Chapter 5*Reactions in Aqueous Solution

Chapter 5

Chapter 5*Compounds in Aqueous SolutionHCl(aq) + NaOH(aq) H2O(l) + NaCl(aq)

This is a reaction in which reactants are in solutionSolution homogeneous mixture composed of two parts:solute the medium which is dissolvedsolvent the medium which dissolves the solute.

Chapter 5

Chapter 5*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

Compounds in Aqueous SolutionNonelectrolyteWeak electrolyteStrong electrolyte

Chapter 5

Chapter 5*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.Compounds in Aqueous Solution

Chapter 5

Chapter 5*Ionic Compounds in WaterCompounds in Aqueous Solution

Chapter 5

Chapter 5*Molecular Compounds in Water(Nonelectrolytes)

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

Compounds in Aqueous Solution

Chapter 5

Chapter 5*Strong and Weak ElectrolytesStrong electrolytes A substance which completely ionizes in water.

For example:

Compounds in Aqueous Solution

Chapter 5

Chapter 5*Strong and Weak Electrolytes

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

For example:Compounds in Aqueous Solution

Chapter 5

Chapter 5*Strong and Weak ElectrolytesCompounds in Aqueous SolutionNotice that the arrow in this reaction has two heads, this indicates that two opposing reactions are occurring simultaneously.

Chapter 5

Chapter 5*Strong and Weak ElectrolytesCompounds in Aqueous SolutionSince both reactions occur at the same time, this is called a chemical equilibrium.

Chapter 5

Chapter 5*

Chapter 5

Chapter 5*

Chapter 5

Chapter 5*Precipitation Reaction

Chapter 5

Chapter 5*A reaction which forms a solid (precipitate)

AgNO3(aq) + NaCl(aq) AgCl(s) + NaNO3(aq)

AgCl is classified as an insoluble substance

Precipitation Reaction

Chapter 5

Chapter 5*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)

Precipitation Reaction

Chapter 5

Chapter 5*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.

Precipitation Reaction

Chapter 5

Chapter 5*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.

Precipitation Reaction

Chapter 5

Chapter 5*Solubility Guidelines for Ionic CompoundsCompounds in Aqueous Solution

Chapter 5

Chapter 5*Solubility Guidelines for Ionic Compounds1. Predict the solubility of the following compounds:PbSO4AgCH3CO2

(NH4)3PO4

KClO4Compounds in Aqueous Solution

Chapter 5

Chapter 5*Solubility Guidelines for Ionic CompoundsPredict the solubility of the following compounds:PbSO4InsolubleAgCH3CO2(NH4)3PO4KClO4Compounds in Aqueous Solution

Chapter 5

Chapter 5*Solubility Guidelines for Ionic CompoundsPredict the solubility of the following compounds:PbSO4InsolubleAgCH3CO2Soluble

(NH4)3PO4KClO4Compounds in Aqueous Solution

Chapter 5

Chapter 5*Solubility Guidelines for Ionic CompoundsPredict the solubility of the following compounds:PbSO4InsolubleAgCH3CO2Soluble

(NH4)3PO4Soluble

KClO4Compounds in Aqueous Solution

Chapter 5

Chapter 5*Solubility Guidelines for Ionic Compounds1. Predict the solubility of the following compounds:PbSO4InsolubleAgCH3CO2Soluble

(NH4)3PO4Soluble

KClO4SolubleCompounds in Aqueous Solution2. Predict the precipitate produced by mixing an Al(NO3)3 solution with NaOH solution. Write the net ionic equation for the reaction.

Chapter 5

Chapter 5*

Chapter 5

Chapter 5*Acids and Bases

Chapter 5

Chapter 5*Acid - substance which ionizes to form hydrogen cations (H+) in solution

Examples: Hydrochloric AcidHCl Nitric AcidHNO3Acetic AcidCH3CO2HSulfuric AcidH2SO4Sulfuric acid can provide two H+s - Diprotic acid, The other acids can provide only one H+ - Monoprotic acid.Acids and Bases

Chapter 5

Chapter 5*Diprotic acid H2SO4 H+ + HSO4-

HSO4- H+ + SO42-Acids and BasesH2SO4 merupakan asam kuat atau elektrolit kuat (tahap ionisasi pertama berlangsung sempurna), tetapi HSO4- adalah asam lemah atau elektrolit lemah (diperlukan panah dua arah untuk menunjukkan ionisasi tak sempurna)

Chapter 5

Chapter 5*Triprotic acid H3PO4 H+ + H2PO4-H2PO4- H+ + HPO42-HPO42- H+ + PO43-Acids and BasesKeberadaan asam triprotik relatif sedikit, sebagai contohnya adalah asam fosfat. H3PO4 , H2PO4- , dan HPO42- merupakan asam lemah (dibutuhkan panah dua arah untuk menunjukkan tahap-tahap ionisasi)

Chapter 5

Chapter 5*Base - substance which ionizes to form OH- ions in solution substance which reacts with H+ ions.

Examples: ammoniaNH3sodium hydroxideNaOHAcids and Bases

Chapter 5

Chapter 5*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 NH4+ + OH-

Acids and Bases

Chapter 5

Chapter 5*Strong and Weak Acids and Bases

Strong acids and bases are strong electrolytes.

Weak acids and bases are weak electrolytes.Acids and Bases

Chapter 5

Chapter 5*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 cant be stored in glass bottles because it reacts with glass (silicon dioxide).Acids and Bases

Chapter 5

Chapter 5*Common Strong Acids and Bases

Common Strong AcidsHydrochloric AcidHClHydrobromic AcidHBrHydroiodic acidHINitric AcidHNO3Perchloric AcidHClO4Sulfuric AcidH2SO4Acids and Bases

Chapter 5

Chapter 5*Common Strong Acids and Bases

Common Strong BasesLithium HydroxideLiOHSodium HydroxideNaOHPotassium HydroxideKOH

Acids and Bases

Chapter 5

Chapter 5*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 Salts

Chapter 5

Chapter 5*Neutralization Reaction

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

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

Total Ionic EquationH+(aq) + Cl-(aq) + Na+(aq) + OH-(aq) H2O(l) + Na+(aq) + Cl-(aq)

Acids, Bases, and Salts

Chapter 5

Chapter 5*Neutralization Reaction

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 Salts

Chapter 5

Chapter 5*Metal Carbonates and AcidGas-Forming Reactions2 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

Chapter 5*Metal Sulfide and AcidGas-Forming Reactions2 HCl(aq) + Na2S(s) H2S(g) + 2 NaCl(aq)

Metal sulfides form a salt and hydrogen sulfide.

Chapter 5

Chapter 5*Metal Sulfite and AcidGas-Forming Reactions2 HCl(aq) + Na2SO3(s) SO2(g) + 2 NaCl(aq) + H2O(l)

Metal sulfites form a salt, sulfur dioxide and water.

Chapter 5

Chapter 5*Ammonium Salt and Strong BaseGas-Forming ReactionsNH4Cl(s) + NaOH(aq) NH3(g) + NaCl(aq) + H2O(l)

This reaction forms ammonia, salt and water

Chapter 5

Chapter 5*Oxidation-Reduction ReactionsReaction 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

Chapter 5*Oxidation-Reduction ReactionsReaction 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

Chapter 5*Oxidation-Reduction ReactionsOxidation NumbersEach 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 (exception in binary compounds of metal-hidrogen, LiH, NaH, CaH2 -1) and O is -2 in most compounds (exception in H2O2 and O22- -1).The sum of the oxidation numbers must equal the charge on the molecule or ion.

Chapter 5

Chapter 5*Oxidation-Reduction ReactionsOxidation Numbers

Examples

PCl5P 1( ) =Cl 5( ) = ________ 0

Chapter 5

Chapter 5*Oxidation-Reduction ReactionsOxidation Numbers

Example

PCl5P 1( ? ) = ?Cl 5(-1) = __-5____ 0

Chapter 5

Chapter 5*Oxidation-Reduction ReactionsOxidation Numbers

Example

PCl5P 1(+5) = +5Cl 5(-1) = __-5____ 0

Chapter 5

Chapter 5*Oxidation-Reduction ReactionsOxidation Numbers

Example

CO32-C 1( ) = O 3( ) = _____ -2

Chapter 5

Chapter 5*Oxidation-Reduction ReactionsOxidation Numbers

Example

CO32-C 1(?) = ?O 3(-2) = __-6__ -2

Chapter 5

Chapter 5*Oxidation-Reduction ReactionsOxidation Numbers

Example

CO32-C 1(+4) = +4O 3(-2) = __-6__ -2

Chapter 5

Chapter 5*Oxidation-Reduction ReactionsOxidation Numbers

Example

K2CrO4K 2( ) =O 4( ) =Cr 1( ) = ________ 0

Chapter 5

Chapter 5*Oxidation-Reduction ReactionsOxidation Numbers

Example

K2CrO4K 2(+1) = +2O 4(-2) = -8Cr 1( ? ) = ___?____ 0

Chapter 5

Chapter 5*Oxidation-Reduction ReactionsOxidation Numbers

Example

K2CrO4K 2(+1) = +2O 4(-2) = -8Cr 1(+6) = ___+6__ 0

Chapter 5

Chapter 5*Oxidation-Reduction Reaction Types

Chapter 5

Chapter 5*Oxidation-Reduction Reactions Types

Chapter 5

Chapter 5*Oxidation-Reduction Reactions TypesHydrogen displacementMetal displacementHalogen displacement

Chapter 5

Chapter 5*Oxidation-Reduction Reactions Types

Chapter 5

Chapter 5*Oxidation-Reduction Reactions Types

Chapter 5

Chapter 5*Oxidation-Reduction Reactions Types

Chapter 5

Chapter 5*

Chapter 5

Chapter 5*Molarity(M)Unit of concentration, moles of solute per liter of solution.Concentration of Solution

Chapter 5

Chapter 5*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

Solutions

Chapter 5

Chapter 5*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

Solutions

Chapter 5

Chapter 5*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

Solutions

Chapter 5

Chapter 5*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

Solutions

Chapter 5

Chapter 5*Example: What is the molarity(M) of a solution which contains 17.51g of NaCl in 751mL of solution?Solutions

Chapter 5

Chapter 5*Example: What is the molarity(M) of a solution which contains 17.51g of NaCl in 751mL of solution?

Solutions

Chapter 5

Chapter 5*Example: What is the molarity(M) of a solution which contains 17.51g of NaCl in 751mL of solution?

Solutions

Chapter 5

Chapter 5*SolutionsMolarity

Chapter 5

SolutionsChapter 5*

Chapter 5

Chapter 5*Dilution

Solutions

Chapter 5

Chapter 5*DilutionExample: What volume of 6.00M NaOH is required to make 500mL of 0.100M NaOH?

Mconcentrated = 6.00MMdilute = 0.100MVconcentrated = ?Vdilute = 500mL

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

Vconcentrated = 8.33mLSolutions

Chapter 5

Chapter 5*

Chapter 5

Chapter 5*pH ScaleConcentration 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

Chapter 5*Solution StoichiometryWe 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

Chapter 5*Solution Stoichiometry2 HCl(aq) + 2 Na(s) H2(g) + 2 NaCl(aq)

Convert quantities to moles

Chapter 5

Chapter 5*Solution Stoichiometry2 HCl(aq) + 2 Na(s) H2(g) + 2 NaCl(aq)

Convert quantities to moles

Chapter 5

Chapter 5*Solution Stoichiometry2 HCl(aq) + 2 Na(s) H2(g) + 2 NaCl(aq)

Convert quantities to moles

Determine limiting reagent

Chapter 5

Chapter 5*Solution Stoichiometry2 HCl(aq) + 2 Na(s) H2(g) + 2 NaCl(aq)

Convert quantities to moles

Determine limiting reagent

Chapter 5

Chapter 5*Solution Stoichiometry2 HCl(aq) + 2 Na(s) H2(g) + 2 NaCl(aq)

Calculate moles of H2

Chapter 5

Chapter 5*Solution Stoichiometry2 HCl(aq) + 2 Na(s) H2(g) + 2 NaCl(aq)

Calculate moles of H2

Chapter 5

Chapter 5*Solution Stoichiometry2 HCl(aq) + 2 Na(s) H2(g) + 2 NaCl(aq)

Calculate moles of H2

Calculate grams of H2

Chapter 5

Chapter 5*Solution Stoichiometry2 HCl(aq) + 2 Na(s) H2(g) + 2 NaCl(aq)

Calculate moles of H2

Calculate grams of H2

Chapter 5

Chapter 5*Solution StoichiometryGravimetric Analysis

Chapter 5

Chapter 5*Solution StoichiometryGravimetric Analysis

Chapter 5

Chapter 5*Solution StoichiometryAcid Base Titrations

Chapter 5

Chapter 5*Solution StoichiometryAcid Base Titrations

Chapter 5

Chapter 5*Solution StoichiometryAcid Base Titrations

Chapter 5

Chapter 5*4,10, 16, 20, 24, 30, 36, 38, 44, 60, 62, 68Practice Problems

Chapter 5

**