View
218
Download
1
Embed Size (px)
Citation preview
1
IONIC COMPOUNDSIONIC COMPOUNDS
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
3
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 it is Acetic acid ionizes only to a small extent, so it is a a weak electrolyte.weak electrolyte.
CHCH33COCO22H(aq) ---> CHH(aq) ---> CH33COCO22--(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) ---> CHH(aq) ---> CH33COCO22--(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
8
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 5.1Guidelines to predict the solubility of ionic compoundsGuidelines to predict the solubility of ionic compounds
9
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
10
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
11
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)
12
The The Nature Nature of Acidsof Acids
The The Nature Nature of Acidsof Acids
HCl
H2O
Cl-
H3O+
hydronium ion
13
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
14
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.
15
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.
18
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
21
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.
22
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)
23
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
25
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 ionic compounds!This is one way to make ionic compounds!
26Gas-Forming ReactionsGas-Forming ReactionsThis is primarily the chemistry of This is primarily the chemistry of metal carbonatesmetal carbonates..
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
27
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 (g) + water (g) + water
(Antacid tablet has citric acid + NaHCO(Antacid tablet has citric acid + NaHCO33))
28
See also: Gas Forming Reactions in Biological SystemsThree of the pioneers in working out the roles of NO forming reactionsshared a Nobel Prize in 1988 for their discoveries.
See also: Gas Forming Reactions in Biological SystemsThree of the pioneers in working out the roles of NO forming reactionsshared a Nobel Prize in 1988 for their discoveries.
29
Quantitative Quantitative Aspects of Aspects of Reactions in Reactions in
SolutionSolutionSections 5.8-5.10Sections 5.8-5.10
30
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
31
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) = [ …]
32
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
33
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
35
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
36
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?
37
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.
38
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?
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
0.15/Volume of final solution = 0.5 M/ 1 L0.15/Volume of final solution = 0.5 M/ 1 L
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
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?
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?
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
45The pH ScaleThe pH ScaleThe pH ScaleThe pH Scale
pH pH = log (1/ [H= log (1/ [H++]) ])
= - log [H= - log [H++]]
Remember : log a = b if 10Remember : log a = b if 10bb=a=aIn 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
What’s the origin of the name of the soda 7up ?What’s the origin of the name of the soda 7up ?
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
48
• 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?
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
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?
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 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?
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
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?
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 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 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?
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?
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)
= 0.0229 mol NaOH= 0.0229 mol NaOH
Step 2: Step 2: Calculate amount of acid titrated.Calculate amount of acid titrated.
0.0229 mol NaOH • 1 mol acid
2 mol NaOH
= 0.0115 mol acid= 0.0115 mol acid
60
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?
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 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.
= 0.0229 mol NaOH= 0.0229 mol NaOH
Step 2: Step 2: Calculate amount of acid titratedCalculate amount of acid titrated
= 0.0115 mol acid= 0.0115 mol acid
61
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?
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.
= 0.0229 mol NaOH= 0.0229 mol NaOH
Step 2: Step 2: Calculate amount of acid titratedCalculate amount of acid titrated
= 0.0115 mol acid= 0.0115 mol acid
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
REDOX REDOX REACTIONSREACTIONS
REDOX REDOX REACTIONSREACTIONS
EXCHANGEEXCHANGEAcid-BaseAcid-BaseReactionsReactions
EXCHANGEEXCHANGEGas-FormingGas-FormingReactionsReactions
EXCHANGEEXCHANGE:: Precipitation Reactions Precipitation Reactions
REACTIONSREACTIONS
63
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
64
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
REDOX REDOX REACTIONSREACTIONS
REDOX REDOX REACTIONSREACTIONS
65
Why Study Redox ReactionsWhy Study Redox ReactionsWhy Study Redox ReactionsWhy Study Redox Reactions
Manufacturing metalsManufacturing metalsManufacturing metalsManufacturing metals
FuelsFuelsFuelsFuels
CorrosionCorrosionCorrosionCorrosion
BatteriesBatteriesBatteriesBatteries
66
OXIDATION NUMBERSOXIDATION NUMBERSOXIDATION NUMBERSOXIDATION NUMBERS
The electric charge an element APPEARS to The electric charge an element APPEARS to have when electrons are counted by some have when electrons are counted by some arbitrary rules: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+
67
OXIDATION NUMBERSOXIDATION NUMBERSOXIDATION NUMBERSOXIDATION NUMBERS
3. F always has an oxidation number of -1 when 3. F always has an oxidation number of -1 when forming compounds with other elements.forming compounds with other elements.
4. Cl, Br and I have oxidation numbers of -1 when 4. Cl, Br and I have oxidation numbers of -1 when forming compounds with other elements, except forming compounds with other elements, except when combined with oxygen and fluorine.when combined with oxygen and fluorine.
5a.5a. O has ox. no. = -2O has ox. no. = -2
(except in peroxides: in H(except in peroxides: in H22OO22, O = -1), O = -1)
68
OXIDATION NUMBERSOXIDATION NUMBERSOXIDATION NUMBERSOXIDATION NUMBERS
5b.5b. 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)
6.6. 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
69
OXIDATION NUMBERSOXIDATION NUMBERSOXIDATION NUMBERSOXIDATION NUMBERS
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?
70
Recognizing a Redox Recognizing a Redox ReactionReaction
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.
71
Recognizing a Redox Recognizing a Redox ReactionReaction
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)
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.
72Recognizing a Redox Recognizing a Redox ReactionReaction
Recognizing 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..
73
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
See Table 5.4See Table 5.4
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
74
Recognizing a Redox Recognizing a Redox ReactionReaction
See Table 5.4See Table 5.4
Recognizing a Redox Recognizing a Redox ReactionReaction
See Table 5.4See Table 5.4
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
75
Examples of Redox Examples of Redox ReactionsReactions
Examples of Redox Examples of Redox ReactionsReactions
Metal + halogenMetal + halogen2 Al + 3 Br2 Al + 3 Br22 ---> Al ---> Al22BrBr66
76
Examples of Redox Examples of Redox ReactionsReactions
Examples of Redox Examples of Redox ReactionsReactions
Metal (Mg) + OxygenMetal (Mg) + Oxygen
Nonmetal (P) + OxygenNonmetal (P) + Oxygen
77
Examples of Redox Examples of Redox ReactionsReactions
Examples of Redox Examples of Redox ReactionsReactions
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