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Reactions in Aqueous Solution. Chapter 5. Compounds in Aqueous Solution. HCl(aq) + NaOH(aq)  H 2 O(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

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compounds in aqueous solution
Compounds 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

slide3

Compounds in Aqueous Solution

Compounds 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

Chapter 5

slide4

Compounds in Aqueous Solution

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.

Chapter 5

compounds in aqueous solution5
Compounds in Aqueous Solution

Ionic Compounds in Water

Chapter 5

compounds in aqueous solution6
Compounds in Aqueous Solution

Molecular Compounds in Water(Nonelectrolytes)

In this case no ions are formed, the molecules just disperse throughout the solvent.

Chapter 5

compounds in aqueous solution7
Compounds in Aqueous Solution

Molecular Compounds in Water(Nonelectrolytes)

There are exceptions to this, some molecules are strongly attracted to water and will react with it.

Chapter 5

slide8

Compounds in Aqueous Solution

Strong and Weak Electrolytes

Strong electrolytes – A substance which completely ionizes in water.

For example:

Chapter 5

slide9

Compounds in Aqueous Solution

Strong and Weak Electrolytes

Weak electrolyte: A substance which partially ionizes when dissolved in water.

For example:

Chapter 5

slide10

Compounds in Aqueous Solution

Strong and Weak Electrolytes

Notice that the arrow in this reaction has two heads, this indicates that two opposing reactions are occurring simultaneously.

Chapter 5

slide11

Compounds in Aqueous Solution

Strong and Weak Electrolytes

Since both reactions occur at the same time, this is called a chemical equilibrium.

Chapter 5

slide13

Precipitation Reaction

  • A reaction which forms a solid (precipitate)
  • AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq)
  • AgCl is classified as an insoluble substance

Chapter 5

slide14

Precipitation Reaction

  • Net 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)

Chapter 5

slide15

Precipitation Reaction

  • Net 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.

Chapter 5

slide16

Precipitation Reaction

  • Net 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.

Chapter 5

compounds in aqueous solution17
Compounds in Aqueous Solution

Solubility Guidelines for Ionic Compounds

  • Most nitrates (NO3-) and acetates (CH3CO2-) are soluble in water.
  • All chlorides are soluble except: Hg+, Ag+, Pb2+, Cu+
  • All sulfates are soluble except: Sr2+, Ba2+, Pb2+
  • Carbonates (CO32-), Phosphates (PO43-), Borates (BO33-),Arsenates (AsO43-), and Arsenites (AsO33-) are insoluble.
  • Hydroxides (OH-) of group Ia and Ba2+ and Sr2+ are soluble.
  • Most sulfides (S2-) are insoluble.

Chapter 5

compounds in aqueous solution18
Compounds in Aqueous Solution

Solubility Guidelines for Ionic Compounds

Predict the solubility of the following compounds:

PbSO4

AgCH3CO2

(NH4)3PO4

KClO4

Chapter 5

compounds in aqueous solution19
Compounds in Aqueous Solution

Solubility Guidelines for Ionic Compounds

Predict the solubility of the following compounds:

PbSO4 Insoluble

AgCH3CO2

(NH4)3PO4

KClO4

Chapter 5

compounds in aqueous solution20
Compounds in Aqueous Solution

Solubility Guidelines for Ionic Compounds

Predict the solubility of the following compounds:

PbSO4 Insoluble

AgCH3CO2 Soluble

(NH4)3PO4

KClO4

Chapter 5

compounds in aqueous solution21
Compounds in Aqueous Solution

Solubility Guidelines for Ionic Compounds

Predict the solubility of the following compounds:

PbSO4 Insoluble

AgCH3CO2 Soluble

(NH4)3PO4 Soluble

KClO4

Chapter 5

compounds in aqueous solution22
Compounds in Aqueous Solution

Solubility Guidelines for Ionic Compounds

Predict the solubility of the following compounds:

PbSO4 Insoluble

AgCH3CO2 Soluble

(NH4)3PO4 Soluble

KClO4 Soluble

Chapter 5

slide23

Acids and Bases

Acid - substance which ionizes to form hydrogen cations (H+) in solution

Examples:

Hydrochloric Acid HCl

Nitric Acid HNO3

Acetic Acid CH3CO2H

Sulfuric Acid H2SO4

Sulfuric acid can provide two H+’s - Diprotic acid,

The other acids can provide only one H+ - Monoprotic acid.

Chapter 5

slide24

Acids and Bases

Diprotic acid

H2SO4 H+ + HSO4-

HSO4-Û H+ + SO42-

Chapter 5

slide25

Acids and Bases

Base - substance which reacts with H+ ions.

Examples:

ammonia NH3

sodium hydroxide NaOH

Chapter 5

slide26

Acids and Bases

  • Acid-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 Û NH4+ + OH-

Chapter 5

slide27

Acids and Bases

Strong and Weak Acids and Bases

Strong acids and bases are strong electrolytes.

Weak acids and bases are weak electrolytes.

Chapter 5

slide28

Acids and Bases

  • Strong Acids
  • The strength of acids and bases are concerned with the ionization (or dissociation) of the substance, not its chemical reactivity.
  • Example:
  • Hydrofluoric acid (HF) is a weak acid, but it is very chemically reactive.
    • this substance can’t be stored in glass bottles because it reacts with glass (silicon dioxide).

Chapter 5

slide29

Acids and Bases

Common Strong Acids and Bases

Common Strong Acids

Hydrochloric Acid HCl

Hydrobromic Acid HBr

Hydroiodic acid HI

Nitric Acid HNO3

Perchloric Acid HClO4

Sulfuric Acid H2SO4

Chapter 5

slide30

Acids and Bases

Common Strong Acids and Bases

Common Strong Bases

Lithium Hydroxide LiOH

Sodium Hydroxide NaOH

Potassium Hydroxide KOH

Chapter 5

slide31

Acids, Bases, and Salts

  • 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.

Chapter 5

slide32

Acids, Bases, and Salts

Neutralization Reaction

HCl(aq) + NaOH(aq)  H2O(l) + NaCl(aq)

- despite the appearance of the equation, the reaction actually takes place between the ions.

Chapter 5

slide33

Acids, Bases, and Salts

Neutralization Reaction

HCl(aq) + NaOH(aq)  H2O(l) + NaCl(aq)

Total Ionic Equation

H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq)  H2O(l) + Na+(aq) + Cl-(aq)

Chapter 5

slide34

Acids, Bases, and Salts

Neutralization Reaction

HCl(aq) + NaOH(aq)  H2O(l) + NaCl(aq)

Total Ionic Equation

H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq)  H2O(l) + Na+(aq) + Cl-(aq)

Chapter 5

slide35

Acids, Bases, and Salts

Neutralization Reaction

HCl(aq) + NaOH(aq)  H2O(l) + NaCl(aq)

Total Ionic Equation

H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq)  H2O(l) + Na+(aq) + Cl-(aq)

Net Ionic Equation

H+(aq) + OH-(aq)  H2O(l)

Chapter 5

slide36

Gas-Forming Reactions

Metal Carbonates and Acid

  • 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

slide37

Gas-Forming Reactions

Metal Sulfide and Acid

  • 2 HCl(aq) + Na2S(s)  H2S(g) + 2 NaCl(aq)
  • Metal sulfides form a salt and hydrogen sulfide.

Chapter 5

slide38

Gas-Forming Reactions

Metal Sulfite and Acid

  • 2 HCl(aq) + Na2SO3(s)  SO2(g) + 2 NaCl(aq) + H2O(l)
  • Metal sulfites form a salt, sulfur dioxide and water.

Chapter 5

slide39

Gas-Forming Reactions

Ammonium Salt and Strong Base

  • NH4Cl(s) + NaOH(aq)  NH3(g) + NaCl(aq) + H2O(l)
  • This reaction forms ammonia, salt and water

Chapter 5

oxidation reduction reactions
Oxidation-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

oxidation reduction reactions41
Oxidation-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

slide42

Oxidation-Reduction Reactions

  • Oxidation Numbers
  • Each atom of a pure element has an oxidation number of zero(0).
  • For monatomic ions, the oxidation number equals the charge on the ion.
  • Fluorine always has an oxidation state of -1 in compounds.
  • Cl, Br, and I always have oxidation numbers of -1, except when combined with oxygen or fluorine.
  • The oxidation number of H is +1 and O is -2 in most compounds.
  • The sum of the oxidation numbers must equal the charge on the molecule or ion.

Chapter 5

slide43

Oxidation-Reduction Reactions

Oxidation Numbers

Examples

PCl5

P 1( ) =

Cl 5( ) = ________

0

Chapter 5

slide44

Oxidation-Reduction Reactions

Oxidation Numbers

Example

PCl5

P 1( ? ) = ?

Cl 5(-1) = __-5____

0

Chapter 5

slide45

Oxidation-Reduction Reactions

Oxidation Numbers

Example

PCl5

P 1(+5) = +5

Cl 5(-1) = __-5____

0

Chapter 5

slide46

Oxidation-Reduction Reactions

Oxidation Numbers

Example

CO32-

C 1( ) =

O 3( ) = _____

-2

Chapter 5

slide47

Oxidation-Reduction Reactions

Oxidation Numbers

Example

CO32-

C 1(?) = ?

O 3(-2) = __-6__

-2

Chapter 5

slide48

Oxidation-Reduction Reactions

Oxidation Numbers

Example

CO32-

C 1(+4) = +4

O 3(-2) = __-6__

-2

Chapter 5

slide49

Oxidation-Reduction Reactions

Oxidation Numbers

Example

K2CrO4

K 2( ) =

O 4( ) =

Cr 1( ) = ________

0

Chapter 5

slide50

Oxidation-Reduction Reactions

Oxidation Numbers

Example

K2CrO4

K 2(+1) = +2

O 4(-2) = -8

Cr 1( ? ) = ___?____

0

Chapter 5

slide51

Oxidation-Reduction Reactions

Oxidation Numbers

Example

K2CrO4

K 2(+1) = +2

O 4(-2) = -8

Cr 1(+6) = ___+6__

0

Chapter 5

solutions
Solutions

Molarity(M)

Unit of concentration, moles of solute per liter of solution.

Chapter 5

solutions53
Solutions

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

Chapter 5

solutions54
Solutions

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

Chapter 5

solutions55
Solutions

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

Chapter 5

solutions56
Solutions

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

Chapter 5

solutions57
Solutions

Example: What is the molarity(M) of a solution which contains 17.51g of NaCl in 751mL of solution?

Chapter 5

solutions58
Solutions

Example: What is the molarity(M) of a solution which contains 17.51g of NaCl in 751mL of solution?

Chapter 5

solutions59
Solutions

Example: What is the molarity(M) of a solution which contains 17.51g of NaCl in 751mL of solution?

Chapter 5

slide60

Solutions

Molarity

Chapter 5

slide61

Solutions

Dilution

MdiluteVdilute = MconcentratedVconcentrated

Chapter 5

slide62

Solutions

Dilution

Example: What volume of 6.00M NaOH is required to make 500mL of 0.100M NaOH?

Mconcentrated = 6.00M Mdilute = 0.100M

Vconcentrated = ? Vdilute = 500mL

0.100M(500mL) = 6.00M(Vconcentrated)

Vconcentrated = 8.33mL

Chapter 5

slide63

pH Scale

Concentration scale for acids and bases.

  • 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.

Chapter 5

slide64

Solution Stoichiometry

  • We can now use molarity to determine stoichiometric quantities.
  • Example
  • How 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

slide65

Solution Stoichiometry

  • 2 HCl(aq) + 2 Na(s)  H2(g) + 2 NaCl(aq)
  • Convert quantities to moles

Chapter 5

slide66

Solution Stoichiometry

  • 2 HCl(aq) + 2 Na(s)  H2(g) + 2 NaCl(aq)
  • Convert quantities to moles

Chapter 5

slide67

Solution Stoichiometry

  • 2 HCl(aq) + 2 Na(s)  H2(g) + 2 NaCl(aq)
  • Convert quantities to moles
  • Determine limiting reagent

Chapter 5

slide68

Solution Stoichiometry

  • 2 HCl(aq) + 2 Na(s)  H2(g) + 2 NaCl(aq)
  • Convert quantities to moles
  • Determine limiting reagent

Chapter 5

slide69

Solution Stoichiometry

  • 2 HCl(aq) + 2 Na(s)  H2(g) + 2 NaCl(aq)
  • Calculate moles of H2

Chapter 5

slide70

Solution Stoichiometry

  • 2 HCl(aq) + 2 Na(s)  H2(g) + 2 NaCl(aq)
  • Calculate moles of H2

Chapter 5

slide71

Solution Stoichiometry

  • 2 HCl(aq) + 2 Na(s)  H2(g) + 2 NaCl(aq)
  • Calculate moles of H2
  • Calculate grams of H2

Chapter 5

slide72

Solution Stoichiometry

  • 2 HCl(aq) + 2 Na(s)  H2(g) + 2 NaCl(aq)
  • Calculate moles of H2
  • Calculate grams of H2

Chapter 5

slide73

Solution Stoichiometry

Titrations

Chapter 5

slide74

Practice Problems

4,10, 16, 20, 24, 30, 36, 38, 44, 60, 62, 68

Chapter 5