Chapter 51 Reactions in Aqueous Solution Chapter 5.
Post on 26-Dec-2015
227 Views
Preview:
Transcript
Chapter 5 1
Reactions in Aqueous Reactions in Aqueous SolutionSolution
Chapter 5Chapter 5
Chapter 5 2
Compounds in Aqueous SolutionCompounds in Aqueous Solution
HCl(aq) + NaOH(aq) H2O(l) + NaCl(aq)
This is a reaction in which reactants are in solution
Solution – homogeneous mixture composed of two parts:
solute – the medium which is dissolved
solvent – the medium which dissolves the solute.
Chapter 5 3
Compounds in WaterCompounds in Water
Some compounds conduct electricity when dissolved in water – electrolytes
Those compounds which do not conduct electricity when dissolved in water are called – nonelectrolytes
Compounds in Aqueous SolutionCompounds in Aqueous Solution
Chapter 5 4
Ionic Compounds in Water (Electrolytes)Ionic Compounds in Water (Electrolytes)
The conductivity of the solution is due to the formation of ions when the compound dissolves in water
These ions are not the result of a chemical reaction, they are the result of a dissociation of the molecule into ions that compose the solid.
)()()( 2 aqClaqNasNaCl OH
Compounds in Aqueous SolutionCompounds in Aqueous Solution
Chapter 5 5
Ionic Compounds in WaterIonic Compounds in Water
Compounds in Aqueous SolutionCompounds in Aqueous Solution
Chapter 5 6
Molecular Compounds in Water(Nonelectrolytes)Molecular Compounds in Water(Nonelectrolytes)
In this case no ions are formed, the molecules just disperse throughout the solvent.
)()( 2 aqsugarssugar OH
Compounds in Aqueous SolutionCompounds in Aqueous Solution
Chapter 5 7
Molecular Compounds in Water(Nonelectrolytes)Molecular Compounds in Water(Nonelectrolytes)
There are exceptions to this, some molecules are strongly attracted to water and will react with it.
ClOHOHHCl
OHNHOHNH
32
423
Compounds in Aqueous SolutionCompounds in Aqueous Solution
Chapter 5 8
Strong and Weak ElectrolytesStrong and Weak ElectrolytesStrong electrolytes – A substance which completely ionizes in water.
For example:
ClOHOHHCl 32
Compounds in Aqueous SolutionCompounds in Aqueous Solution
Chapter 5 9
Strong and Weak ElectrolytesStrong and Weak Electrolytes
Weak electrolyte: A substance which partially ionizes when dissolved in water.
For example:
OHCOCHOHHCOCH 323223
Compounds in Aqueous SolutionCompounds in Aqueous Solution
Chapter 5 10
Strong and Weak ElectrolytesStrong and Weak Electrolytes
OHCOCHOHHCOCH 323223
Compounds in Aqueous SolutionCompounds in Aqueous Solution
Notice that the arrow in this reaction has two heads, this indicates that two opposing reactions are occurring simultaneously.
Chapter 5 11
Strong and Weak ElectrolytesStrong and Weak Electrolytes
OHHCOCHOHCOCH
and
OHCOCHOHHCOCH
223323
323223
Compounds in Aqueous SolutionCompounds in Aqueous Solution
Since both reactions occur at the same time, this is called a chemical equilibrium.
Chapter 5 12
Precipitation ReactionPrecipitation Reaction
Chapter 5 13
- A reaction which forms a solid (precipitate)
AgNO3(aq) + NaCl(aq) AgCl(s) + NaNO3(aq)
- AgCl is classified as an insoluble substance
Precipitation ReactionPrecipitation Reaction
Chapter 5 14
Net Ionic EquationNet Ionic Equation
AgNO3(aq) + NaCl(aq) AgCl(s) + NaNO3(aq)
- AgNO3 and NaNO3 are electrolytes in solution so they actually occur as free ions.
Ag+(aq) + NO3-(aq) + Na+(aq) + Cl-(aq)
AgCl(s) + Na+(aq) + NO3-
(aq)
Precipitation ReactionPrecipitation Reaction
Chapter 5 15
Net Ionic EquationNet Ionic Equation
Ag+(aq) + NO3-(aq) + Na+(aq) + Cl-(aq)
AgCl(s) + Na+(aq) + NO3-
(aq)
- Notice that NO3-(aq) and Na+(aq) occur in both the left
and right side of the equation.-These are called spectator ions.
Precipitation ReactionPrecipitation Reaction
Chapter 5 16
Net Ionic EquationNet Ionic Equation
Ag+(aq) + Cl-(aq) AgCl(s)
- With the spectator ions removed, the resulting equation shows only the ions involved in the reaction remain.
- This is a net ionic equation.
Precipitation ReactionPrecipitation Reaction
Chapter 5 17
Solubility Guidelines for Ionic CompoundsSolubility Guidelines for Ionic Compounds1. Most nitrates (NO3
-) and acetates (CH3CO2-) are
soluble in water.2. All chlorides are soluble except: Hg+, Ag+, Pb2+, Cu+
3. All sulfates are soluble except: Sr2+, Ba2+, Pb2+
4. Carbonates (CO32-), Phosphates (PO4
3-), Borates (BO3
3-),Arsenates (AsO43-), and Arsenites (AsO3
3-) are insoluble.
5. Hydroxides (OH-) of group Ia and Ba2+ and Sr2+ are soluble.
6. Most sulfides (S2-) are insoluble.
Compounds in Aqueous SolutionCompounds in Aqueous Solution
Chapter 5 18
Solubility Guidelines for Ionic CompoundsSolubility Guidelines for Ionic CompoundsPredict the solubility of the following compounds:
PbSO4
AgCH3CO2
(NH4)3PO4
KClO4
Compounds in Aqueous SolutionCompounds in Aqueous Solution
Chapter 5 19
Solubility Guidelines for Ionic CompoundsSolubility Guidelines for Ionic CompoundsPredict the solubility of the following compounds:
PbSO4 Insoluble
AgCH3CO2
(NH4)3PO4
KClO4
Compounds in Aqueous SolutionCompounds in Aqueous Solution
Chapter 5 20
Solubility Guidelines for Ionic CompoundsSolubility Guidelines for Ionic CompoundsPredict the solubility of the following compounds:
PbSO4 Insoluble
AgCH3CO2 Soluble
(NH4)3PO4
KClO4
Compounds in Aqueous SolutionCompounds in Aqueous Solution
Chapter 5 21
Solubility Guidelines for Ionic CompoundsSolubility Guidelines for Ionic CompoundsPredict the solubility of the following compounds:
PbSO4 Insoluble
AgCH3CO2 Soluble
(NH4)3PO4 Soluble
KClO4
Compounds in Aqueous SolutionCompounds in Aqueous Solution
Chapter 5 22
Solubility Guidelines for Ionic CompoundsSolubility Guidelines for Ionic CompoundsPredict the solubility of the following compounds:
PbSO4 Insoluble
AgCH3CO2 Soluble
(NH4)3PO4 Soluble
KClO4 Soluble
Compounds in Aqueous SolutionCompounds in Aqueous Solution
Chapter 5 23
Acid - substance which ionizes to form hydrogen cations (H+) in solution
Examples: Hydrochloric Acid HCl Nitric Acid HNO3
Acetic Acid CH3CO2HSulfuric Acid H2SO4
Sulfuric acid can provide two H+’s - Diprotic acid, The other acids can provide only one H+
- Monoprotic acid.
Acids and BasesAcids and Bases
Chapter 5 24
Diprotic acidDiprotic acid
H2SO4 H+ + HSO4-
HSO4- H+ + SO4
2-
Acids and BasesAcids and Bases
Chapter 5 25
Base - substance which reacts with H+ ions.
Examples: ammonia NH3
sodium hydroxide NaOH
Acids and BasesAcids and Bases
Chapter 5 26
Acid-Base ReactionAcid-Base Reaction
H+ + OH- H2O
- It is clear that the metal hydroxides (NaOH for example) provide OH- by disassociation.
- Bases like ammonia make OH- by reacting with water (ionization)
NH3 + H2O NH4+ + OH-
Acids and BasesAcids and Bases
Chapter 5 27
Strong and Weak Acids and BasesStrong and Weak Acids and Bases
Strong acids and bases are strong electrolytes.
Weak acids and bases are weak electrolytes.
Acids and BasesAcids and Bases
Chapter 5 28
Strong AcidsStrong Acids
- The strength of acids and bases are concerned with The strength of acids and bases are concerned with the ionization (or dissociation) of the substance, the ionization (or dissociation) of the substance, notnot its its chemical reactivity.chemical reactivity.
Example:Example:Hydrofluoric acid (HF) is a weak acid, but it is very Hydrofluoric acid (HF) is a weak acid, but it is very chemically reactive.chemically reactive.
- this substance can’t be stored in glass bottles this substance can’t be stored in glass bottles because it reacts with glass (silicon dioxide).because it reacts with glass (silicon dioxide).
Acids and BasesAcids and Bases
Chapter 5 29
Common Strong Acids and BasesCommon Strong Acids and Bases
Common Strong AcidsCommon Strong AcidsHydrochloric AcidHydrochloric Acid HClHClHydrobromic AcidHydrobromic Acid HBrHBrHydroiodic acidHydroiodic acid HIHINitric AcidNitric Acid HNOHNO33
Perchloric AcidPerchloric Acid HClOHClO44
Sulfuric AcidSulfuric Acid HH22SOSO44
Acids and BasesAcids and Bases
Chapter 5 30
Common Strong Acids and BasesCommon Strong Acids and Bases
Common Strong BasesCommon Strong BasesLithium HydroxideLithium Hydroxide LiOHLiOHSodium HydroxideSodium Hydroxide NaOHNaOHPotassium HydroxidePotassium Hydroxide KOHKOH
Acids and BasesAcids and Bases
Chapter 5 31
Neutralization Reaction Neutralization Reaction - Reaction between an acid and a base.
HCl(aq) + NaOH(aq) H2O(l) + NaCl(aq)
“The neutralization between acid and metal hydroxide produces water and a salt”
Salt – an ionic compound whose cation comes from a base and anion from an acid.
Acids, Bases, and SaltsAcids, Bases, and Salts
Chapter 5 32
Neutralization Reaction Neutralization Reaction
HCl(aq) + NaOH(aq) H2O(l) + NaCl(aq)
- despite the appearance of the equation, the reaction actually takes place between the ions.
Acids, Bases, and SaltsAcids, Bases, and Salts
Chapter 5 33
Neutralization Reaction Neutralization Reaction
HCl(aq) + NaOH(aq) H2O(l) + NaCl(aq)
Total Ionic EquationH+(aq) + Cl-(aq) + Na+(aq) + OH-(aq)
H2O(l) + Na+(aq) + Cl-
(aq)
Acids, Bases, and SaltsAcids, Bases, and Salts
Chapter 5 34
Neutralization Reaction Neutralization Reaction
HCl(aq) + NaOH(aq) H2O(l) + NaCl(aq)
Total Ionic EquationH+(aq) + Cl-(aq) + Na+(aq) + OH-(aq)
H2O(l) + Na+(aq) + Cl-
(aq)
Acids, Bases, and SaltsAcids, Bases, and Salts
Chapter 5 35
Neutralization Reaction Neutralization Reaction
HCl(aq) + NaOH(aq) H2O(l) + NaCl(aq)
Total Ionic EquationH+(aq) + Cl-(aq) + Na+(aq) + OH-(aq)
H2O(l) + Na+(aq) + Cl-
(aq)
Net Ionic EquationH+(aq) + OH-(aq) H2O(l)
Acids, Bases, and SaltsAcids, Bases, and Salts
Chapter 5 36
Metal Carbonates and AcidMetal Carbonates and Acid
Gas-Forming ReactionsGas-Forming Reactions
2 HCl(aq) + Na2CO3(aq) 2 NaCl(aq) + H2O(l) + CO2(g)
- Metal carbonates (or bicarbonates) always form a salt, water and carbon dioxide
Chapter 5 37
Metal Sulfide and AcidMetal Sulfide and Acid
Gas-Forming ReactionsGas-Forming Reactions
2 HCl(aq) + Na2S(s) H2S(g) + 2 NaCl(aq)
- Metal sulfides form a salt and hydrogen sulfide.
Chapter 5 38
Metal Sulfite and AcidMetal Sulfite and Acid
Gas-Forming ReactionsGas-Forming Reactions
2 HCl(aq) + Na2SO3(s) SO2(g) + 2 NaCl(aq) + H2O(l)
- Metal sulfites form a salt, sulfur dioxide and water.
Chapter 5 39
Ammonium Salt and Strong BaseAmmonium Salt and Strong Base
Gas-Forming ReactionsGas-Forming Reactions
NH4Cl(s) + NaOH(aq) NH3(g) + NaCl(aq) + H2O(l)
- This reaction forms ammonia, salt and water
Chapter 5 40
Oxidation-Reduction ReactionsOxidation-Reduction Reactions
- Reaction where electrons are exchanged.
2 Na(s) + 2 H2O(l) 2 NaOH(aq) + H2(g)
Na(s) Na+(aq) + 1 e-
oxidation – loss of electrons
2 H+(g) + 2 e- H2(g)
reduction – gain of electrons
Chapter 5 41
Oxidation-Reduction ReactionsOxidation-Reduction Reactions
- Reaction where electrons are exchanged.
2 Na(s) + 2 H2O(l) 2 NaOH(aq) + H2(g)
- An alternate approach is to describe how one reagent effects another.- Reducing Agent, a substance that causes another substance
to be reduced.Na(s) Na+(aq) + 1 e-
- Oxidizing Agent, a substance that causes another substnace to be oxidized.
2 H+(g) + 2 e- H2(g)
Chapter 5 42
Oxidation-Reduction ReactionsOxidation-Reduction Reactions
Oxidation NumbersOxidation Numbers1. Each atom of a pure element has an oxidation number of
zero(0).2. For monatomic ions, the oxidation number equals the
charge on the ion.3. Fluorine always has an oxidation state of -1 in compounds.4. Cl, Br, and I always have oxidation numbers of -1, except
when combined with oxygen or fluorine.5. The oxidation number of H is +1 and O is -2 in most
compounds.6. The sum of the oxidation numbers must equal the charge
on the molecule or ion.
Chapter 5 43
Oxidation-Reduction ReactionsOxidation-Reduction Reactions
Oxidation NumbersOxidation Numbers
Examples
PCl5
P 1( ) =Cl 5( ) = ________
0
Chapter 5 44
Oxidation-Reduction ReactionsOxidation-Reduction Reactions
Oxidation NumbersOxidation Numbers
Example
PCl5
P 1( ? ) = ?Cl 5(-1) = __-5____
0
Chapter 5 45
Oxidation-Reduction ReactionsOxidation-Reduction Reactions
Oxidation NumbersOxidation Numbers
Example
PCl5
P 1(+5) = +5Cl 5(-1) = __-5____
0
Chapter 5 46
Oxidation-Reduction ReactionsOxidation-Reduction Reactions
Oxidation NumbersOxidation Numbers
Example
CO32-
C 1( ) = O 3( ) = _____
-2
Chapter 5 47
Oxidation-Reduction ReactionsOxidation-Reduction Reactions
Oxidation NumbersOxidation Numbers
Example
CO32-
C 1(?) = ?O 3(-2) = __-6__
-2
Chapter 5 48
Oxidation-Reduction ReactionsOxidation-Reduction Reactions
Oxidation NumbersOxidation Numbers
Example
CO32-
C 1(+4) = +4O 3(-2) = __-6__
-2
Chapter 5 49
Oxidation-Reduction ReactionsOxidation-Reduction Reactions
Oxidation NumbersOxidation Numbers
Example
K2CrO4
K 2( ) =O 4( ) =Cr 1( ) = ________
0
Chapter 5 50
Oxidation-Reduction ReactionsOxidation-Reduction Reactions
Oxidation NumbersOxidation Numbers
Example
K2CrO4
K 2(+1) = +2O 4(-2) = -8Cr 1( ? ) = ___?____
0
Chapter 5 51
Oxidation-Reduction ReactionsOxidation-Reduction Reactions
Oxidation NumbersOxidation Numbers
Example
K2CrO4
K 2(+1) = +2O 4(-2) = -8Cr 1(+6) = ___+6__
0
Chapter 5 52
Molarity(M)Molarity(M)Unit of concentration, moles of solute per liter of solution.
solutionofLiters
soluteofMolesMolarity
SolutionsSolutions
Chapter 5 53
Example: What is the molarity(M) of a solution which contains 17.51g of NaCl in 751mL of solution?
F.W. (NaCl): 58.45g/mol
molg
gNaClmoles
/45.58
51.17
SolutionsSolutions
Chapter 5 54
Example: What is the molarity(M) of a solution which contains 17.51g of NaCl in 751mL of solution?
F.W. (NaCl): 58.45g/mol
molmolg
gNaClmoles 2996.0
/45.58
51.17
SolutionsSolutions
Chapter 5 55
Example: What is the molarity(M) of a solution which contains 17.51g of NaCl in 751mL of solution?
F.W. (NaCl): 58.45g/mol
Solution volume
molmolg
gNaClmoles 2996.0
/45.58
51.17
mL
LmLsolutionofLiters
1000
1751
SolutionsSolutions
Chapter 5 56
Example: What is the molarity(M) of a solution which contains 17.51g of NaCl in 751mL of solution?
F.W. (NaCl): 58.45g/mol
Solution volume
molmolg
gNaClmoles 2996.0
/45.58
51.17
LmL
LmLsolutionofLiters 751.0
1000
1751
SolutionsSolutions
Chapter 5 57
Example: What is the molarity(M) of a solution which contains 17.51g of NaCl in 751mL of solution?
solutionofLiters
soluteofMolesMolarity
SolutionsSolutions
Chapter 5 58
Example: What is the molarity(M) of a solution which contains 17.51g of NaCl in 751mL of solution?
L
molmolarity
751.0
2996.0
SolutionsSolutions
Chapter 5 59
Example: What is the molarity(M) of a solution which contains 17.51g of NaCl in 751mL of solution?
M
L
molmolarity
399.0
751.0
2996.0
SolutionsSolutions
Chapter 5 60
SolutionsSolutionsMolarity
Chapter 5 61
DilutionDilution
MdiluteVdilute = MconcentratedVconcentrated
SolutionsSolutions
Chapter 5 62
DilutionDilution
Example: What volume of 6.00M NaOH is required to make 500mL of 0.100M NaOH?
Mconcentrated = 6.00M Mdilute = 0.100MVconcentrated = ? Vdilute = 500mL
0.100M(500mL) = 6.00M(Vconcentrated)
Vconcentrated = 8.33mL
SolutionsSolutions
Chapter 5 63
pH ScalepH ScaleConcentration scale for acids and bases.
]log[ HpH
• The square brackets around the H+ indicate that the concentration of H+ is in molarity.
• So, a change of 1 pH unit indicates a 10X change in H+ concentration.
pHH 10][
Chapter 5 64
Solution StoichiometrySolution Stoichiometry
- We can now use molarity to determine stoichiometric quantities.
ExampleHow many grams of hydrogen gas are produced when 20.0 mL of 1.75M HCl is allowed to react with 15.0g of sodium metal?
2 HCl(aq) + 2 Na(s) H2(g) + 2 NaCl(aq)
Chapter 5 65
Solution StoichiometrySolution Stoichiometry
2 HCl(aq) + 2 Na(s) H2(g) + 2 NaCl(aq)
- Convert quantities to moles
molg
gmoles
LMmoles
Na
HCl
/0.23
0.15
0200.075.1
Chapter 5 66
Solution StoichiometrySolution Stoichiometry
2 HCl(aq) + 2 Na(s) H2(g) + 2 NaCl(aq)
- Convert quantities to moles
molmolg
gmoles
molLMmoles
Na
HCl
652.0/0.23
0.15
0350.00200.075.1
Chapter 5 67
Solution StoichiometrySolution Stoichiometry
2 HCl(aq) + 2 Na(s) H2(g) + 2 NaCl(aq)
- Convert quantities to moles
- Determine limiting reagent
molmolg
gmoles
molLMmoles
Na
HCl
652.0/0.23
0.15
0350.00200.075.1
2
652.02
0350.0
Na
HCl
Chapter 5 68
Solution StoichiometrySolution Stoichiometry
2 HCl(aq) + 2 Na(s) H2(g) + 2 NaCl(aq)
- Convert quantities to moles
- Determine limiting reagent
molmolg
gmoles
molLMmoles
Na
HCl
652.0/0.23
0.15
0350.00200.075.1
326.02
652.0
0175.02
0350.0
Na
HCl
Chapter 5 69
Solution StoichiometrySolution Stoichiometry
2 HCl(aq) + 2 Na(s) H2(g) + 2 NaCl(aq)
- Calculate moles of H2
mol
x
HCl
H
0350.02
1 2
Chapter 5 70
Solution StoichiometrySolution Stoichiometry
2 HCl(aq) + 2 Na(s) H2(g) + 2 NaCl(aq)
- Calculate moles of H2
2
2
0175.0
0350.02
1
Hmolx
mol
x
HCl
H
Chapter 5 71
Solution StoichiometrySolution Stoichiometry
2 HCl(aq) + 2 Na(s) H2(g) + 2 NaCl(aq)
- Calculate moles of H2
- Calculate grams of H2
2
2
0175.0
0350.02
1
Hmolx
mol
x
HCl
H
molgmolHg /02.20175.02
Chapter 5 72
Solution StoichiometrySolution Stoichiometry
2 HCl(aq) + 2 Na(s) H2(g) + 2 NaCl(aq)
- Calculate moles of H2
- Calculate grams of H2
2
2
0175.0
0350.02
1
Hmolx
mol
x
HCl
H
gmolgmolHg 0353.0/02.20175.02
Chapter 5 73
Solution StoichiometrySolution Stoichiometry
TitrationsTitrations
Chapter 5 74
4,10, 16, 20, 24, 30, 36, 38, 44, 60, 62, 68
Practice ProblemsPractice Problems
top related