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Chapter 4 Chapter 4 Types of Chemical Types of Chemical Reactions Reactions And Solution And Solution Stoichiometry Stoichiometry

Chapter 4

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Page 1: Chapter 4

Chapter 4Chapter 4

Types of Chemical Types of Chemical ReactionsReactions

And Solution And Solution StoichiometryStoichiometry

Page 2: Chapter 4

Water, The Common Water, The Common SolventSolventHydration of an ionic compound will occur when the partial positive end of a water becomes attracted to the anions in the compound; likewise for the partial negative center of the water and the cations.

Solubility depends on the strength of the intermolecular attractions between the ions and water, as well as the intramolecular attractions of the cations and anions of the compound.

NHNH44NONO3(s)3(s) NH NH44++

(aq)(aq) + NO + NO33--(aq)(aq)

Page 3: Chapter 4

What can dissolve in HWhat can dissolve in H22O?O?

SolubleSoluble AlcoholsAlcohols

ex: Cex: C22HH55OHOH SugarsSugars ex: Cex: C66HH1212OO66

Ionic compoundsIonic compounds

ex: NaCl, KOH, LiBrex: NaCl, KOH, LiBr

InsolubleInsoluble FatsFats

ex: bacon greaseex: bacon grease OilsOils

ex: cooking oilex: cooking oil Non-Polar SubstancesNon-Polar Substances

ex: turpentineex: turpentine

Because of intermolecular forces: the OH group on the sugars and alcohols is particularly attractive to a water molecule.

Generally speaking: “Like Dissolves Like”

Page 4: Chapter 4

Strong and Weak ElectrolytesStrong and Weak Electrolytes

Solute + Solvent = SolutionSolute + Solvent = Solution Strong electrolytes conduct electricityStrong electrolytes conduct electricity Weak electrolytes barely conduct electricityWeak electrolytes barely conduct electricity Conductivity depends upon ionizationConductivity depends upon ionization

Page 5: Chapter 4

Strong ElectrolytesStrong Electrolytes Soluble saltsSoluble salts Strong acidsStrong acids Strong basesStrong bases

All of these dissociate All of these dissociate completelycompletely in water. in water.

Weak ElectrolytesWeak Electrolytes Weak acidsWeak acids Weak basesWeak bases

All of these All of these partiallypartially dissociate in waterdissociate in water

HCl H+ + Cl-

NaOH Na+ + OH-

HC2H3O2 H+ + C2H3O2

Non-ElectrolytesNon-Electrolytes are completely molecular substances in water (not even a little dissociation); Non polar substances.

Page 6: Chapter 4

Composition of SolutionsComposition of Solutions Concentration is measured in molarity, Concentration is measured in molarity,

molality, and many others.molality, and many others. Concentration DOES NOT directly Concentration DOES NOT directly

express the number of ions present in a express the number of ions present in a solution.solution.

M= moles solute liters solution

M= moles solute liters solution

MgClMgCl22 Mg Mg2+2+ + 2Cl + 2Cl--

1.0 M1.0 M 1.0 M 2.0 1.0 M 2.0 MM

Page 7: Chapter 4

Sample ProblemsSample Problems

Calculate the number of moles of ClCalculate the number of moles of Cl-- ions in 1.75 ions in 1.75 L of 1.0 x 10L of 1.0 x 10-3-3 M ZnCl M ZnCl22..

A chemist needs 1.0 L of 0.20 M KA chemist needs 1.0 L of 0.20 M K22CrCr22OO77 solution. solution. How much solid KHow much solid K22CrCr22OO77 must be weighed out to must be weighed out to make this solution?make this solution?

ClmolxL

molxxL 3

3

105.31

100.275.1

722120.294

722

722120.0

8.5820.0

20.00.1

722 OCrKgxOCrKmol

OCrKmolxL

molOCrgK

Lmol

Page 8: Chapter 4

Standard Solution: a solution whose Standard Solution: a solution whose concentration is accurately known.concentration is accurately known. Example: 0.1022 M HCl; 1.003 M NaOHExample: 0.1022 M HCl; 1.003 M NaOH

Creating dilutionsCreating dilutions Chemical analysis of a compoundChemical analysis of a compound Theoretical CalculationsTheoretical Calculations

ANSWER ANSWER NOWNOW

What would you do to What would you do to prepare a standard solution? prepare a standard solution? In your answer, include In your answer, include specific pieces of glassware, specific pieces of glassware, techniques, or equipment techniques, or equipment you should use.you should use.

Page 9: Chapter 4

DilutionsDilutions Dilution is the process used to make the Dilution is the process used to make the

solution less concentrated.solution less concentrated.

moles before dilution = moles after dilutionmoles before dilution = moles after dilutionBecause M =mol/L,Because M =mol/L,

VV11(M(M11) = V) = V22(M(M22))

Lab Technique: Use a pipet to deliver the Lab Technique: Use a pipet to deliver the correct amount of original solution to a correct amount of original solution to a

volumetric flask. Add some water, swirl. Fill volumetric flask. Add some water, swirl. Fill to line, invert. to line, invert.

Page 10: Chapter 4

DO DO NOWNOW

You have a large quantity of 0.9 You have a large quantity of 0.9 M NaOH solution available. M NaOH solution available. Dilute this to 100.0mL of a 0.05 Dilute this to 100.0mL of a 0.05 M solution. Submit your M solution. Submit your calculations and solution to your calculations and solution to your teacher. teacher.

Page 11: Chapter 4

Types of Chemical Types of Chemical ReactionsReactions

There are more than just these few There are more than just these few types, but in this chapter we will types, but in this chapter we will cover…cover…

PrecipitationPrecipitationAcid-baseAcid-baseOxidation-ReductionOxidation-Reduction

Page 12: Chapter 4

Precipitation ReactionsPrecipitation Reactions

Precipitation Reactions Precipitation Reactions (double (double displacement)displacement)

Forms a solid precipitate from aqueous Forms a solid precipitate from aqueous reactants.reactants.

Color of precipitate can help in identificationColor of precipitate can help in identification Solubility rules help BUNCHESSolubility rules help BUNCHES

MORE…

Page 13: Chapter 4

Solubility RULESSolubility RULES

All compounds containing All compounds containing alkali metalalkali metal cations and the cations and the ammoniumammonium ion are soluble. ion are soluble.

All compounds containing All compounds containing NONO33--, ClO, ClO44

--, ClO, ClO33--, and C, and C22HH33OO22

-- anions are soluble.anions are soluble.

All All chlorides, bromides, and iodideschlorides, bromides, and iodides are soluble except are soluble except those containing Agthose containing Ag++, Pb, Pb2+2+, and Hg, and Hg2+2+..

All All sulfatessulfates are soluble except those containing Hg are soluble except those containing Hg2+2+, Pb, Pb2+2+, , SrSr2+2+, Ca, Ca2+2+, and Ba, and Ba2+2+..

All hydroxides are only slightly soluble, except those All hydroxides are only slightly soluble, except those containing an alkali metal, Cacontaining an alkali metal, Ca2+2+, Ba, Ba2+2+,and Sr,and Sr2+2+. . NaOHNaOH and and KOHKOH are the most soluble hydroxides. are the most soluble hydroxides.

All compounds containing POAll compounds containing PO443-3-, S, S2-2-, CO, CO33

2-2-, and SO, and SO332-2- are only are only

slightly soluble except for those containing alkali metals or slightly soluble except for those containing alkali metals or the ammonium ion.the ammonium ion.

Page 14: Chapter 4

Practice PredictingPractice Predicting

Potassium nitrate and barium Potassium nitrate and barium chloridechloride

Sodium sulfate and lead (II) nitrateSodium sulfate and lead (II) nitrate

Potassium hydroxide and iron (III) Potassium hydroxide and iron (III) nitratenitrate

Page 15: Chapter 4

ALL REACTIONS ALL REACTIONS SHOULD BE SHOULD BE WRITTEN IN WRITTEN IN NET IONIC NET IONIC

FORMFORM

Page 16: Chapter 4

Stoichiometry of Precipitation Stoichiometry of Precipitation ReactionsReactions

Stoichiometry in a precipitation Stoichiometry in a precipitation reaction is performed just like reaction is performed just like stoichiometry for a molecular stoichiometry for a molecular reaction. reaction.

You need to know which ion comes You need to know which ion comes from which molecular formula.from which molecular formula.

Page 17: Chapter 4

Sample problem Sample problem

Calculate the mass of solid NaCl needed Calculate the mass of solid NaCl needed to add to 1.5 L of 0.1 M silver nitrate to add to 1.5 L of 0.1 M silver nitrate solution to precipitate all Agsolution to precipitate all Ag++ ions in the ions in the form of AgCl.form of AgCl.

Net Ionic Eq: Ag+ + Cl- AgCl

NaClgxNaClmol

NaClmolClmolxAgmol

AgmolxL

NaClmolNaClg

Ag

Cl

LAgM

77.8150.0

15.015.0150.0

150.05.1

145.58

1

1

110.0

Page 18: Chapter 4

General FormatGeneral Format

Write the Net Ionic EquationWrite the Net Ionic Equation Calculate the moles presentCalculate the moles present Identify the Limiting Reactant*Identify the Limiting Reactant* Use Mole Ratio(s)Use Mole Ratio(s) Fancy-fy your answer (put in correct Fancy-fy your answer (put in correct

units)units)

Page 19: Chapter 4

Try Me!Try Me!

What mass of precipitate will be What mass of precipitate will be produced when 50.0 mL of 0.200M produced when 50.0 mL of 0.200M aluminum nitrate is added to 200.0 aluminum nitrate is added to 200.0 mL of 0.100 M potassium hydroxide?mL of 0.100 M potassium hydroxide?

Page 20: Chapter 4

Acid-Base ReactionsAcid-Base Reactions

Definitions of acid Definitions of acid and base vary.and base vary.

Arrhenius and Arrhenius and Bronsted/Lowry Bronsted/Lowry are common are common theories.theories.

Acid-Base rxns Acid-Base rxns are called are called NEUTRALIZATIONNEUTRALIZATIONSS

Acids yield H+

Bases yield OH -

Acids are proton

donors

Bases are proton

acceptors

Page 21: Chapter 4

Strong Acid-Strong BaseStrong Acid-Strong Base

(HCl) (NaOH)(HCl) (NaOH)Both dissociate completelyBoth dissociate completely

HH++ + OH + OH-- H H22OO

NaNa++ and Cl and Cl-- are spectators. are spectators.

Weak Acid - Strong BaseWeak Acid - Strong Base

(HC(HC22HH33OO22) (KOH)) (KOH)

Acetic acid will not dissociateAcetic acid will not dissociate

KOH will completelyKOH will completely

HCHC22HH33OO22 + OH + OH-- H H22O + O + CC22HH33OO22--

KK++ is a spectator. is a spectator.

Page 22: Chapter 4

Stoichiometry sampleStoichiometry sample

What volume of 0.100 M HCl is What volume of 0.100 M HCl is needed to neutralize 25 mL of 0.35 M needed to neutralize 25 mL of 0.35 M NaOH?NaOH?

H+ + OH- H2O

HClLx

xxNaOHxL molmLHCl

molOHmolH

LNaOHmolOH

2

100.1

11

135.0

1075.8

025.0

Page 23: Chapter 4

Titrations…define me!Titrations…define me!

Volumetric analysisVolumetric analysis TitrationTitration TitrantTitrant AnalyteAnalyte Equivalence pointEquivalence point IndicatorIndicator EndpointEndpoint

Page 24: Chapter 4

To complete a successful To complete a successful titration…titration…

1.1. The reaction between the titrant and the The reaction between the titrant and the analyte should be known (analyte should be known (you should you should know WHAT substances you haveknow WHAT substances you have))

2.2. The equivalence point should be marked The equivalence point should be marked accurately (accurately (you should use the right you should use the right indicatorindicator))

3.3. Volume of the titrant needed to reach the Volume of the titrant needed to reach the equivalence point should be recorded equivalence point should be recorded accurately (accurately (you should use a buret!)you should use a buret!)

Page 25: Chapter 4

Effective Indicator RangesEffective Indicator Ranges

Page 26: Chapter 4

Titration Try Me Calc 1Titration Try Me Calc 1

A 50.0 mL sample of a sodium hydroxide A 50.0 mL sample of a sodium hydroxide solution is to be standardized. 1.3009 solution is to be standardized. 1.3009 M of KHP (potassium hydrogen M of KHP (potassium hydrogen phthalate, KHCphthalate, KHC88HH44OO44) is used as the ) is used as the titrant. KHP has one acidic hydrogen. titrant. KHP has one acidic hydrogen. 41.20 mL of the KHP solution is used 41.20 mL of the KHP solution is used to titrate the sodium hydroxide solution to titrate the sodium hydroxide solution to the endpoint. What is the resulting to the endpoint. What is the resulting concentration of the analyte?concentration of the analyte?

Page 27: Chapter 4

Titration Try Me Calc 2Titration Try Me Calc 2

How many milliliters of a 0.610 M How many milliliters of a 0.610 M sodium hydroxide solution are sodium hydroxide solution are needed to neutralize 20.0 mL of a needed to neutralize 20.0 mL of a 0.245 M sulfuric acid solution?0.245 M sulfuric acid solution?

Page 28: Chapter 4

Norton TutorialNorton Tutorial

Go to the website Go to the website http://www.wwnorton.com/college/chemistry/chemistry3/ch/17/chemtours.aspx

Find the tutorial on Acid/Base Find the tutorial on Acid/Base ionization.ionization.

Complete the tutorial question form.Complete the tutorial question form.

Page 29: Chapter 4

Redox ReactionsRedox ReactionsWhat is it??What is it??

-A reaction that occurs in conjunction with a -A reaction that occurs in conjunction with a transfer of electrons.transfer of electrons.

We assign oxidation states to individual atoms We assign oxidation states to individual atoms in a reaction to observe the change in in a reaction to observe the change in electrons. electrons.

Oxidation states are written with

the +/- sign before the quantity.

Ion charges are Ion charges are written with the written with the +/- sign behind +/- sign behind

the quantity.the quantity.

Page 30: Chapter 4

The Oxidation State The Oxidation State of…of…

Quantity of Oxid. Quantity of Oxid. StateState

ExamplesExamples

An atom in element form

Zero Na(s), O2(g)

A monatomic ion Equal to the charge on the ion

Na+, Cl-

Fluorine in a compound -1 , always HF, PF3

Oxygen in a compound -2, except in peroxide where it is -1

H2O, CO2, H2O2

Hydrogen in a compound

+1, always H2O, HCl, NH3

Page 31: Chapter 4

Oxidation= an increase in the Oxidation= an increase in the oxidation stateoxidation state

Reduction = a decrease in the Reduction = a decrease in the oxidation stateoxidation state

2Na(s) + Cl2(g) 2NaCl(s)

oxidation

reduction

Page 32: Chapter 4

The metal is The metal is oxidized and the oxidized and the

other substance is other substance is reduced.reduced.

Metal Atom

Metal Ion

Other Ion

Other Atom

e-

Oxidized Substance:Oxidized Substance:•Loss of electronsLoss of electrons•Oxidation state increasesOxidation state increases•Gets SmallerGets Smaller•Called the Reducing AgentCalled the Reducing Agent

Reduced Substance:Reduced Substance:•Gain of electronsGain of electrons•Oxidation state decreasesOxidation state decreases•Gets BiggerGets Bigger•Called the Oxidizing AgentCalled the Oxidizing Agent

Page 33: Chapter 4

Section 4.10: Section 4.10: Balancing Balancing RedoxRedox

1.1. Write the ½ reactions Write the ½ reactions

2.2. Balance the non-H and non-O atomsBalance the non-H and non-O atoms

3.3. Balance O by adding HBalance O by adding H22O where neededO where needed

4.4. Balance H by adding HBalance H by adding H++ where needed where needed

5.5. Balance charge using e-Balance charge using e-

6.6. Multiply by coefficients until both e- are Multiply by coefficients until both e- are equal for each ½ reactionequal for each ½ reaction

7.7. Add the ½ reactions together (cancel stuff)Add the ½ reactions together (cancel stuff)

Page 34: Chapter 4

Redox Sample Problem Redox Sample Problem

Balance:Balance:

MnOMnO44-- + Fe + Fe2+ 2+ Fe Fe3+3+ + Mn + Mn2+2+

OHMnMnOH

OHMnMnO

MnMnO

22

4

22

4

24

48

4

32 FeFe

Check Charges!8+ 1- 2+ 0

OHMnMnOHe 22

4 485

2+ 3+

eFeFe 32

x5!

OHMnFeHMnOFe 223

42 4585

+7 +2

+2 +3

x5!

Page 35: Chapter 4

Redox Try Me ProblemRedox Try Me Problem

AsAs22OO3(s)3(s) + NO + NO33-- H H33AsOAsO44 + NO + NO(g)(g)

Page 36: Chapter 4

1.1. Repeat steps from old Repeat steps from old procedureprocedure

2.2. Cancel out HCancel out H++ by adding by adding OHOH-- ions ions

3.3. Re-write HRe-write H++ and OH and OH-- as as waterwater

4.4. Add ½ reactions together Add ½ reactions together (and cancel stuff)(and cancel stuff)

Page 37: Chapter 4

Redox Try Me Problem 2Redox Try Me Problem 2

Balance, in base:Balance, in base:

AgAg(s)(s) + CN + CN-- + O + O22 Ag(CN) Ag(CN)22--