1/13/2015 1 George Mason University General Chemistry 211 Chapter 4 Three Major Classes of Chemical...

1/13/2015 1

George Mason UniversityGeneral Chemistry 211

Chapter 4Three Major Classes of Chemical Reactions

AcknowledgementsCourse Text: Chemistry: the Molecular Nature of Matter and

Change, 7th edition, 2011, McGraw-Hill

Martin S. Silberberg & Patricia Amateis

The Chemistry 211/212 General Chemistry courses taught at George Mason are intended for those students enrolled in a science /engineering oriented curricula, with particular emphasis on chemistry, biochemistry, and biology The material on these slides is taken primarily from the course text but the instructor has modified, condensed, or otherwise reorganized selected material.Additional material from other sources may also be included. Interpretation of course material to clarify concepts and solutions to problems is the sole responsibility of this instructor.

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Chapter 4Classes of Chemical Reactions

The Role of Water as a Solvent Polar Nature of Water Ionic Compounds Covalent Compounds

Writing Equations for Aqueous Ionic Reactions Precipitation Reactions

Formation of a Solid Predicting a Precipitate

Acid-Base Reactions Formation of Water Acid-Base Titrations Proton Transfer in Acid-Base Reactions

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Chapter 4Classes of Chemical Reactions

Oxidation-Reduction (REDOX) Reactions Movement of Electrons Redox Terminology Oxidation Number Balancing Redox Equations Redox Titrations

Elements in Redox Reactions Reaction Reversibility and the Equilibrium State

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Water as a Solvent Nearly all reactions in the

environment and, especially in organisms, take place in water

Water is a covalent molecule where the shared pair of electrons is attracted more strongly toward the oxygen nucleus forming a partially negative charged “pole” near the Oxygen nucleus and a partially positively charged pole near the hydrogen proton nucleus

The H-O-H arrangement forms an angle

(104.5o)

The bent shape and polar bonds produce a polar molecule

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Solubility Substances (solutes) soluble in water

In some cases, the force of the attraction between the ions’ solid substance form mixed with water is so strong that it cannot be overcome by the interaction of the ions with the polarized water molecules.

These materials will be insoluble in water The interaction with water depends on the

structure of the molecule. If the interaction between ion and water is

strong, the substance will be soluble If the interaction is weak. the substance will

not be very soluble

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Solubility Substances (solutes) soluble in water

If a covalent molecule contains polar groups, they will interact well with the polar solvent water, i.e., they will dissolve!

A few covalent molecules, such as HCl, dissociate completely into ions

In general, covalent compounds which produce ions in an aqueous solution interact with the water molecules to form either:● H+(aq) or H3O+ (aq) ions (acidic solutions)

or● OH–(aq) ions (basic or alkaline solutions)

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Ions in Aqueous Solution Ionic Compounds as Electrolytes

In an Ionic solid, the oppositely charged ions (Cations & Anions) are held together by electrostatic attraction

In aqueous solution, the electrostatic attraction is replaced with Water molecules

Many ionic compounds dissociate into independent ions when dissolved in water

Compounds that “freely” dissociate into independent ions in aqueous solution are called electrolytes because the solutions are good conductors of electricity

2 + -NaCl(s) Na (aq) + Cl (aq)H O

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Ions in Aqueous Solution Molecular Compounds as Electrolytes

Not all electrolytes are ionic compoundsMolecular compounds consist of individual

molecules whose physical state depends on intermolecular forces (to be discussed later), not ionic attractions

Some molecular compounds dissociate into ions

The resulting solution is electrically conducting, and is called an electrolyte

+ -HCl(aq) H (aq) + Cl (aq)

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Ions in Aqueous Solution Molecular Compounds as Nonelectrolytes

Some molecular compounds dissolve but do not dissociate into ions

These compounds contain their own polar bonds, which interact with the polar bonds of water

They do not conduct an electric current, i.e., they produce a nonconducting solution

The compounds are referred to as nonelectrolytes

2 H O

6 12 6 6 12 6C H O (s) (glucose) C H O (aq)

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Ions in Aqueous Solution Ionic Theory of Solutions

Strong and Weak Electrolytes

● A strong electrolyte is an electrolyte that exists in solution almost entirely as ions

● Most ionic solids that dissolve in water do so almost completely as ions, so they are strong electrolytes

H O + -2NaCl(s) Na (aq) + Cl (aq)

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Ions in Aqueous Solution Ionic Theory of Solutions

Strong and Weak Electrolytes

● A weak electrolyte is an electrolyte that dissolves in water to give a relatively small percentage of ions

● Most soluble molecular compounds are either:

nonelectrolytes or weak electrolytes

+ -4 4NH OH(aq) NH (aq) + OH (aq)

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Ions in Aqueous Solution Molecular and Ionic Equations

A molecular equation is one in which the reactants and products are written as if they were molecules, even though they may actually exist in solution as ions

Note that Ca(OH)2, Na2CO3, and NaOH are all soluble compounds, but CaCO3 is a solid, i.e., a precipitate

2 2 3 3Ca(OH) (aq) + Na CO (aq) CaCO (s) + 2NaOH(aq)

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An ionic equation, however, represents strong electrolytes as separate independent ions. This is a more accurate representation of the way electrolytes behave in solution

2+ - + 2-3Ca (aq) + 2OH (aq) + 2Na (aq) + CO (aq)

+ -3CaCO (s) + 2Na (aq) + 2OH (aq)

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Complete and net ionic equations● A complete ionic equation is a

chemical equation in which strong electrolytes (such as soluble ionic compounds) are written as separate ions in solution

2+ - + 2-3 3Ca (aq) + 2NO (aq) + 2K (aq) + CO (aq)

+ -3 3CaCO (s) + 2K (aq) + 2NO (aq)

3 2 2 3 3 3Ca(NO ) (aq) + K CO (aq) CaCO (s) + 2KNO (aq)

strong strong insoluble strong

Molecular Equation

Complete Ionic Equation

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Complete and net ionic equations (con’t)

● A net ionic equation is a chemical equation from which the spectator ions have been removed

● A spectator ion is an ion in an ionic equation that does not take part in the reaction

Ex. Reaction between Calcium Nitrate [Ca(NO3)2] & Potassium Carbonate [K2CO3]

Potassium (K+) and Nitrate (NO3-) are spectator ions that

do not participate in the reaction

2+ - + 2-3 3Ca (aq) + 2NO (aq) + 2K (aq) + CO (aq)

+ -3 3CaCO (s) + 2K (aq) + 2NO (aq)

2+ 2-3 3Ca (aq) + CO (aq) CaCO (s)

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Practice ProblemLimestone (Calcium Carbonate – CaCO3) is insoluble in water but dissolves when Hydrochloric Acid (HCl) is added. Write balanced total and net ionic equations

Ans:

Calcium Carbonate dissolves in HCl(aq) because the Carbonate ion, a base, reacts with the acid to form H2CO3 which decomposes into CO2(g) and H2O(l).

CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2CO3(aq)

Total ionic equation:

CaCO3(s) + 2H+(aq) + 2 Cl–(aq) → Ca2+(aq) + 2 Cl–

(aq)

+ H2O(l) + CO2(g)

Net ionic equation:

CaCO3(s) + 2H+(aq) → Ca2+(aq) + H2O(l) + CO2(g)

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Types of Chemical Reactions Most of the reactions we will study fall into

one of the following categories

Precipitation Reactions

Acid-Base Reactions

Oxidation-Reduction Reactions

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Types of Chemical Reactions Precipitation Reactions

A precipitation reaction occurs in aqueous solution because one product is insoluble

● A precipitate is an insoluble solid compound formed during a chemical reaction in solution

● For example, the reaction of Sodium Chloride with Silver Nitrate forms AgCl(s), an insoluble precipitate

3 3NaCl(aq) + AgNO (aq) AgCl(s) + NaNO (aq)

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Predicting Precipitation Reactions

To predict whether a precipitate will form, we need to look at potential insoluble products.

The following table lists eight solubility rules for ionic compounds. These rules apply to the most common ionic compounds

ClO4-

Solubility Rules for Ionic Compounds

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Solubility RulesNegative Ions

(Anions) + Positive Ions(Cations) =

Solubility of Compounds

In WaterExample

Any Anion + Alkali Ions(Li+, Na+, K+, Rb+, Cs+, Fr+) = Soluble NaF, KNO3

Any Anion + Hydrogen Ion[H+(aq) = Soluble HCl

Any Anion + Ammonium Ion(NH4

+) = Soluble NH4Cl

Nitrate (NO3) + Any Cation = Soluble Ca(NO3)2

Acetate (CH3COO-) + Any Cation = Soluble CH3COONa

Halides (Cl-, Br-, I-)+ Ag+, Pb2+, Hg2+, Cu+, Tl+ = Insoluble AgCl, PbBr2

+ Any Other Cation = Soluble KBr, CaI2

Sulfate (SO42-)

+ Ca2+, Sr2+, Ba2+, Ag+, Pb2+, Ra2+ = Insoluble BaSO4

+ Any Other Cation = Soluble CuSO4

Sulfide (S2-)+

Alkali Ions – Li+, Na+, K+, Rb+, Cs+, Fr+

Alkali Earth Metals – Be2+, Mg2+, Ca2+, Sr2+, Ba2+, Ra2+, H+(aq), NH4

+= Soluble H2S, MgS, (NH4)2S

+ Any Other Cation = Insoluble ZnS

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Solubility Rules (con’t)Negative

Ions(Anions)

+ Positive Ions(Cations) =

Solubility of Compounds

In WaterExample

Hydroxide (OH-)+ Alkali Ions - Li+, Na+, K+, Rb+, Cs+, Fr+,

H+(aq), NH4+, Sr2+, Ba2+, Ra2+, Tl+ = Soluble Sr(OH)2

+ Any Other Cation = Insoluble AgOH

Phosphate (PO43-)

Carbonate (CO32-)

+ Alkali Ions - Li+, Na+, K+, Rb+, Cs+, Fr+,H+(aq), NH4

+ = Soluble (NH4)3PO4

+ Any Other Cation = Insoluble MgCO3

General Solubility Ruleso All Compounds of the Ammonium Ion (NH4

+) and the

Alkali Metal (Group IA) cations are “Soluble”o All Nitrates and Acetates (Ethanoates) are “Soluble”o All Chlorides, Bromides and Iodides are “Soluble” except those of

Ag, Pb, Hgo All Sulfates are “Soluble” except those of Ag, Pb, Hg(I), Ba, Sr, Cao All Carbonates, Sulfites and Phosphates are “Insoluble” except

those ofAmmonium (NH4

+), and Alkali metal (Group IA) cationso All Hydroxides are “Insoluble” except those of NH4

+, Alkali Metal

(Group IA) cationso All Sulfides are “Insoluble” except those of NH4

+, Alkali Metal

(Group Ia) cations and Alkali Earth metal (Group II) cationso All Oxides are “Insoluble” except those of Calcium, Barium, and

Alkali Metal(group IA) cations, which actually react with water to form hydroxides

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Predicting Precipitation Reactions

● Suppose you mix together solutions of Nickel(II) Chloride, NiCl2, and Sodium Phosphate, Na3PO4

● How can you tell if a reaction will occur, and if it does, what products to expect?

2 3 4NiCl + Na PO ?

Con’t

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Types of Chemical Reactions Precipitation Reactions (Con’t)

Predicting Precipitation Reactions (Con’t)

● Precipitation reactions have the form of an “exchange reaction”

● An exchange (or metathesis) reaction is a reaction between compounds that, when written as a molecular equation, appears to involve an exchange of cations and anions

2 3 4 3 4 2NiCl + Na PO Ni (PO ) + NaCl

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Types of Chemical Reactions■ Precipitation Reactions (Con’t)

■Predicting Precipitation Reactions (Con’t)

● Now that we have predicted potential products, we must balance the equation

● We must verify that NiCl2 and Na3PO4 are soluble and then check the solubilities of the products

2 3 4 3 4 23NiCl + 2Na PO Ni (PO ) + 6NaCl

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● The Table of Solubility (slides 19-21) indicates that the reactants, Nickel(II) Chloride and Sodium Phosphate are both soluble

● Looking at the potential products we find that Nickel(II) Phosphate is not soluble

● Sodium Chloride is soluble

2 3 4 3 4 23 NiCl (aq) + 2 Na PO (aq) Ni (PO ) (s) + 6 NaCl(aq)

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● To see the reaction that occurs on the ionic level, we must rewrite the molecular equation as an ionic equation

● We predict that a reaction occurs because Nickel(II) Phosphate is insoluble and precipitates from the reaction mixture

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Predicting Precipitation Reactions (con’t)

● First, write strong electrolytes (the soluble ionic compounds) in the form of ions to obtain the complete ionic equation

2+ - + 3-43 Ni (aq) + 6 Cl (aq) + 6 Na (aq) + 2 PO (aq)

+ -3 4 2Ni (PO ) (s) + 6 Na (aq) + 6 Cl (aq)

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Predicting Precipitation Reactions (con’t)

● After canceling the spectator ions, you obtain the net ionic equation – The “Essential Equation”

2+ - + 3-43 Ni (aq) + 6 Cl (aq) + 6 Na (aq) + 2 PO (aq)

+ -3 4 2Ni (PO ) (s) + 6 Na (aq) + 6 Cl (aq)

2+ 3-4 3 4 23 Ni (aq) + 2 PO (aq) Ni (PO ) (s)

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Practice ProblemWhen solutions of Barium Chloride and Sodium Sulfate are mixed, the spectator ions in the resulting reaction are

a. only Ba2+

b. only SO42–

c. only Na+

d. only Cl–

e. both Na+ and Cl–

Ans: e

BaCl2 + Na2SO4 BaSO4 (s) + 2Na+ + 2Cl-

BaSO4 (s) Precipitate does participate in reaction

Na+ & Cl- Spectator ions do not participate in reaction

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Practice ProblemPredict whether a chemical reaction occurs when Potassium Fluoride (KF) is mixed with Strontium Nitrate (Sr(NO3)2 )

Molecular Equation:

2KF(aq) + Sr(NO3)2 SrF2(s) + 2KNO3(aq)

Total Ionic Equation:

2K+(aq) + 2F- (aq) + Sr2+(aq) +2NO3-(aq) SrF2(s) + 2K+(aq) + 2NO3

-

(aq)

Net Ionic Equation:

Sr2+(aq) + 2F-(aq) SrF2(s)

Strontium Fluoride is a solid, therefore a reaction has occurred

K+ & NO3- are spectator ions and do not participate in

the reaction

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Types of Chemical Reactions Acid-Base Reactions

The Arrhenius Concept

● The Arrhenius concept defines acids as: substances that produce Hydrogen ions, H+, when dissolved in water

● An example is Nitric Acid, HNO3, a molecular substance that dissolves in water to give H+ and NO3

-

2 H O + -

3 3HNO (aq) H (aq) + NO (aq)

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The Arrhenius Concept

● The Arrhenius concept defines bases as substances that produce Hydroxide ions, OH-, when dissolved in water

● An example is Sodium Hydroxide, NaOH, an ionic substance that dissolves in water to give sodium ions and hydroxide ions

2 H O + -NaOH(s) Na (aq) + OH (aq)

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The Arrhenius Concept● The molecular substance Ammonia, NH3, is a

base in the Arrhenius view

● It yields Hydroxide ions when it reacts with water

+ -3 2 4NH (aq) + H O(l) NH (aq) + OH (aq)

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The Brønsted-Lowry Concept

● The Brønsted-Lowry concept of acids and bases involves the transfer of a proton (H+) from the acid to the base

● In this view, acid-base reactions are:

proton-transfer reactions

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● In the reaction of Ammonia with Water

● The H2O molecule is the acid because it donates a proton

● The NH3 molecule is a base, because it accepts a proton


H+

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● The H+(aq) ion associates itself with Water to form H3O+(aq)

● this “mode of transportation” for the H+ ion is called the “Hydronium” ion

+ +2 3H (aq) + H O(l) H O (aq)

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H+



● The dissolution of Nitric Acid, HNO3, in water is therefore a proton-transfer reaction

● Where HNO3 is an acid (proton donor) and H2O is a base (proton acceptor)

- +3 2 3 3HNO (aq) + H O(l) NO (aq) + H O (aq)

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Strong and Weak Acids and Bases

● A strong acid is an acid that ionizes completely in water; it is a strong electrolyte

- +3 2 3 3HNO (aq) + H O(l) NO (aq) + H O (aq)

- +2 3HCl(aq) + H O(l) Cl (aq) + H O (aq)

Common Strong Acids:

HCl, HBr, HI, HNO3, H2SO4, HClO4

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● A weak acid is an acid that only partially ionizes in water; it is a weak electrolyte

● The Hydrogen Cyanide molecule, HCN, reacts with water to produce a small percentage of ions in solution

- +2 3HCN(aq) + H O(l) CN (aq) + H O (aq)

H+

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● A strong base is a base that is present entirely as ions, one of which is OH-; it is a strong electrolyte

● The hydroxides of Group IA and IIA elements, except for Beryllium Hydroxide, are strong bases

2H O + -NaOH(s) Na (aq) + OH (aq)

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● A weak base is a base that is only partially ionized in water; it is a weak electrolyte

● Ammonia, NH3, is an example


H+

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● It is important to be able to identify an acid or base as strong or weak

● In an ionic equation, strong acids and bases are represented as

Separate Ions

● Weak acids and bases are represented as:

Undissociated “molecules” in

ionic equations

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Reactions of Weak Acids● Acetic Acid is a weak acid (dissociates very

little) Molecular:

Total Ionic:

Net Ionic:

3 3 2CH COOH(aq) + NaOH(aq) CH COONa + H O(l)

+ -3

- +3 2

CH COOH(aq) + Na (aq) + OH (aq)

CH COO (aq) + Na (aq) + H O(l)

- -3 3 2CH COOH(aq) + OH (aq) CH COO (aq) + H O(l)

Note: The Na+ ions are called “Spectator ions” and cancel out

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Neutralization (Acid-Base) Reactions

● One of the chemical properties of acids and bases is that they neutralize one another

● A neutralization reaction is a reaction of an acid and a base that results in an ionic compound and water

● The ionic compound that is the product of a neutralization reaction is called a salt 2HCN(aq) + KOH(aq) KCN(aq) + H O(l)

acid base salt

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Neutralization (Acid-Base) Reactions

● The net ionic equation for each acid-base neutralization reaction involves a transfer of a proton

● Consider the reaction of the strong acid, HCl(aq) and a strong base, LiOH(aq)

2HCl(aq) + KOH(aq) KCl(aq) + H O(l)

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Neutralization Reactions

● Writing the strong electrolytes in the form of ions gives the complete ionic equation

+ - + -

+ -2

H (aq) + Cl (aq) + K (aq) + OH (aq)

K (aq) + Cl (aq) + H O(l)

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H+



● Canceling the spectator ions results in the net ionic equation. Note the proton transfer

+ - + -H (aq) + Cl (aq) + K (aq) + OH (aq) + -

2K (aq) + Cl (aq) + H O(l)

+ -2H (aq) + OH (aq) H O(l)

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H+



● In a reaction involving HCN(aq), a weak acid, and KOH(aq), a strong base, the product is KCN, a strong electrolyte

● The net ionic equation for this reaction is:

- -2HCN(aq) + OH (aq) CN (aq) + H O(l)

Note the proton transfer

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Practice ProblemComplete and balance the following neutralization equation (in aqueous solution); include phase labels

Then write the net ionic equation

Ba(OH)2 + HC2H3O2

Strong Base Weak Acid

Ans:

Molecular: Ba(OH)2 + 2HC2H3O2 Ba(C2H3O2)2 (aq) + 2H2O(l)

Total Ionic: Ba+2(aq) + 2OH-(aq) + 2C2H3OOH(aq) Ba+2 (aq) + 2C2H3OO- (aq) + 2H2O(l)

Net Ionic: 2OH-(aq) + 2C2H3OOH(aq) 2C2H3OO- (aq) + H2O(l)

(Ba+2 is the spectator ion)

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Acid-Base Reactions with Gas Formation(Displacement Reactions)

● Carbonates react with acids to form CO2, Carbon Dioxide gas and water

● Sulfites react with acids to form SO2, Sulfur Dioxide gas and water

2 3 2Na CO + 2HCl 2NaCl + H O + CO(g)

2 3 2 2Na SO + 2HCl 2NaCl + H O + SO (g)

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Acid-Base Reactions with Gas Formation

● Sulfides react with acids to form H2S, Hydrogen Sulfide gas

2 2Na S + 2 HCl 2 NaCl + H S(g)

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Acid-base reactions with gas formation sometimes involve unstable chemical species such as H2CO3 and H2SO3

Note the unstable spiecies are enclosed in parentheses

2 3 2 2H CO (aq) H O(l) + CO (g)

2 3 2 2H SO (aq) H O(l) + SO (g)

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Types of Chemical Reactions Oxidation-Reduction Reactions

Oxidation-Reduction reactions involve the transfer of electrons from one species to another

Oxidation is defined as the “loss” of electrons

Reduction is defined as the “gain” of electrons

Oxidation and reduction always occur simultaneously

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The reaction of an iron nail with a solution of Copper(II) Sulfate, CuSO4, is an

oxidation - reduction reaction

The molecular equation for this reaction is

4 4Fe(s) + CuSO (aq) FeSO (aq) + Cu(s)

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The net ionic equation shows the reaction of iron metal with Cu2+

(aq) to produce Iron(II) ion and Copper metal

2+ 2+Fe(s) + Cu (aq) Fe (aq) + Cu(s)

Loss of 2 e-1 (oxidation)

Gain of 2 e-1 (reduction)

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Oxidation Numbers● A book keeping system to monitor

which atom loses electron charge and which atom gains electron charge

● The oxidation number (or oxidation state) of an atom in a molecule (or formula unit) is the actual charge the atom would have if all the electrons it was sharing were transferred completely, not shared

● Alternatively, the charge on the central atom of a covalent compound if all ligands attached to the central atom along with the shared electron pairs were removed

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Oxidation Number Rules

Rule Applies to

Statement

1 ElementsThe oxidation number of an atom in an element is zero.

2 Monatomicions

The oxidation number of an atom in a monatomic ion equals the charge of the ion.

3 Oxygen

–2 in oxides, e.g. ZnO, CO2, H2O–1 in peroxides, e.g. H2O2

–(1/2) in superoxides, e.g. KO2

–(1/3) in inorganic ozonides, e.g. RbO3

0 in O2

+(1/2) in dioxygenyl, e.g. dioxygenyl hexafluoroarsenate O2

+ [AsF6]−

+1 in O2F2

+2 in OF2

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Oxidation Number RulesRule Applies to Statement

4 Hydrogen The oxidation number of hydrogen is+1 in most of its compounds

5 HalogensFluorine is –1 in all its compounds. The other

halogens are –1 unless the other element is another halogen or oxygen

6 Compounds and ions

The sum of the oxidation numbers of the atoms in a neutral compound is zero. The sum in a polyatomic

ion equals the charge on the ion

7Carbon inOrganic

Compounds

The oxidation number of Carbon in Organic Compounds has an average value of 0

The O.N. of other elements in carbon compounds (H+1, O-2, N+3, Cl-1, etc.) still apply

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Oxidation Number ExamplesZinc Chloride (ZnCl2)

O.N. of Zn+2 ion = +2

O.N. of Cl- ion = -1 x 2 = -2 (total for 2 Cl)

Neutral molecule – sum of O.N.s must = 0

+2 + (-2) = 0Sulfur Trioxide (SO3)

O.N. of Oxygen = -2 x 3 = -6 (total for 3 Oxygen)

Neutral molecule – sum of O.N.s must = 0

O.N. of Sulfur = -6 + S = 0

S = +6

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Oxidation Number ExamplesNitric Acid (HNO3)

O.N. Hydrogen (H) = +1O.N. atoms in NO3 = -1 (NO3

- ion charge)

O.N. Oxygen = - 2 x 3 = -6 (total)

O.N. Nitrogen = -6 + N = -1

N = +5Dichromate Ion - Cr2O7

2-

O.N. each Oxygen = -2 x 7 = -14 (total)

O.N. Cr (total) = -14 + Cr(2) = -2 (ion charge)

Cr(2) = +12 (total for 2 Cr)

Cr = +12 / 2 = +6

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Oxidation Number ExamplesOxalate Ion - C2O4

2-

O.N. each Oxygen = -2 x 4 = -8 (total)

O.N. C (total) = -8 + C(2) = -2 (ion charge)

C(2) = +6 (total for 2 Carbons)

C = +6 / 2 = +3

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Describing Oxidation-Reduction Reactions

Look again at the reaction of

Iron with Copper(II) Sulfate

2+ 2+Fe(s) + Cu (aq) Fe (aq) + Cu(s)

This reaction can be written in terms of two

half-reactions

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Describing Oxidation-Reduction Reactions● A half-reaction is one of the two parts

of an oxidation-reduction reaction● One involves the loss of electrons

(oxidation)● The other involves the gain of electrons

(reduction)2+ -Fe(s) Fe (aq) + 2e

2+ -Cu (aq) + 2e Cu(s)

oxidation half-reaction

(Loss of 2 e-)

reduction half-reaction

(Gain of 2 e-)

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Describing Oxidation-Reduction Reactions● An oxidizing agent (oxidant) is a

species that oxidizes another species; it is an electron acceptor (gains e-) and is thus reduced

● A reducing agent (reductant) is a species that reduces another species; it is an electron donor (loses electrons) and is thus oxidized

2+ 2+Fe(s) + Cu (aq) Fe (aq) + Cu(s)

oxidizing agent

reducing agentLoss of 2 e- oxidation

Gain of 2 e- reduction22

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Some Common Oxidation-Reduction Reactions● Most of the oxidation-reduction

reactions fall into one of the following simple categories

Combination Reaction

Decomposition Reactions

Displacement Reactions

Combustion Reactions

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Combination Reactions

● A combination reaction is a reaction in which two substances combine to form a third substance

● Examples

Metal and nonmetal form binary ionic or covalent compounds

0 0 +1 -122 Na (s) + Cl (g) 2 Na Cl (s)

Metal nonmetal Binary ionic

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● Two nonmetals form a covalent compound

0 0 -3 +12 2 3N (g) + 3H (g) 2N H (g)

0 0 +3 -14 2 3P (s) + 6Cl (g) 4P Cl (l)

0 0 2 -22 2N O(g) O (g) 2N O (g)

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● Combining a compound and an element

+2 -2 0 +4 -22 22N O (g) + O (g) 2N O (g)

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Decomposition Reactions

● A decomposition reaction is a reaction in which a single compound reacts to give two or more substances

● Ex. Decomposition reaction of Mercury(II) oxide

+2 -2 o o22 Hg O (s) 2 Hg (l) + O (g)

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Displacement Reactions● A displacement reaction (also called

a single- replacement reaction) is a reaction in which an element reacts with a compound, displacing an element from it

The results of reactions between a metal and water, acids, metal-ion solutions form the basis of the

Activity Series of the Metals● Elements higher on the list are stronger

reducing agents; thus they are oxidized (lose electrons) causing the ions of other elements lower on the list to be reduced as they gain these electrons

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Displacement Reactions

● Activity Series of the Metals

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Displacement Reactions (Con’t)● The Activity Series (Con’t)

Note: The element higher on the list usually exists as the native metal, which when oxidized as the reducing agent, loses electrons to form a cation (positively charge ion)

The element lower on the list, relative to the metal higher on the list, usually exists in solution as a cation, thus suitable for gaining the electrons (reduced) lost by the element higher in the list.

The reduced metal is usually displaced as the native metal or in the case of hydrogen ions, as hydrogen gas (H2)gas

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Types of Chemical Reactions Oxidation-Reduction Reactions (Con’t)

Displacement Reactions (Con’t)

● The Activity Series (Con’t)

Ex.

Zn is higher on the activity list than Hydrogen

The Zinc metal oxidizes (loses electrons) that are gained by the Hydrogen Ion (H+) forming H2 which is displaced as Hydrogen gas from the solution

0 +1 -1 +2 1 02 2Zn (s) + 2 H Cl (aq) Zn Cl (aq) + H (g)

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Displacement Reactions (Con’t)● The Activity Series (Con’t)

Ex.

Aluminum is higher on the list than Hydrogen (in water as steam)

Alo is oxidized to Al+3; Hydrogen Ion (H+1) is reduced and displaced as Hydrogen (H2)

Ex.

Zinc is higher in the activity list than CopperThus, Zinc acts as a reducing agent causing the Cu+2 ion to gain electrons (it is reduced) displacing it from solution as Copper metal

0 1 -2 +3 -2 +1 1 02 3 22Al (s) + 6H O (g) 2Al (O H ) (s) + 3H (g)

-2 -2+2 +6 2 0 0 +2 +6 24 4Cu (aq) + S O + Zn (s) Cu (s) + Zn (aq) + S O (aq)

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Combustion ReactionsA combustion reaction is a reaction in

which a substance reacts with Oxygen, usually with the rapid release of heat to produce a flame, i.e., it burns

Ex. Combustion reaction of Iron

0 0 3 22 2 34 Fe (s) + 3 O (g) 2 Fe O (s)

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Practice ProblemIdentify the Oxidizing agent and Reducing agent in the following reaction:

2AL(s) + 3H2SO4(aq) Al2SO4)3(aq) + 3H2(g)

Assign Oxidation Numbers

2AL(s) + 3H2SO4(aq) Al2(SO4)3(aq) + 3H2(g)

The O.N. of Aluminum (Al) increased from 0 to +3

Aluminum lost electrons, it was oxidized, it is a reducing agent

The O.N. of Hydrogen (H) decreased from +1 to 0Hydrogen gained electrons, it was reduced, H2SO4 is an oxidizing agent

O +1+6

-2 +3+6

-2 O

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Balancing Simple Oxidation-Reduction Reactions

At first glance, the equation representing the reaction of Zinc (Zno) metal with Silver(Ag+1) ions might appear to be balanced

However, a balanced equation must have a charge balance as well as a mass balance

The previous reaction can be broken into two reactions, i.e., half-reactions, one representing the oxidation process (losing e-) and one representing the reduction process (gaining e-)

0 +1 +2 0Zn (s) + Ag (aq) Zn (aq) + Ag (s)

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Types of Chemical Reactions Half-Reactions

The number of electrons lost in the oxidation half-reaction must equal the number of electrons gained in the reduction half-reaction

The reaction involving the reduction of the Silver must be doubled0 +2 -Zn (s) Zn (aq) + 2e

+1 - 02Ag aq + 2 e 2Ag s

oxidation half-reactionZinc loses 2 e- (oxidized)

reduction half-reactionSilver gains 2 e- (reduced)

0 +1 +2 0Zn (s) + 2Ag (aq) Zn (aq) + 2Ag (s)

0 +2 -Zn (s) Zn (aq) Oxidation (lose e )

+1 0 -2Ag aq 2Ag s Reduction (gain e )

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The balancing process for incomplete equations may involve the addition of components other than electrons

Consider the unbalanced reaction:__ (Cr2

6+O7)2- + __ Cuo __Cr3+ + __ Cu2+

Balance charges for Cr2O72- 2Cr3+ Half Reaction

Cr2O72- 2Cr3+ + 7H2O Add H20 to balance Oxygen

14 H+ + Cr2O72- 2Cr3+ + 7H2O Add H+ ions on left to

balance 14 H on right6 e- + 14 H+ + Cr2O7

2- 2Cr3+ + 7H2O Add 6 e- on left to balance

-6 +14 (+12 -14) (-2) = +6 reaction charges

6 electrons gained – Reduction reaction

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Types of Chemical ReactionsBalance charges for Cuo Cu2+ No need to added H2O or H+

Cuo Cu2+ + 2e- Add 2 e- on right to balancereaction charges

Multiply each half-reaction by appropriate integer to balance e-

6 e- + 14 H+ + Cr2O72- 2Cr3+ + 7H2O

3Cuo 3Cu2+ + 6e- Add 2 reactions

6 e- + 14 H+ + Cr2O72- + 3Cuo 2 Cr3+ + 7H2O + 3Cu2+ +

6e-

14 H+ + Cr2O72- + 3Cuo 2 Cr3+ + 3Cu2+ + 7H2O

14 + 1 + 3 + 2 + 3 + 7 = 30 Total Coefficients

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Practice ProblemObtain the Oxidation Number for the element noted in each of the following:

(a). Nb in NbO2 (b). Mn in KMnO4

(c). P in Na3PO4 (d). I in IO3-

(e). S in Na2S4O6

Ans:

a. O 2(-2) = - 4 -4 = -4 = Nb+4

b. K 1(+1) = +1 O 4(-2) = -8 +1 -8 = -7 = Mn+7

c. Na 3(+1) = +3 O 4(-2) = -8 +3 -8 = -5 = P+5

d. O 3(-2) = -6 IO3-1 =-1 -6 -1 = -7 = I+7

e. Na2(+1) = +2 O 6(-2) = -12 (+2 -12)/4 = -2.5 =

S+2.5

Charge Deficit

Charge Required

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Practice ProblemIn each of the following reactions, label the oxidizing and reducing agent.

(a). ZnO(s) + C(s) Zn(g) + CO(g)(b). 8 Fe(s) + S8(s) 8 FeS(s)

Ans: (a) Zn+2O-2

(s) + C0 Zn0(g) + C+2O(g)

Zn gains 2 e- (it is reduced; it is the oxidizing agent)

C loses 2 e- (it is oxidized; it is the reducing agent)

(b) 8 Fe0(s) + S8

0 8 Fe+2S-2(s)

Fe loses 2 e- (it is oxidized; it is the reducing agent)

S gains 2 e- (it is reduced; it is the oxidizing agent)

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Practice ProblemBalance the following redox reaction by the half-reaction method.

FeI3(aq) + Mg(s) Fe(s) + MgI2(aq)

Ans:

Reduction half-cell reaction

Fe+3 + 3 e- Fe0 Fe gains 3 e- (reduced)

Oxidation half-cell reaction

Mg0 Mg+2 + 2 e- Mg loses 2 e- (oxidized)

Balance half-cells (total electrons gained & lost) and combine:

2 Fe+3 I-13 + 6 e- 2 Fe0

3 Mg0 3Mg+2I (-1)2 + 6 e-

2 Fe+3 I-13 (aq) + 3 Mg (s) 2 Fe(s) + 3Mg +2 I-12(aq)

2FeI3(aq) + 3 Mg(s) 2 Fe(s) + 3 MgI2 (molecular formula)

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Oxidation Nos. -Practice ProblemWhat is the total number of electrons transferred when the equation below is balanced? I2 + ClO3

- + H2O IO3- + Cl- + H+ (in acid)

Multiply by factors to make e- lost equal to e- gained and use the factors as coefficients

3I2 + 5 (ClO3)- + H2O 6 (IO3)- + 5 Cl- + H+

Complete balancing by inspection: Need total of 18 Oxygen on each side; Add 3 Oxygen to left side using 3 moles H2

Add 6 Hydrogen on right to balance 6 water Hydrogen 3I2 + 5 (ClO3)- + 3 H2O 6 (IO3)- + 5 Cl- + 6 H+

Total electrons transferred (gained = lost) = 30

Each Iodine loses 5 e- x 2 atoms Iodine = 10 e- lost for I2

Each Chlorine gains 6 e-

Need a total of 30 e- lost (6 x 5 = 30)

SetupOxidation Nos

Adjusting for 2 atoms of Iodine - Total of 30 e- gained (2 x 5 x 3 = 30)

-1 -10 +5 -2 +5 -2 -1 +1

2 3 2 3I + Cl O + H O I O + Cl + H

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Practice ProblemWhich of the following chemical reactions is an oxidation–reduction reaction?

a. Zn(s) + S(s) ZnS(s)

b. H2SO4 + 2NaOH Na2SO4 + 2H2O

c. CO2+ H2O H2CO3

d. AgNO3 + NaCl AgCl + NaNO3

e. NaOH + HCl NaCl + H2O

Ans: a

Zinc loses 2 e- it is oxidized

Sulfur gains 2 e- it is reduced

None of the other reactionsinvolves a change of “oxidation number”!!

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Working with Solutions Molar Concentration

When we dissolve a substance in a liquid, we call the substance the solute and the liquid the solvent

The general term concentration refers to the quantity of solute in a standard quantity of solution

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Molar concentration, or molarity (M), is defined as the moles of solute dissolved in one liter (cubic decimeter) of solution

moles of solute molMolarity (M) = =

liter of solution L

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Example.

● A sample of 0.0341 mol Iron(III) Chloride, FeCl3, was dissolved in Water to give 25.0 mL of solution. What is the Molarity (M) of the solution?

Since:

Then:

3moles of FeClMolarity (M) =

liter of solution

33

0.0341 mole of FeClM = = 1.36 M FeCl

0.0250 liter of solution

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The molarity of a solution and its volume are inversely proportional

Therefore, adding water (dilution) makes the solution less concentrated

● This inverse relationship takes the form of:

● So, as water is added, the final volume, Vf, increases and the final molarity, Mf, decreases

● This is a mass balance equation for solute:

moli = molf

i i f fM × V = M × V

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Practice ProblemHow many moles of Sulfate ions are there in a 0.20-L solution of 0.030 molar Al2(SO4)3?

0.0030 b. 0.0060 c. 0.012 d. 0.018 e. 0.024

Ans: d

Note: Each mole of Al2(SO4)3 contains 3 moles of SO42-

ions

2 4 3 44

2 4 3

0.030 mol Al (SO ) 3 mol ions SO 0.20 L = 0.018 mole ions SO

L mol Al (SO )

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Quantitative Analysis Analytical chemistry deals with the determination

of composition of materials – that is, the analysis of materials

Quantitative analysis involves the determination of the amount of a substance or species present in a material

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Practice ProblemIf 53.63 mL of 0.2970 M AgNO3 is required to titrate 25.00 mL of Seawater for Chloride (Cl-) ion, what is the mass percent (%) of Cl- in the seawater?

The density of seawater is 1.024 g/mL

Ans:

- -- Mass Cl 0.56465 g Cl

Mass % Cl = x 100% = x 100% = 2.20566 = 2.206 % ClMass Sample 25.60 g Sample

- --3

-3

0.2970 M AgNO1 L 1 mol Cl 35.45 g Cl 53.63 mL = 0.56465 g Cl

1000 mL L 1 mol Ag NO 1 mol Cl

1.024 gMass of Sample = 25.00 mL = 25.60 g sample

mL

- -3 3AgNO (aq) + Cl (aq) AgCl(s) + NO (aq)

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Quantitative Analysis Gravimetric Analysis

Gravimetric analysis is a type of quantitative analysis in which the amount of a species in a material is determined by converting the species into a product that can be isolated and weighed

Precipitation reactions are often used in gravimetric analysis

The precipitate from these reactions is then filtered, dried, and weighed.

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Quantitative Analysis Gravimetric Analysis

Consider the problem of determining the amount of lead in a sample of drinking water● Adding Sodium Sulfate (Na2SO4) to the sample

will precipitate Lead(II) Sulfate

● The PbSO4 can then be filtered, dried, and weighed

2+ +2 4 4Na SO (aq) + Pb (aq) 2 Na (aq) + PbSO (s)

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Practice ProblemSuppose a 1.00 L sample of polluted water was analyzed for Lead(II) ion, Pb2+, by adding an excess of Sodium Sulfate

The mass of Lead(II) Sulfate that precipitated was 229.8 mg.

What is the mass of Lead in a liter of the water?

Express the answer as mg of Lead per liter of solution

Ans:

Thus, [Pb2+] = 157.0 mg/1.00 L = 157 mg/L

2+ +2 4 4Na SO (aq) + Pb (aq) 2 Na (aq) + PbSO (s)

2+ 2+4

4 2+4 4

1 mmol PbSO 1 mmol Pb 207.2 mg Pb229.8 mg PbSO × × ×

303.3 mg PbSO 1 mmol PbSO 1 mmol Pb

2+= 157.0 mg Pb (Mass of Lead in the 1.0 L sample)

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Practice ProblemA soluble Silver compound was analyzed for the percentage of Silver by adding Sodium Chloride solution to precipitate the Silver ion as Silver Chloride

If 1.583 g of Silver compound gave 1.788 g of Silver Chloride (FW = 143.321), what is the mass percentage of Silver (AW = 107.868 g/mol) in the compound?

Ans:

107.868 g / mol% Silver = x 100 = 85.07 % Ag

126.8 g / mol

+ - + -(aq) (aq) (aq) (s) (aq)AgX + Na + Cl = AgCl + Na + X ( )aq

1 mol Ag Cl1.788 g AgCl = 0.01248 mol AgCL = 0.01248 mol AgX

143.321 g AgCl

11.583 g AgX = 126.8 g AgX / 1 mol AgX = Mol Wgt AgX

0.01248 mol AgX

Mol Wgt Ag = 107.868 g / mol Mol Wgt AgX = 126.8 g / mol

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Quantitative Analysis Volumetric Analysis

An important method for determining the amount of a particular substance is based on measuring the volume of the reactant solution

● Titration is a procedure for determining the amount of substance A by adding a carefully measured volume of a solution with known concentration of B until the reaction of A and B is just complete

● Volumetric analysis is a method of analysis based on titration

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Practice ProblemHow many mL of 0.250 M KMnO4 are needed to react with 3.36 g of Iron(II) Sulfate (FW = 151.92)?

The reaction is as follows:

10 FeSO4(aq) + 2 KMnO4(aq) + 8 H2SO4(aq)

5 Fe2(SO4)3(aq) + 2 MnSO4(aq) + K2SO4(sq) + 8 H2O(l)

Ans:

4 44

4 4 4

1 mol FeSO 2 mol KMnO 1 L 1000 mL3.36 g FeSO x = 17.7 mL

151.92 g FeSO 10 mol FeSO 0.250 Mol KMnO 1 L

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Practice ProblemA sample of Calcium Oxalate is titrated with a 2.05 mL of a 4.88 x 10-4 M solution of Potassium Permanganate (an Oxidizing Agent)

Calculate the amount of Calcium (Ca2+) (moles) in the sample

2KMnO4(aq) + 5CaC2O4(s) + 8H2SO4(aq)

2MnSO4(aq) + K2SO4(aq) + 5CaSO4(s) + 10CO2(g) + 8H2O(l)-4

44

-64

4.88 x 10 mol KMnO1LMoles of KMnO = 2.05 mL soln ×

1000 mL 1 L soln

= 1.00 x 10 mol KMnO

Convert moles of KMnO4 to moles of CaC2O4 titrated

-6 2 2 42 4 4

4

-62 2 4

5 mol Ca C OMoles of CaC O = 1.00×10 mol KMnO ×

2 mol KMnO

= 2.5 10 mol Ca C O

Calculate amount of Ca2+ present2+

2+ -6 -6 2+2 4

2 4

1 mol CaMoles of Ca = 2.5 x 10 mol CaC O = 2.5 x 10 mol Ca

1 mol CaC O

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Practice ProblemFind the concentrations of two monoprotic acids (HA & HB):

Flask A – 43.5 mL HA

Flask B – Total Volume – 50 ml (37.2 mL HA + 12.8 mL HB)

Titrate Flask A with 87.3 mL of 0.0906 M NaOH

Titrate Flask B with 96.4 mL of 0.0906 M NaOH

The balanced chemical equations for HA or HB with sodium hydroxide are the same. For HA it is:

HA(aq) + NaOH(aq) → NaA(aq) + H2O(l)

The concentration of the first HA solution is:

Volume of NaOH reacting with HA in flask B has same ratio as in A:

-3

(HA) -3

10 L 0.0906 mol NAOH 1mol HA 1 1 mLM = (87.3mL)

1 mL L 1 mol NaOH 43.5 mL 10 L

(HA)M = 0.182 mol / L HA

x = (87.3 mL NaOH) (37.2 mL HA / 43.5 mL HA) = 74.7 mL NaOH Con’t

43.5 mL 37.2(A) = B

87.3 mL x

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Practice Problem (con’t)Volume of NaOH reacting with HB in Flask B

Molarity of HB

96.4 mL - 74.7 mL = 21.74 mL NaOH

-3

-3

10 L 0.0906 mol NaOH 1 mol HB 1 1 mLMolarity HB = (21.74 mL)

1 mL L 1 mol NaOH 12.8 mL 10 L

Molarity HB = 0.154 mol HB / L

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Equation Summary

moles of solute molMolarity (M) = =

liter of solution L

i i f fM × V = M × V

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