chapter 51 reactions in aqueous solution chapter 5

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