acidsof Acids

The The Nature Nature of Acidsof Acids

HCl

H2O

Cl-

H3O+

hydronium ion

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Weak AcidsWeak AcidsWeak AcidsWeak AcidsWEAK ACIDS = weak WEAK ACIDS = weak

electrolyteselectrolytes

CHCH33COCO22HH

acetic acidacetic acid

HH22COCO33 carbonic acidcarbonic acid

HH33POPO44 phosphoric phosphoric acidacid

HFHF

hydrofluoric acidhydrofluoric acid

Acetic acid

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ACIDSACIDSACIDSACIDSNonmetal oxides can be acidsNonmetal oxides can be acids

COCO22(aq) + H(aq) + H22O(liq) O(liq) ---> --->

HH22COCO33(aq)(aq)

SOSO33(aq) + H(aq) + H22O(liq) O(liq) ---> H---> H22SOSO44(aq)(aq)

and can come from burning and can come from burning coal and oil.coal and oil.

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Base ---> OHBase ---> OH-- in water in waterBase ---> OHBase ---> OH-- in water in water

BASESBASESsee Screen 5.9 and Table 5.2see Screen 5.9 and Table 5.2

BASESBASESsee Screen 5.9 and Table 5.2see Screen 5.9 and Table 5.2

NaOH(aq) ---> NaNaOH(aq) ---> Na++(aq) + OH(aq) + OH--(aq)(aq)

NaOH is NaOH is a strong a strong basebase

16Ammonia, NHAmmonia, NH33

An Important Base

17

BASESBASESBASESBASES

Metal oxides are Metal oxides are basesbases

CaO(s) + HCaO(s) + H22O(liq) O(liq)

--> Ca(OH)--> Ca(OH)22(aq)(aq)

CaO in water. Indicator shows solution is basic.

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Know the strong Know the strong acids & bases!acids & bases!

19Net Net Ionic Ionic

EquatioEquationsns

Net Net Ionic Ionic

EquatioEquationsns

Mg(s) + 2 HCl(aq) --> HMg(s) + 2 HCl(aq) --> H22(g) + MgCl(g) + MgCl22(aq)(aq)

We really should writeWe really should write

Mg(s) + 2 HMg(s) + 2 H++(aq) + 2 Cl(aq) + 2 Cl--(aq) ---> (aq) ---> HH22(g) + Mg(g) + Mg2+2+(aq) + 2 Cl(aq) + 2 Cl--(aq)(aq)

The two ClThe two Cl-- ions are ions are SPECTATOR IONSSPECTATOR IONS — — they do not participate. Could have used NOthey do not participate. Could have used NO33

--..

20Net Ionic Net Ionic EquationsEquationsNet Ionic Net Ionic EquationsEquations

Mg(s) + 2 HCl(aq) Mg(s) + 2 HCl(aq) --> H--> H22(g) + MgCl(g) + MgCl22(aq)(aq)

Mg(s) + 2 HMg(s) + 2 H++(aq) + 2 Cl(aq) + 2 Cl--(aq) (aq) ---> H---> H22(g) + Mg(g) + Mg2+2+(aq) + 2 Cl(aq) + 2 Cl--(aq)(aq)

We leave the spectator ions out —We leave the spectator ions out —Mg(s) + 2 HMg(s) + 2 H++(aq) ---> H(aq) ---> H22(g) + Mg(g) + Mg2+2+(aq)(aq)

to give the NET IONIC EQUATIONNET IONIC EQUATION

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Chemical Reactions in WaterChemical Reactions in WaterSections 5.2 & 5.4-5.6—CD-ROM Ch. 5Sections 5.2 & 5.4-5.6—CD-ROM Ch. 5

We will look at We will look at EXCHANGE EXCHANGE REACTIONSREACTIONS

AX + BY AY + BX

Pb(NOPb(NO33) ) 22(aq) + 2 KI(aq)(aq) + 2 KI(aq)

----> PbI----> PbI22(s) + 2 KNO(s) + 2 KNO33 (aq) (aq)

The anions The anions exchange places exchange places between cations.between cations.

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Precipitation ReactionsPrecipitation ReactionsPrecipitation ReactionsPrecipitation Reactions

The “driving force” is the formation of an The “driving force” is the formation of an insoluble compound — a precipitate.insoluble compound — a precipitate.

Pb(NOPb(NO33))22(aq) + 2 KI(aq) ----->(aq) + 2 KI(aq) ----->

2 KNO2 KNO33(aq) + PbI(aq) + PbI22(s)(s)

Net ionic equationNet ionic equation

PbPb2+2+(aq) + 2 I(aq) + 2 I--(aq) ---> PbI(aq) ---> PbI22(s)(s)

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Acid-Base ReactionsAcid-Base Reactions• The “driving force” is the formation of water.The “driving force” is the formation of water.

NaOH(aq) + HCl(aq) --->NaOH(aq) + HCl(aq) --->

NaCl(aq) + HNaCl(aq) + H22O(liq)O(liq)

• Net ionic equationNet ionic equation

OHOH--(aq) + H(aq) + H++(aq) (aq) ---> H---> H22O(liq)O(liq)

• This applies to ALL reactions of This applies to ALL reactions of STRONGSTRONG

acids and bases.acids and bases.

24Acid-Base ReactionsAcid-Base ReactionsCCR, page 162CCR, page 162

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Acid-Base ReactionsAcid-Base Reactions• A-B reactions are sometimes called A-B reactions are sometimes called

NEUTRALIZATIONSNEUTRALIZATIONS because the solution is because the solution is

neither acidic nor basic at the end.neither acidic nor basic at the end.

• The other product of the A-B reaction is a The other product of the A-B reaction is a SALTSALT, MX., MX.

HHXX + + MMOH ---> OH ---> MMXX + H + H22OO

MMn+n+ comes from comes from base base && XXn-n- comes from comes from acidacid

This is one way to make compounds!This is one way to make compounds!

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Gas-Forming ReactionsGas-Forming Reactions

This is primarily the chemistry of This is primarily the chemistry of metal metal

carbonatescarbonates..

COCO22 and water ---> H and water ---> H22COCO33

HH22COCO33(aq) + Ca(aq) + Ca2+2+ ---> --->

2 H2 H++(aq) + CaCO(aq) + CaCO33(s) (limestone)(s) (limestone)

Adding acid reverses this reaction.Adding acid reverses this reaction.

MCOMCO33 + acid ---> CO + acid ---> CO22 + salt + salt

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Gas-Gas-Forming Forming

ReactionsReactions

CaCOCaCO33(s) + H(s) + H22SOSO44(aq) ---> (aq) --->

2 CaSO2 CaSO44(s) + H(s) + H22COCO33(aq) (aq)

Carbonic acid is unstable and forms COCarbonic acid is unstable and forms CO22 & H & H22OO

HH22COCO33(aq) ---> CO(aq) ---> CO22 + water + water

(Antacid tablet has citric acid + NaHCO(Antacid tablet has citric acid + NaHCO33))

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Quantitative Quantitative Aspects of Aspects of Reactions in Reactions in

SolutionSolutionSections 5.8-5.10Sections 5.8-5.10

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TerminologyTerminologyTerminologyTerminology

In solution we need to define the In solution we need to define the --

•SOLVENTSOLVENT

the component whose the component whose physical state is physical state is preserved when preserved when solution formssolution forms

•SOLUTESOLUTE

the other solution componentthe other solution component

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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

Concentration (M) = [ …]

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1.0 L of water 1.0 L of water was used to was used to

make 1.0 L of make 1.0 L of solution. Notice solution. Notice

the water left the water left over.over.

CCR, page 177

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PROBLEM: Dissolve 5.00 g of NiClPROBLEM: Dissolve 5.00 g of NiCl22•6 •6 HH22O in enough water to make 250 mL O in enough water to make 250 mL of solution. Calculate molarity.of solution. Calculate molarity.

PROBLEM: Dissolve 5.00 g of NiClPROBLEM: Dissolve 5.00 g of NiCl22•6 •6 HH22O in enough water to make 250 mL O in enough water to make 250 mL of solution. Calculate molarity.of solution. Calculate 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

34The Nature of a CuClThe Nature of a CuCl22 Solution Solution

Ion ConcentrationsIon Concentrations

CuClCuCl22(aq) --> (aq) -->

CuCu2+2+(aq) + (aq) + 22 Cl Cl--(aq)(aq)

If [CuClIf [CuCl22] = 0.30 M, then ] = 0.30 M, then

[Cu[Cu2+2+] = 0.30 M] = 0.30 M

[Cl[Cl--] = 2 x 0.30 M] = 2 x 0.30 M

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USING MOLARITYUSING MOLARITYUSING MOLARITYUSING MOLARITY

What mass of oxalic acid, HWhat mass of oxalic acid, H22CC22OO44, is , is required to make 250. mL of a required to make 250. mL of a 0.0500 M solution?0.0500 M solution?

Because Because Conc (M) = moles/volume = mol/VConc (M) = moles/volume = mol/V

this means thatthis means that

moles = M•Vmoles = M•V

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Step 1: Step 1: Calculate moles of acid required.Calculate moles of acid required.

(0.0500 mol/L)(0.250 L) = 0.0125 mol(0.0500 mol/L)(0.250 L) = 0.0125 mol

Step 2: Step 2: Calculate mass of acid required.Calculate mass of acid required.

(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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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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PROBLEM: You have 50.0 mL of PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M 3.0 M NaOH and you want 0.50 M NaOH. What do you do?NaOH. What do you do?

PROBLEM: You have 50.0 mL of PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M 3.0 M NaOH and you want 0.50 M NaOH. What do you do?NaOH. What do you do?

Add water to the 3.0 M solution to lower Add water to the 3.0 M solution to lower its concentration to 0.50 M its concentration to 0.50 M

Dilute the solution!Dilute the solution!

39

PROBLEM: You have 50.0 mL of 3.0 M PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. NaOH and you want 0.50 M NaOH. What do you do?What do you do?


3.0 M NaOH 0.50 M NaOH

H2O

Concentrated Dilute

But how much water But how much water do we add?do we add?

40

PROBLEM: You have 50.0 mL of 3.0 M PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. NaOH and you want 0.50 M NaOH. What do you doWhat do you do??

PROBLEM: You have 50.0 mL of 3.0 M PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. NaOH and you want 0.50 M NaOH. What do you doWhat do you do??

How much water is added?How much water is added?

The important point is that --->The important point is that --->

moles of NaOH in ORIGINAL solution = moles of NaOH in ORIGINAL solution = moles of NaOH in FINAL solutionmoles of NaOH in FINAL solution

41

PROBLEM: You have 50.0 mL of 3.0 M NaOH PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do?and you want 0.50 M NaOH. What do you do?PROBLEM: You have 50.0 mL of 3.0 M NaOH PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do?and you want 0.50 M NaOH. What do you do?

Amount of NaOH in original solution = Amount of NaOH in original solution =

M • VM • V = =

(3.0 mol/L)(0.050 L) = 0.15 mol NaOH(3.0 mol/L)(0.050 L) = 0.15 mol NaOH

Amount of NaOH in final solution must also = Amount of NaOH in final solution must also = 0.15 mol NaOH0.15 mol NaOH

Volume of final solution =Volume of final solution =

(0.15 mol NaOH)(1 L/0.50 mol) = 0.30 L(0.15 mol NaOH)(1 L/0.50 mol) = 0.30 L

or or 300 mL300 mL

42



Conclusion:Conclusion:

add 250 mL add 250 mL of waterof water to to 50.0 mL of 50.0 mL of 3.0 M NaOH 3.0 M NaOH to make 300 to make 300 mL of 0.50 M mL of 0.50 M NaOH. NaOH.

3.0 M NaOH 0.50 M NaOH

H2O

Concentrated Dilute

43

A shortcutA shortcut

CCinitialinitial • V • Vinitialinitial = C = Cfinalfinal • V • Vfinalfinal

Preparing Solutions Preparing Solutions by Dilutionby Dilution

Preparing Solutions Preparing Solutions by Dilutionby Dilution

44pH, a Concentration pH, a Concentration ScaleScale

pH, a Concentration pH, a Concentration ScaleScale

pH: a way to express acidity -- the concentration of HpH: a way to express acidity -- the concentration of H++ in solution.in solution.

Low pH: high [HLow pH: high [H++]] High pH: low [HHigh pH: low [H++]]

Acidic solutionAcidic solution pH < 7pH < 7 Neutral Neutral pH = 7pH = 7 Basic solution Basic solution pH > 7pH > 7

Acidic solutionAcidic solution pH < 7pH < 7 Neutral Neutral pH = 7pH = 7 Basic solution Basic solution pH > 7pH > 7

45

The pH ScaleThe pH ScaleThe pH ScaleThe pH Scale

pH pH = log (1/ [H= log (1/ [H++]) ])

= - log [H= - log [H++]]In a In a neutralneutral solution, solution,

[H[H++] = [OH] = [OH--] = 1.00 x 10] = 1.00 x 10-7-7 M at 25 M at 25 ooCC

pH = - log [HpH = - log [H++] = -log (1.00 x 10] = -log (1.00 x 10-7-7) ) = - (-7) = - (-7)

= 7= 7

See CD Screen 5.17 for a tutorialSee CD Screen 5.17 for a tutorial

46

[H[H++] and pH] and pH[H[H++] and pH] and pHIf the [HIf the [H++] of soda is 1.6 x 10] of soda is 1.6 x 10-3-3 M, M,

the pH is ____?the pH is ____?

Because pH = - log [HBecause pH = - log [H++] ]

thenthen

pH= - log (1.6 x 10pH= - log (1.6 x 10-3-3) )

pH = - (-2.80)pH = - (-2.80)

pH = 2.80pH = 2.80

47

pH and [HpH and [H++]]pH and [HpH and [H++]]

If the pH of Coke is 3.12, it is ____________.If the pH of Coke is 3.12, it is ____________.

Because pH = - log [HBecause pH = - log [H++] then] then

log [Hlog [H++] = - pH] = - pH

Take antilog and getTake antilog and get

[H[H++] = 10] = 10-pH-pH

[H[H++] = 10] = 10-3.12-3.12 = = 7.6 x 107.6 x 10-4-4 M M

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• Zinc reacts with Zinc reacts with acids to produce Hacids to produce H22 gas. gas.

• Have 10.0 g of ZnHave 10.0 g of Zn

• What volume of What volume of 2.50 M HCl is 2.50 M HCl is needed to convert needed to convert the Zn completely?the Zn completely?

SOLUTION SOLUTION STOICHIOMETRYSTOICHIOMETRY

Section 5.10Section 5.10

49GENERAL PLAN FOR GENERAL PLAN FOR STOICHIOMETRY CALCULATIONSSTOICHIOMETRY CALCULATIONS

GENERAL PLAN FOR GENERAL PLAN FOR STOICHIOMETRY CALCULATIONSSTOICHIOMETRY CALCULATIONS

Mass zinc

StoichiometricfactorMoles

zincMoles HCl

Mass HCl

VolumeHCl

50

Step 1: Step 1: Write the balanced equationWrite the balanced equation

Zn(s) + 2 HCl(aq) --> ZnClZn(s) + 2 HCl(aq) --> ZnCl22(aq) + H(aq) + H22(g)(g)

Step 2: Step 2: Calculate amount of ZnCalculate amount of Zn

10.0 g Zn • 1.00 mol Zn65.39 g Zn

= 0.153 mol Zn10.0 g Zn • 1.00 mol Zn65.39 g Zn

= 0.153 mol Zn

Zinc reacts with acids to produce HZinc reacts with acids to produce H22 gas. If you gas. If you have 10.0 g of Zn, what volume of 2.50 M HCl is have 10.0 g of Zn, what volume of 2.50 M HCl is needed to convert the Zn completely?needed to convert the Zn completely?


Step 3: Step 3: Use the stoichiometric factorUse the stoichiometric factor

51

Step 3: Step 3: Use the stoichiometric factorUse the stoichiometric factor

0.153 mol Zn • 2 mol HCl1 mol Zn

= 0.306 mol HCl0.153 mol Zn • 2 mol HCl1 mol Zn

= 0.306 mol HCl

0.306 mol HCl • 1.00 L

2.50 mol = 0.122 L HCl0.306 mol HCl •

1.00 L2.50 mol

= 0.122 L HCl



Step 4: Step 4: Calculate volume of HCl req’dCalculate volume of HCl req’d

52

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

53Setup for titrating an acid with a baseSetup for titrating an acid with a base

CCR, page 186

54

TitratioTitrationn

TitratioTitrationn

1. Add solution from the 1. Add solution from the buret.buret.

2. Reagent (base) reacts 2. Reagent (base) reacts with compound (acid) in with compound (acid) in solution in the flask.solution in the flask.

3. Indicator shows when 3. Indicator shows when exact stoichiometric exact stoichiometric reaction has occurred.reaction has occurred.

4. Net ionic equation4. Net ionic equation

HH++ + OH + OH-- --> H --> H22OO

5. At equivalence point 5. At equivalence point moles Hmoles H++ = moles OH = moles OH--

55

1.065 g of H1.065 g of H22CC22OO44 (oxalic (oxalic

acid) requires 35.62 mL of acid) requires 35.62 mL of

NaOH for titration to an NaOH for titration to an

equivalence point. What is equivalence point. What is

the concentra-tion of the the concentra-tion of the

NaOH?NaOH?

LAB PROBLEM #1: Standardize a LAB PROBLEM #1: Standardize a solution of NaOH — i.e., accurately solution of NaOH — i.e., accurately determine its concentration.determine its concentration.

LAB PROBLEM #1: Standardize a LAB PROBLEM #1: Standardize a solution of NaOH — i.e., accurately solution of NaOH — i.e., accurately determine its concentration.determine its concentration.

56

1.065 g of H1.065 g of H22CC22OO44 (oxalic acid) requires 35.62 (oxalic acid) requires 35.62 mL of NaOH for titration to an equivalence mL of NaOH for titration to an equivalence point. What is the concentration of the NaOH?point. What is the concentration of the NaOH?


Step 1: Step 1: Calculate amount of HCalculate amount of H22CC22OO44

1.065 g • 1 mol

90.04 g = 0.0118 mol1.065 g •

1 mol90.04 g

= 0.0118 mol

0.0118 mol acid • 2 mol NaOH1 mol acid

= 0.0236 mol NaOH

Step 2: Step 2: Calculate amount of NaOH req’dCalculate amount of NaOH req’d

57



Step 1: Step 1: Calculate amountCalculate amount of Hof H22CC22OO44

= 0.0118 mol acid= 0.0118 mol acid

Step 2: Step 2: Calculate amount of NaOH req’dCalculate amount of NaOH req’d

= 0.0236 mol NaOH= 0.0236 mol NaOH

Step 3: Step 3: Calculate concentration of NaOHCalculate concentration of NaOH

0.0236 mol NaOH0.03562 L

= 0.663 M0.0236 mol NaOH

0.03562 L= 0.663 M

[NaOH] = 0.663 M[NaOH] = 0.663 M

58

LAB PROBLEM #2: LAB PROBLEM #2: Use standardized NaOH to determine Use standardized NaOH to determine the amount of an acid in an unknown.the amount of an acid in an unknown.

LAB PROBLEM #2: LAB PROBLEM #2: Use standardized NaOH to determine Use standardized NaOH to determine the amount of an acid in an unknown.the amount of an acid in an unknown.

Apples contain malic acid, CApples contain malic acid, C44HH66OO55..

CC44HH66OO55(aq) + 2 NaOH(aq) --->(aq) + 2 NaOH(aq) --->

NaNa22CC44HH44OO55(aq) + 2 H(aq) + 2 H22O(liq)O(liq)

76.80 g of apple requires 34.56 mL of 0.663 M 76.80 g of apple requires 34.56 mL of 0.663 M

NaOH for titration. What is weight % of malic NaOH for titration. What is weight % of malic

acid?acid?

59

76.80 g of apple requires 34.56 mL of 76.80 g of apple requires 34.56 mL of 0.663 M NaOH for titration. What is 0.663 M NaOH for titration. What is weight % of malic acid?weight % of malic acid?


Step 1: Step 1: Calculate amount of NaOH used.Calculate amount of NaOH used.

C • V = (0.663 M)(0.03456 L) C • V = (0.663 M)(0.03456 L)


Step 2: Step 2: Calculate amount of acid titrated.Calculate amount of acid titrated.

0.0229 mol NaOH • 1 mol acid

2 mol NaOH


60



Step 3: Step 3: Calculate mass of acid titrated.Calculate mass of acid titrated.

0.0115 mol acid • 134 gmol

= 1.54 g

Step 1: Step 1: Calculate amount of NaOH used. Calculate amount of NaOH used.


Step 2: Step 2: Calculate amount of acid titratedCalculate amount of acid titrated


61



Step 1: Step 1: Calculate amount of NaOH used. Calculate amount of NaOH used.


Step 2: Step 2: Calculate amount of acid titratedCalculate amount of acid titrated


Step 3: Step 3: Calculate mass of acid titrated.Calculate mass of acid titrated.

= 1.54 g acid= 1.54 g acid

Step 4: Step 4: Calculate % malic acid.Calculate % malic acid.1.54 g76.80 g

• 100% = 2.01%

62

Oxidation-Reduction Oxidation-Reduction ReactionsReactions


Oxidation-Reduction Oxidation-Reduction ReactionsReactions


Thermite reactionThermite reaction

FeFe22OO33(s) + 2 Al(s)(s) + 2 Al(s)

---->---->

2 Fe(s) + Al2 Fe(s) + Al22OO33(s)(s)

63

REDOX REDOX REACTIONSREACTIONS


EXCHANGEEXCHANGEAcid-BaseAcid-BaseReactionsReactions

EXCHANGEEXCHANGEGas-FormingGas-FormingReactionsReactions

EXCHANGEEXCHANGE:: Precipitation Reactions Precipitation Reactions

REACTIONSREACTIONS

64

REDOX REACTIONSREDOX REACTIONSOxidation—Oxidation—

2 H2 H22(g) + O(g) + O22(g) ---> 2 H(g) ---> 2 H22O(liq)O(liq)

Mg(s) + 2 HCl(aq) --> MgClMg(s) + 2 HCl(aq) --> MgCl22(aq) + H(aq) + H22(g)(g)

All corrosion reactions are oxidations.All corrosion reactions are oxidations.

2 Al(s) + 3 Cu2 Al(s) + 3 Cu2+2+(aq) ---> 2 Al(aq) ---> 2 Al3+3+(aq) + 3 Cu(s)(aq) + 3 Cu(s)

Reduction—Reduction—

FeFe22OO33(s) + 2 Al(s) --> 2 Fe(s) + Al(s) + 2 Al(s) --> 2 Fe(s) + Al22OO33(s)(s)

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Cu(s) + 2 AgCu(s) + 2 Ag++(aq) (aq)

---> Cu---> Cu2+2+(aq) + 2 Ag(s) (aq) + 2 Ag(s)

In all reactions if In all reactions if something has been something has been oxidized then oxidized then something has also something has also been reducedbeen reduced



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Why Study Redox ReactionsWhy Study Redox ReactionsWhy Study Redox ReactionsWhy Study Redox Reactions

Manufacturing metalsManufacturing metalsManufacturing metalsManufacturing metals

FuelsFuelsFuelsFuels

CorrosionCorrosionCorrosionCorrosion

BatteriesBatteriesBatteriesBatteries

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Redox reactions are characterized byRedox reactions are characterized by ELECTRON TRANSFERELECTRON TRANSFER between an between an electron donor and electron acceptor.electron donor and electron acceptor.

Transfer leads to— Transfer leads to—

1. 1. increase in oxidation numberincrease in oxidation number of some element = of some element = OXIDATIONOXIDATION

2.2. decrease in oxidation numberdecrease in oxidation number of some element = of some element = REDUCTIONREDUCTION

REDOX REACTIONSREDOX REACTIONSREDOX REACTIONSREDOX REACTIONS

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OXIDATION NUMBERSOXIDATION NUMBERSOXIDATION NUMBERSOXIDATION NUMBERS

The electric charge an element The electric charge an element APPEARS to have when electrons are APPEARS to have when electrons are counted by some arbitrary rules:counted by some arbitrary rules:

1.1. Each atom in free element has ox. no. = 0.Each atom in free element has ox. no. = 0.

Zn OZn O22 I I22 S S88

2.2. In simple ions, ox. no. = charge on ion.In simple ions, ox. no. = charge on ion.

-1 for Cl-1 for Cl-- +2 for Mg +2 for Mg2+2+

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3.3. O has ox. no. = -2O has ox. no. = -2

(except in peroxides: in H(except in peroxides: in H22OO22, O = -1), O = -1)

4.4. Ox. no. of H = +1Ox. no. of H = +1

(except when H is associated with a metal as in (except when H is associated with a metal as in NaH where it is -1)NaH where it is -1)

5.5. Algebraic sum of oxidation numbers Algebraic sum of oxidation numbers

= 0 for a compound = 0 for a compound

= overall charge for an ion= overall charge for an ion

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NHNH33 N = N =

ClOClO-- Cl = Cl =

HH33POPO44 P = P =

MnOMnO44- - Mn = Mn =

CrCr22OO772-2- Cr = Cr =

CC33HH88 C = C =

Oxidation Oxidation number of F number of F in HF?in HF?

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Recognizing a Redox Recognizing a Redox ReactionReaction


Corrosion of aluminumCorrosion of aluminum

2 Al(s) + 3 Cu2 Al(s) + 3 Cu2+2+(aq) --> 2 Al(aq) --> 2 Al3+3+(aq) + 3 Cu(s)(aq) + 3 Cu(s)

Al(s) --> AlAl(s) --> Al3+3+(aq) + 3 e(aq) + 3 e--

• Ox. no. of Al increases as eOx. no. of Al increases as e-- are donated by the metal. are donated by the metal.

• Therefore, Therefore, Al is OXIDIZED Al is OXIDIZED

• Al Al is the is the REDUCING AGENTREDUCING AGENT in this balanced in this balanced half-half-reaction.reaction.

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Corrosion of aluminumCorrosion of aluminum

2 Al(s) + 3 Cu2 Al(s) + 3 Cu2+2+(aq) --> 2 Al(aq) --> 2 Al3+3+(aq) + 3 Cu(s)(aq) + 3 Cu(s)

CuCu2+2+(aq) + 2 e(aq) + 2 e- - --> Cu(s) --> Cu(s)

• Ox. no. of Cu decreases as eOx. no. of Cu decreases as e-- are accepted by the ion. are accepted by the ion.

• Therefore, Therefore, Cu is REDUCED Cu is REDUCED

• CuCu is the is the OXIDIZING AGENTOXIDIZING AGENT in this balanced in this balanced half-half-reaction.reaction.

73Recognizing a Redox Recognizing a Redox ReactionReaction


Notice that the 2 half-reactions add up to Notice that the 2 half-reactions add up to give the overall reaction give the overall reaction —if we use 2 mol of Al and 3 mol of Cu—if we use 2 mol of Al and 3 mol of Cu2+2+..

2 Al(s) --> 2 Al2 Al(s) --> 2 Al3+3+(aq) + 6 e(aq) + 6 e--

3 Cu3 Cu2+2+(aq) + 6 e(aq) + 6 e- - --> 3 Cu(s) --> 3 Cu(s)

----------------------------------------------------------------------------------------------------------------------

2 Al(s) + 3 Cu2 Al(s) + 3 Cu2+2+(aq) ---> 2 Al(aq) ---> 2 Al3+3+(aq) + 3 Cu(s)(aq) + 3 Cu(s)

Final eqn. is balanced for Final eqn. is balanced for massmass and and chargecharge..

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Common Oxidizing and Common Oxidizing and Reducing AgentsReducing Agents

See Table 5.4See Table 5.4

Common Oxidizing and Common Oxidizing and Reducing AgentsReducing Agents


HNOHNO33 is an is an

oxidizing oxidizing agentagent

2 K + 2 H2 K + 2 H22O --> O -->

2 KOH + H2 KOH + H22

Metals Metals (Na, K, (Na, K, Mg, Fe) Mg, Fe) are are reducing reducing agentsagents

Metals Metals (Cu) are (Cu) are reducing reducing agentsagents

Cu + HNOCu + HNO33 --> -->

CuCu2+2+ + NO + NO22

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In terms of oxygenIn terms of oxygen gaingain lossloss

In terms of halogenIn terms of halogen gaingain lossloss

In terms of electronsIn terms of electrons lossloss gaingain

Reaction TypeReaction Type OxidationOxidation ReductionReduction

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Examples of Redox Examples of Redox ReactionsReactions


Metal + halogenMetal + halogen2 Al + 3 Br2 Al + 3 Br22 ---> Al ---> Al22BrBr66

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Metal (Mg) + OxygenMetal (Mg) + Oxygen

Nonmetal (P) + OxygenNonmetal (P) + Oxygen

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Metal + acidMetal + acidMg + HClMg + HClMg = reducing agentMg = reducing agentHH++ = oxidizing agent = oxidizing agent

Metal + acidMetal + acidCu + HNOCu + HNO33

Cu = reducing agentCu = reducing agentHNOHNO3 3 = oxidizing agent= oxidizing agent

1 ionic compounds compounds in aqueous solution many reactions involve ionic compounds, especially...

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