Types of Chemical Reactions and Solution Stoichiometry

Types of Chemical Types of Chemical ReactionsReactions

and Solution Stoichiometryand Solution Stoichiometry

Classification of MatterClassification of MatterSolutions are homogeneous mixtures

SoluteA solute is the dissolved substance in a solution.

A solvent is the dissolving medium in a solution.

SolvenSolventt

Salt in salt water Sugar in soda drinks

Carbon dioxide in soda drinks

Water in salt water Water in soda

Saturation of SolutionsA solution that contains the maximum

amount of solute that may be dissolved under existing conditions is saturated.

A solution that contains less solute than a saturated solution under existing conditions is unsaturated.

A solution that contains more dissolved solute than a saturated solution under the same conditions is supersaturated.

The ammeter measures the flow of electrons (current) through the circuit.

If the ammeter measures a current, and the bulb glows, then the solution conducts. If the ammeter fails to measure a current, and the bulb does not glow, the solution is non-conducting.

Electrolytes vs. Nonelectrolytes

An electrolyte is:

A substance whose aqueous solution conducts an electric current.

A nonelectrolyte is:

A substance whose aqueous solution does not conduct an electric current.

Try to classify the following substances as electrolytes or nonelectrolytes…

Definition of Electrolytes and Nonelectrolytes

1.Pure water2.Tap water3.Sugar solution4.Sodium chloride solution5.Hydrochloric acid solution6.Lactic acid solution7.Ethyl alcohol solution8.Pure, solid sodium chloride

Electrolytes?

ELECTROLYTES: NONELECTROLYTES:

Tap water (weak)

NaCl solution

HCl solution

Lactate solution (weak)

Pure water

Sugar solution

Ethanol solution

Pure, solid NaCl

But why do some compounds conduct electricity insolution while others do not…?

We must understand the nature of water

Answers…

The Nature of WaterWater molecules are BENT (105⁰<)H & O share electrons UNEVENLY

O pulls harder on e-s, so the O end of the molecule is slightly negative

H pulls less hard on e-s, so the H end of the molecule is slightly positive

This makes H2O a POLAR molecule (oppositely charged ends), like a little magnet

AKA “a dipole”

The Nature of WaterThe polarity of water gives it great ability to

dissolve compounds

1. Ionic Compounds Ionize in Solution• + ions associate with the - (oxygen)

end of the water dipole.

• - ions associate with the + (hydrogen) end of the water dipole.

•

Ions tend to stay in solution where they can conduct a current rather than reforming a solid.

• IONIC CPDS ARE ELECTROLYTES!• Dissociation o

f sodium chloride

Many Ionic Compounds Dissociate Many Ionic Compounds Dissociate (break apart)(break apart)

NaCl(s)

AgNO3(s) MgCl2(s)

Na2SO4(s)

AlCl3(s)

Na+(aq) + Cl-(aq)

Ag+(aq) + NO3-(aq)

Mg2+(aq) + 2 Cl-(aq)

2 Na+(aq) + SO42-

(aq)Al3+(aq) + 3 Cl-(aq)

Note: write “H2O” above each arrow to show that the ionic compound is placed in water, NOT reacting with water.

H2O

H2O

H2O

H2O

H2O

Practice Dissociation EquationsStudy Guide p 106 #1, 2

•Again, b/c water is polar…and also b/c covalent acids are also polar!

•For instance, hydrogen chloride molecules, which are polar, give up their hydrogens to water,

2. Covalent (molecular) acids IONIZE in solution

•forming chloride ions (Cl-) •and hydronium ions (H3O+).

Examples of strong acids include:

Hydrochloric acid, HCl Sulfuric acid, H2SO4

Nitric acid, HNO3

Hydroiodic acid, HI Perchloric acid, HClO4

Strong acids are completely ionized in solution (Strong Electrolytes)

In general, we can assume nearly all other acids are weak

Many of these weaker acids are “organic” acidsthat contain a “carboxyl” group.

The carboxyl group does not easily give up itshydrogen.

Weak acids ionize only slightly (Weak Electrolytes)HC2H3O2 (acetic acid or

vinegar)is a weak acid

Other organic acids and their sources include:

o Citric acid – citrus fruito Malic acid – appleso Butyric acid – rancid buttero Amino acids – proteino Nucleic acids – DNA and RNAo Ascorbic acid – Vitamin C

This is an enormous group of compounds; these are only a few examples.

FYI: Because of the carboxyl group, organic acids are sometimes called “carboxylic acids”.

3.Some Other Polar Covalent Compounds are Weak Electrolytes

They ionize very slightly in water

Ex: Ammonia, NH3

Only about 1% of the molecules dissociate!

•Sugar (sucrose – C12H22O11),

•and ethanol (ethyl alcohol – C2H5OH)

4. MOST covalent compounds do not ionize at all in solution. (Nonelectrolyes)

NOTE: These molecular compounds DISSOLVE (the water pulls the molecules away from each other, but do not IONIZE (the molecules are not charged.)

Practice Identifying ElectrolytesStudy Guide, p 106 # 4 & 5

MolarityThe concentration of a solution measured in moles of solute per liter of solution.

mol = M L

Preparation of Molar Solutions

Problem: How many grams of sodium chloride are needed to prepare 1.50 liters of 0.500 M NaCl solution?

Step #1: Ask “How Much?” (What volume to prepare?)

1.500 L

Step #2: Ask “How Strong?” (What molarity?)

0.500 mol

1 L

Step #3: Ask “What does it weigh?” (Molar mass is?)

58.44 g

1 mol= 43.8 g

Practice calculating mass needed to make a solution

How many grams of NaOH are needed to make 3.5 L of a 2.5 molar solution?

Given3.5 L x

Unknown=?g NaOH2.5 mol NaoH x

1 L of soln39.9g NaoH1 mol NaOH

Serial DilutionIt’s not practical to keep solutions of many different concentrations on hand, so chemists prepare more dilute solutions from a more concentrated “stock” solution.

Problem: What volume of stock (11.6 M) hydrochloric acid is needed to prepare 250. mL of 3.0 M HCl solution? (Note: must convert volumes to L) MstockVstock = MdiluteVdilute OR

M1V1=M2V2(11.6 M)(x Liters) = (3.0 M)(0.250 Liters)

x Liters = (3.0 M)(0.250 Liters) 11.6 M

= 0.065 L

Molarity of Ions in SolutionWhen an ionic compound ionizes in

solution, the number of ions formed may be different than the moles of compound.

H2O(l)

Ex: CrCl3 (s) Cr3+(aq) + 3Cl-(aq)

Say “1 mole of CrCl3 ionizes to form 1 mole of Cr3+ ions and 3 moles Cl- ions”

Calculating Molarity of Ions in Solution0.25 M CrCl3 What is the molarity of Cr3+

ions? Cl- ions? CrCl3 (s) Cr3+(aq) + 3Cl-(aq)

ANSWER Molarity of Cr3+ = molarity of CrCl3 = 0.25 M

Molarity of Cl- = 3 x molarity of CrCl3

= 3 x 0.25M = 0.75M

Practice Dilutions & Molarity CalcsStudy Guide, p 106 # 7, 9, 11, 13,

Types of Reactions5 basicPrecipitationNet ionicOxidation-reduction (redox)Acid-Base

1. Single Replacement Reactions

Replacement of:

Metals by another metal Hydrogen in water by a metal Hydrogen in an acid by a metal Halogens by more active halogens

A + BX AX + B

BX + Y BY + X

The Activity Series of the Metals Lithium Potassium Calcium Sodium* Magnesium Aluminum Zinc Chromium Iron Nickel Lead HydrogenHydrogen Bismuth Copper Mercury Silver Platinum Gold

Metals can replace other metalsprovided that they are above themetal that they are trying to replace.

Metals above hydrogen can replace hydrogen in acids.

*Metals from sodium upward canreplace hydrogen in water

The Activity Series of the Halogens

Fluorine Chlorine Bromine Iodine

Halogens can replace other halogens in compounds, providedthat they are above the halogenthat they are trying to replace.

2NaCl(s) + F2(g) 2NaF(s) + Cl2(g)

MgCl2(s) + Br2(g) ??? No Reaction

???

2.Double Replacement Reactions

The ions of two compounds exchange places in anaqueous solution to form two new compounds.

AX + BY AY + BX

One of the compounds formed is usually • a precipitate (an insoluble solid), • an insoluble gas that bubbles out of solution,• or a molecular compound, usually water.

Highly Soluble Ionic Compounds (& their exceptions) MEMORIZE THESE RULES!

IonIon SolubilitSolubility y

ExceptionsExceptions

NONO33-- SolubleSoluble NoneNone

ClOClO44-- SolubleSoluble NoneNone

NaNa++ SolubleSoluble NoneNone

KK++ SolubleSoluble NoneNone

NHNH44++ SolubleSoluble NoneNone

ClCl--, I, I-- SolubleSoluble PbPb2+2+, Ag, Ag++, Hg, Hg222+2+

SOSO442-2- SolubleSoluble CaCa2+2+, Ba, Ba2+2+, Sr, Sr2+2+, Pb, Pb2+2+, Ag, Ag++, ,

HgHg2+2+

Slightly Soluble (usually considered insoluble) Ionic Compounds MEMORIZE THESE RULES!

IonIon Solubility Solubility ExceptionsExceptions

COCO332-2- Slightly Slightly

solublesolubleGroup 1 (IA) and NHGroup 1 (IA) and NH44

++

POPO443-3- “ Group 1 (IA) and NHGroup 1 (IA) and NH44

++

OHOH-- “ Group 1 (IA) and CaGroup 1 (IA) and Ca2+2+, Ba, Ba2+2+, , SrSr2+2+

SS2-2- “ Groups 1 (IA), 2 (IIA), and Groups 1 (IA), 2 (IIA), and NHNH44

++

CrOCrO4 4 2-2- “ nonenone

Completing a Double replacement Rxn:

Ex: Pb(NO3)2(aq) + 2KI(aq) PbI2 + 2KNO3

1. Determine the products formed when the ions are exchanged, & balance

2. Decide if the products are soluble (aq) or insoluble (s)

PbI2 Insoluble, so it will be a solid

KNO3 Soluble, so it will be in aqueous solution

(s) (aq)

Complete Ionic Equation

Shows all soluble compounds as aqueous ions

Shows all insoluble compounds as a unit, use the symbol (s) to show it is a solid.

Pb2+(aq) + 2 NO3-(aq) + 2 K+(aq) +2 I-(aq) 2K+(aq) + 2

NO3-(aq)

PbI2(s) +

Pb(NO3)2(aq)

+ 2KI 2KNO3 (aq)

PbI2(s) +

Net Ionic EquationEliminates all “spectator” ions (ions that appear identically on both sides of

equation)

PbI2(s) +

PbPb2+2+(aq) + 2 I(aq) + 2 I--(aq) (aq) PbI PbI22(s)(s)

Practice Complete Ionic Equation & Net Ionic Equation

CuSO4 + Na2S CuS (insoluble) + Na2SO4 (soluble)

Complete Ionic

Net Ionic

Cu2+(aq) +SO42-(aq) +2Na+(aq) +S2-(aq) CuS(s) + 2Na+(aq)+SO4

2- (aq)

Cu2+(aq) + S2-(aq) CuS(s)

Practice ProblemsStudy Guide: Please Note: show all work in

your notebook, not your study guide!

Pp 108 #28

Stoichiometry of Precipitation Rxns

It is helpful to be able to predict the amount of precipitant formed b/c it is often collected & used.

Ex: Calculate the mass of solid NaCl that must be added to 1.50L of a 0.100 M AgNO3 solution to precipitate all the Ag+ ions in the form of AgCl.

First, we need a balanced equation: NaCl(s) + AgNO3(aq) AgCl(s) + NaNO3(aq)

___g NaCl8.77

Stoichiometry of Precipitation Rxns, cont.Please note: you will also encounter limiting

reactant problems with precip. Rxns!Remember, set up 2 separate equations to

see which reactant forms the LEAST product. This is your limiting reactant. The equations will be identical to those on the prior page.

Stoichiometry of Precipitation Rxns, cont.

HINT: label all equations with your givens & unknown.

Example: When aqueous solutions of Na2SO4 and Pb(NO3)2 are mixed, PbSO4 precipitates. Calculate the mass of PbSO4 formed when 1.25 L of 0.0500 M Pb(NO3)2 and 2.00 L of 0.0250 M Na2SO4 are mixed.

Na2SO4 (aq) + Pb(NO3)2 (aq) PbSO4 (s) + 2NaNO3 (aq)

V=2.00 L M=0.0250mol/L

V=1.25 L M=0.0500mol/L

m=? g

Practice Problems:Study Guide, pp __________ Q # _________Please show all your work in your notebook,

not your study guide.

Oxidation and Reduction Reactions Oxidation and Reduction Reactions (Redox)(Redox)

Def: Rxns in which electrons are transferred

Ex: Na(s) + Cl2(g) NaCl(s)

An electron transfers from the Na atom to the Cl atom.

What about molecular compounds?

Non-ionic compounds can also be formed from redox reactions. Even though e-s aren’t FULLY transferred, they can be assumed to involve a transfer…let’s see how!

Oxidation and Reduction Reactions Oxidation and Reduction Reactions (Redox)(Redox)

Ex:CH4(g) + 2O2(g) CO2(g) + 2H2O + energy

Carbon is less EN than oxygen, so we assume there is a transfer of e-s from C to O.

NOTE: We will study electronegativity (EN) in Chapter 8.

•EN is the strength with which an atom in a bond pulls on electrons.(Check Figure 8.3 on p 353.)

•This shows us the value for EN of O is 3.5 and the EN value for C is 2.5.

Oxidation and Reduction Rxns, Oxidation and Reduction Rxns, cont.cont.

Oxidation States (Oxidation #s) provide a way to track e-s in redox reactions, especially in molecular substances.

Rules for Assigning Oxidation NumbersRules for Assigning Oxidation NumbersRules 1 & 2Rules 1 & 2

1. The oxidation number of any uncombined element is zero

2. The oxidation number of a monatomic ion equals its charge

11

2

00

22

ClNaClNa

Rules for Assigning Oxidation NumbersRules for Assigning Oxidation NumbersRules 3 & 4Rules 3 & 4

3. The oxidation number of oxygen in compounds is -2 (except in peroxides, in which the oxidation number is -1)

4. The oxidation number of hydrogen in compounds is +1

2

1

2

1

OH21

2

OH

Rules for Assigning Oxidation Number Rules for Assigning Oxidation Number Rule 5Rule 5

5. The sum of the oxidation numbers in the formula of a compound is 02

2

1

OH2(+1) + (-2) = 0 H O

2

122

)(

HOCa(+2) + 2(-2) + 2(+1) = 0 Ca O H

Rules for Assigning Oxidation NumbersRules for Assigning Oxidation NumbersRule 6Rule 6

6. The sum of the oxidation numbers in the formula of a polyatomic ion is equal to its charge

3

2?

ONX + 3(-2) = -1N O

24

2?

OS

X = +5 X = +6

6. The sum of the oxidation numbers in the formula of a polyatomic ion is equal to its chargeX + 4(-2) = -2S O

X = +6

Practice TogetherAssign oxidation

states to all atoms in the following:

1.CO2

2.SF6

3.NO2-

1. C+4 O-2

2. S+6 F-1

3. N+3 O-2

Check the math on #3 (NO2-)

+3 + 2(-2) = -1

Practice Assigning Oxidation Numbers

Study Guide, p 110Q# 50-52, Please Note: show all work in your

notebook, not your study guide!

Oxidation and Reduction

GGainain EElectronslectrons = = RReductioneduction

An old memory device for An old memory device for oxidation and reduction goes oxidation and reduction goes like this…like this… LEOLEO says says GERGER

LLoseose EElectronslectrons = = OOxidationxidation

Using Half Reactions to understand Redox Using Half Reactions to understand Redox ReactionsReactions

11

2

00

22

ClNaClNaEach sodium atom loses one electron:

Each chlorine atom gains one electron:

eNaNa10

10 CleCl

LEO says GERLEO says GER : :

eNaNa10

Lose Electrons = Oxidation

Sodium is oxidized

Gain Electrons = Reduction

10 CleCl Chlorine is reduced

Reducing Agents and Oxidizing AgentsReducing Agents and Oxidizing Agents

The substance reduced is the oxidizing agent The substance oxidized is the reducing agent

eNaNa10

10 CleCl

Sodium is oxidized – it is the reducing agent

Chlorine is reduced – it is the oxidizing agent

Using Half-reactions to Balance Redox Equations

It is helpful to separate the rxn into two half-reactions:1.Oxidation2.Reduction

Example: Ce4+ (aq) + Sn2+ (aq) Ce3+ (aq) + Sn4+ (aq)

1. Oxidation: Sn2+ (aq) + 2 e- Sn4+ (aq)

2. Reduction: Ce4+ (aq) Ce3+ (aq) + e-

NOTE: If you combine these equations, the e-s and charges don’t balance. So, multiply the 2nd equation by 2.

Using Half-reactions to Balance Redox Equations

1. Oxidation: Sn2+ (aq) + 2 e- Sn4+ (aq)

2. Reduction: 2(Ce4+ (aq) Ce3+ (aq) + e-)

Now, combine (add) the two equations, cancelling items that appear identically on both sides of the arrow.

2Ce4+ (aq) 2Ce3+ (aq) + 2 e-=

Practice ProblemsStudy Guide, p 110, Q # 53-54Show all your work in your notebook (not

your study guide!)

Trends in Oxidation and ReductionTrends in Oxidation and Reduction

Active metals: Lose electrons easily Are easily oxidized Are strong reducing

agents

Active nonmetals: Gain electrons easily Are easily reduced Are strong oxidizing agents

Redox Reaction Prediction #1

Important Oxidizers

Formed in reaction

MnO4- (acid solution)

MnO4- (basic solution)

MnO2 (acid solution)

Cr2O72- (acid)

CrO42-

HNO3, concentrated

HNO3, dilute

H2SO4, hot conc

Metallic IonsFree Halogens

HClO4

Na2O2

H2O2

Mn(II)MnO2

Mn(II)Cr(III)Cr(III)NO2

NOSO2

Metallous IonsHalide ionsCl-

OH-

O2

Redox Reaction Prediction #2

Important Reducers Formed in reaction

Halide IonsFree MetalsMetalous IonsNitrite IonsSulfite IonsFree Halogens (dil, basic

sol)Free Halogens (conc, basic

sol)C2O4

2-

HalogensMetal IonsMetallic ionsNitrate IonsSO42-Hypohalite ionsHalate ionsCO2

Not All Reactions are Redox ReactionsNot All Reactions are Redox Reactions

Reactions in which there has been no change in oxidation number are not redox rxns.

Examples:

)()()()( 3

2511111

3

251

aqONNasClAgaqClNaaqONAg

)()()()(22

2

1

4

26

2

1

4

26

2

1121

lOHaqOSNaaqOSHaqHONa

Acid-Base ReactionsArrhenius definition:

Acid is a proton (hydrogen ion) donorBase is a proton (hydrogen ion) acceptor

Acid-Base Reactions:Strong Acid/Strong Base

To predict results of rxn, focus on the species present in mixed solution.

Ex#1 : HCl(aq) + NaOH(aq) ???

Species Present before rxn occurs: H+ (aq) + Cl- (aq) + Na+ (aq) +

OH- (aq) (b/c HCl is a strong acid and NaOH is a strong

base=complete dissociation of both acid & base)

NaCl is soluble in water (check your table) so Na+ & Cl- are spectator ions.

Acid-Base Reactions:Strong Acid/Strong Base

Ex#1(continued): HCl(aq) + NaOH(aq) ???

H+ (aq) + OH- (aq) are the only other species present.

Water is a nonelectrolyte, so H+ (aq) + OH- will NOT exist in solution together. Therefore, they must combine to form H2O.

Our net ionic equation for this reaction is therefore:

H+ (aq) + OH- (aq) H2O (l)

Acid-Base Reactions:Weak Acid/Strong Base

Again, to predict results of rxn, focus on the species present in mixed solution.

Ex#2 : HC2H3O2(aq) + KOH(aq) ???

Species Present before rxn occurs: HC2H3O2(aq) + K+ (aq) + OH- (aq)

b/c HC2H3O2(aq) is a weak acid (almost no dissociation = 1%) &

b/c KOH is a strong base (complete dissociation)

Acid-Base Reactions:Weak Acid/Strong Base

Note: OH- is such a strong base that it will strip H+ ions from the HC2H3O2(acetic acid)

we can assume it reacts completely with any weak acid it will be combined with

By the same logic, we can assume that any STRONG ACID will react completely with any WEAK BASE

Acid-Base ReactionsOften called “neutralization reactions” Complete equation shows formation of

a salt and water

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

Stoichiometry of Neutralization Reactions

1. List the species present in the combined solution before any reaction occurs & decide what will occur.

2. Write balanced net ionic equation.3. Set up your calculation, identifying Given,

Unknown, and conversion factors that cancel the necessary units. Or identify the proper formula and solve for unknown.

Stoichiometry of Neutralization ReactionsExample: What volume of a 0.100 M HNO3 solution is

needed to neutralize 25.0 mL of 0.350 M NaOH?

1. List the species present in the combined solution before any reaction occurs & decide what will occur.

H+ (aq) + NO3- (aq) + Na+ (aq) + OH- (aq)

(HNO3 is a strong acid and NaOH is a strong base, so both completely ionize in aqueous solution)

• NaNO3 is soluble, so the Na+ and NO3- ions are spectators

• H+ and OH- do not coexist, so H2O will form

Stoichiometry of Neutralization Reactions

2. Write the balanced net ionic equation for this reaction.

H+ (aq) + OH- (aq) H2O (l)

Stoichiometry of Neutralization Reactions3. Set up your calculation, identifying Given, Unknown, and conversion factors that cancel the necessary units. Or

identify the proper formula and solve for the unknown.

• Analysis: Careful! Even though there are 2 Ms & 2 Vs, this is not a dilution problem where we use M1V1 = M2V2!

• Instead, set up Given & Unknown & map your path.

M =0.100 mol/L V= ??? L

M=0.350 mol/LV = 0.025 L

H+ (aq) + OH- (aq) H2O (l)

Given0.025L OH-

Unknown= L H+x 0.350 mol OH-

1 L OH- x 1 mol H+ 1 mol OH-

x 1L H+ 0.100 mol H+

8.75 x 10-3

Practice Acid-Base Stoichiometry ProblemsStudy Guide, p 110, Q # 55-57Show all your work in your notebook (not

your study guide!)