Chem chapter 04 LEC

8/13/2019 Chem chapter 04 LEC

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Chapter 4

Chemical

Quantities andAqueous

Reactions

2008, Prentice Hall

Chemistry: A Molecular Approach, 1stEd.

Nivaldo Tro

Roy Kennedy

Massachusetts Bay Community College

Wellesley Hills, MA


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Tro, Chemistry: A Molecular Approach 2

Reaction Stoichiometry

the numerical relationships between chemical amountsin a reaction is called stoichiometry the coefficients in a balanced chemical equation

specify the relative amounts in moles of each of the

substances involved in the reaction2 C8H18(l) + 25 O2(g) 16 CO2(g) + 18 H2O(g)

2 molecules of C8H18react with 25 molecules of O2

to form 16 molecules of CO2

and 18 molecules of H2

O

2 moles of C8H18react with 25 moles of O2to form 16 moles of CO2and 18 moles of H2O

2 mol C8H18: 25 mol O2: 16 mol CO2: 18 mol H2O


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Predicting Amounts from Stoichiometry

the amounts of any other substance in a chemicalreaction can be determined from the amount ofjust one substance

How much CO2 can be made from 22.0 moles ofC8H18in the combustion of C8H18?

2 C8H18(l) + 25 O2(g) 16 CO2(g) + 18 H2O(g)

2 moles C8H18: 16 moles CO2

2188

2188 COmoles176

HCmol2

COmol16HCmoles22.0


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ExampleEstimate the mass of CO2produced in

2004 by the combustion of 3.4 x 1015g gasoline

assuming that gasoline is octane, C8H18, the equationfor the reaction is:

2 C8H18(l) + 25 O2(g) 16 CO2(g) + 18 H2O(g)

the equation for the reaction gives the mole relationshipbetween amount of C8H18and CO2, but we need to

know the mass relationship, so the Concept Plan will

be:

g C8H18 mol CO2 g CO2mol C8H18


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ExampleEstimate the mass of CO2produced in

2004 by the combustion of 3.4 x 1015g gasoline

since 8x moles of CO2as C8H18, but the molar mass of C8H18is

3x CO2, the number makes sense

1 mol C8H18= 114.22g, 1 mol CO2= 44.01g, 2 mol C8H18= 16 mol CO2

3.4 x 10

15

g C8H18g CO2

Check:

Solution:

Concept Plan:

Relationships:

Given:Find:

g114.22

mol1

216

2

2

188

2

188

188188

15

COg101.0

COmol1

COg44.01

HCmol2

COmol16

HCg114.22

HCmol1HCg10.43

188

2

HCmol2

COmol61

g C8H18 mol CO2 g CO2mol C8H18

mol1

g44.01


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Practice

According to the following equation, howmany milliliters of water are made in thecombustion of 9.0 g of glucose?

C6H

12O

6(s) + 6 O

2(g) 6 CO

2(g) + 6 H

2O(l)

1. convert 9.0 g of glucose into moles (MM 180)

2. convert moles of glucose into moles of water

3. convert moles of water into grams (MM 18.02)

4. convert grams of water into mLa) How? what is the relationship between mass and

volume?

density of water = 1.00 g/mL


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Practice

OHmL5.4

OHg1.00

OHmL1x

OHmole1

OHg18.0x

OHCmole1

OHmole6x

g10x80.1

OHCmole1xOHCg0.9

2

2

2

2

2

6126

2

2

61266126

According to the following equation, how manymilliliters of water are made in the combustion of

9.0 g of glucose?

C6H12O6(s)+ 6 O2(g)6 CO2(g)+ 6 H2O(l)


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Limiting Reactant for reactions with multiple reactants, it is likely that

one of the reactants will be completely used before theothers

when this reactant is used up, the reaction stops and nomore product is made

the reactant that limits the amount of product is calledthe limiting reactant sometimes called the limiting reagent

the limiting reactant gets completely consumed

reactants not completely consumed are called excessreactants the amount of product that can be made from the

limiting reactant is called the theoretical yield


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Things Dont Always Go as Planned!

many things can happen during the course of anexperiment that cause the loss of product

the amount of product that is made in a reactionis called the actual yield

generally less than the theoretical yield, never more!

the efficiency of product recovery is generally

given as the percent yield

%100yieldltheoretica

yieldactualYieldPercent


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Limiting and Excess Reactants in the

Combustion of MethaneCH4(g) + 2O2(g) CO2(g) + 2H2O(g)

Our balanced equation for the combustion of methane

implies that every 1 molecule of CH4reacts with 2molecules of O2

H

H

C

H

H +

O

O

C +

OO

OO

+

OH H

OH H

+


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11


Combustion of Methane

If we have 5 molecules of CH4and 8 molecules

of O2, which is the limiting reactant?

H

HC

H

H

+

OO

OO

OO

OO

OO

OO

OO

OO

?H

HC

H

H

H

HC

H

H

H

HC

H

H H

HC

H

H

CH4(g) + 2O2(g) CO2(g) + 2H2O(g)


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12


Combustion of Methane

H

HC

H

H

H

HC

H

H

+

OO

OO

OO

OO

OO

OO

OO

OO

H

HC

H

H

H

HCH

H

H

HC

H

H

24

24 COmolecules16

CHmolecules1

COmolecules2CHmolecules8

CH4(g) + 2O2(g) CO2(g) + 2H2O(g)

2

2

22 COmolecules10

Omolecules2

COmolecules2Omolecules10

since less CO2

can be made

from the O2thanthe CH4,the O2

is the limiting

reactant


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Example 4.4

Finding Limiting Reactant,

Theoretical Yield, andPercent Yield


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Example:

When 28.6 kg of C are allowed to react with 88.2 kg ofTiO2in the reaction below, 42.8 kg of Ti are obtained.Find the Limiting Reactant, Theoretical Yield, and

Percent Yield.

(g)(s)(s)(s) CO2TiC2TiO 2


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Example:

When 28.6 kg of C reacts with 88.2

kg of TiO2, 42.8 kg of Ti are

obtained. Find the Limiting

Reactant, Theoretical Yield, and

Percent Yield.

TiO2(s) + 2 C(s) Ti(s) + 2 CO(g)

Write down the given quantity and its units.Given: 28.6 kg C

88.2 kg TiO2

42.8 kg Ti produced


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Write down the quantity to find and/or its units.Find: limiting reactant

theoretical yield

percent yield

Example:

Find the Limiting

Reactant, Theoretical

Yield, and Percent Yield.

TiO2(s) + 2 C(s)

Ti(s) + 2 CO(g)

Information

Given: 28.6 kg C, 88.2 kg TiO2, 42.8 kg Ti


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17

Write a Concept Plan:

Information


Find: Lim. Rct., Theor. Yld., % Yld.

Example:

Find the Limiting



TiO2(s) + 2 C(s)

Ti(s) + 2 CO(g)

kgTiO2

kg

C

2

2

TiOg9.877

TiOmol1

Cg.0112

Cmol1

mol

C

molTiO2

mol

Ti

molTi

2TiOmol1

Timol1

Cmol2

Timol1 }smallest

amount is

from

limitingreactant

gTiO2

g

Ckg1

g1000

kg1

g1000

smallest

mol Tig Ti

Timol1

g87.47 % Yieldyieldtheor.

yieldact.

yield%

kg Ti

T.Y.g0001kg1


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18

Collect Needed Relationships:1000 g = 1 kg

Molar Mass TiO2= 79.87 g/mol

Molar Mass Ti = 47.87 g/mol

Molar Mass C = 12.01 g/mol

1 mole TiO2: 1 mol Ti (from the chem. equation)

2 mole C 1 mol Ti (from the chem. equation)

Information


Find: Lim. Rct., Theor. Yld., % Yld.

CP: kg rct g rct mol rct mol Tipick smallest mol Ti TY kg Ti %Y Ti

Example:

Find the Limiting



TiO2(s) + 2 C(s)

Ti(s) + 2 CO(g)


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Timol104301.1TiOmol1

Timol1TiOg79.87TiOmole1

kg1g0001TiOkg8.28 3

22

22

Apply the Concept Plan:

Information

Given: 28.6 kg C, 88.2 kg TiO2, 42.8 kg TiFind: Lim. Rct., Theor. Yld., % Yld.

CP: kg rct g rct mol rct mol Ti

pick smallest mol Ti TY kg Ti %Y TiRel: 1 mol C=12.01g; 1 mol Ti =47.87g;1 mol TiO2= 79.87g; 1000g = 1 kg;

1 mol TiO2: 1 mol Ti; 2 mol C : 1 mol Ti

Example:

Find the Limiting


Yield, and Percent

Yield.

TiO2(s) + 2 C(s)

Ti(s) + 2 CO(g)

Timol100791.1Cmol2

Timol1

Cg12.01

Cmole1

kg1

g0001Ckg8.62 3

smallest moles of TiLimiting Reactant


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Tikg52.9g1000

kg1

mol1

Tig47.87Timol104301.1 3

Theoretical Yield

Information





Example:

Find the Limiting


Yield, and Percent

Yield.

TiO2(s) + 2 C(s)

Ti(s) + 2 CO(g)


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YieldPercent100YieldlTheoretica

YieldActual %

%9.08%100Tikg52.9

Tikg42.8

Information





Example:

Find the Limiting


Yield, and Percent

Yield.

TiO2(s) + 2 C(s)

Ti(s) + 2 CO(g)


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Check the Solutions:

Limiting Reactant = TiO2

Theoretical Yield = 52.9 kg

Percent Yield = 80.9%

Since Ti has lower molar mass than TiO2, the T.Y. makes sense

The Percent Yield makes sense as it is less than 100%.

Information





Example:

Find the Limiting


Yield, and Percent

Yield.

TiO2(s) + 2 C(s)

Ti(s) + 2 CO(g)


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PracticeHow many grams of N2(g) can be made from

9.05 g of NH3reacting with 45.2 g of CuO?

2 NH3(g) + 3 CuO(s) N2(g) + 3 Cu(s) + 3 H2O(l)

Practice How many grams of N (g) can be made from 9 05 g of


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PracticeHow many grams of N2(g) can be made from 9.05 g of

NH3reacting with 45.2 g of CuO?

2 NH3(g) + 3 CuO(s) N2(g) + 3 Cu(s) + 3 H2O(l)

1 mol NH3= 17.03g, 1 mol CuO = 79.55g, 1 mol N2= 28.02 g

2 mol NH3= 1 mol N2, 3 mol CuO = 1 mol N2

9.05 g NH3

, 45.2 g CuO

g N2

Concept Plan:

Relationships:

Given:

Find:

g17.03

mol1

3

2

NHmol2

Nmol1

g NH3 mol N2mol NH3

g28.02

mol1

g CuO mol N2mol CuO

CuOmol3

Nmol1 2

g79.55

mol1

g N2smallest moles N2


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Solutions

when table salt is mixed with water, it seems to disappear,or become a liquidthe mixture is homogeneous

the salt is still there, as you can tell from the taste, or simply

boiling away the water

homogeneous mixtures are called solutions the component of the solution that changes state is called

the solute

the component that keeps its state is called the solvent if both components start in the same state, the major component

is the solvent


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Describing Solutions

since solutions are mixtures, the composition canvary from one sample to another

pure substances have constant composition

salt water samples from different seas or lakes havedifferent amounts of salt

so to describe solutions accurately, we must

describe how much of each component is presentwe saw that with pure substances, we can describethem with a single name because all samples identical


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Solution Concentration qualitatively, solutions are often

described as dilute orconcentrated

dilute solutionshave a smallamount of solute compared tosolvent

concentrated solutionshave alarge amount of solutecompared to solvent

quantitatively, the relativeamount of solute in the solutionis called the concentration


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Solution Concentration

Molarity moles of solute per 1 liter of solution

used because it describes how many moleculesof solute in each liter of solution

L)(insolutionofamount

moles)(insoluteofamount

Mmolarity,


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Preparing 1 L of a 1.00 M NaCl Solution

Example 4 5 Find the molarity of a solution that


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Example 4.5Find the molarity of a solution that

has 25.5 g KBr dissolved in 1.75 L of solution

since most solutions are between 0 and

18 M, the answer makes sense

1 mol KBr = 119.00 g,

M = moles/L

25.5 g KBr, 1.75 L solution

Molarity, M

Check: Check

Solution: Follow theConcept Plan

to Solvetheproblem

Concept Plan:

Relationships:

Strategize

Given:

Find:

SortInformation

g119.00

mol1

M0.122L1.75

KBrmol29421.0solutionL

KBrmolesMmolarity,

KBrmol2940.21KBrg119.00

KBrmol1KBrg5.52

g KBr mol KBr

L soln

ML

molM


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Using Molarity in Calculations

molarity shows the relationship between themoles of solute and liters of solution

If a sugar solution concentration is 2.0 M, then

1 liter of solution contains 2.0 moles of sugar2 liters = 4.0 moles sugar

0.5 liters = 1.0 mole sugar

1 L solution : 2 moles sugar

solutionL1

sugarmol2

sugarmol2

solutionL1

Example 4 6 How many liters of 0 125 M NaOH


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Example 4.6How many liters of 0.125 M NaOH

contains 0.255 mol NaOH?

since each L has only 0.125 mol NaOH,

it makes sense that 0.255 mol should

require a little more than 2 L

0.125 mol NaOH = 1 L solution

0.125 M NaOH, 0.255 mol NaOH

liters, L

Check: Check


to Solvetheproblem

Concept Plan:

Relationships:

Strategize

Given:

Find:

SortInformation

NaOHmol0.125

solutionL1

solutionL2.04

NaOHmol0.125

solutionL1NaOHmol552.0

mol NaOH L soln


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Dilution

often, solutions are stored as concentrated stocksolutions

to make solutions of lower concentrations from thesestock solutions, more solvent is added

the amount of solute doesnt change, just the volume ofsolution

moles solute in solution 1 = moles solute in solution 2

the concentrations and volumes of the stock and newsolutions are inversely proportional

M1V1= M2V2

Example 4 7 To what volume should you dilute


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Example 4.7To what volume should you dilute

0.200 L of 15.0 M NaOH to make 3.00 M NaOH?

since the solution is diluted by a factor

of 5, the volume should increase by a

factor of 5, and it does

M1V1= M2V2

V1= 0.200L, M1= 15.0 M, M2= 3.00 M

V2, L

Check: Check


to Solvetheproblem

Concept Plan:

Relationships:

Strategize

Given:

Find:

SortInformation

2

2

11 V

M

VM

L1.00

Lmol3.00

L200.0L

mol15.0

V1, M1, M2 V2


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Solution Stoichiometry

since molarity relates the moles of solute to theliters of solution, it can be used to convert

between amount of reactants and/or products ina chemical reaction

Example 4.8 What volume of 0.150 M KCl is required to


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Example 4.8 What volume of 0.150 M KCl is required to

completely react with 0.150 L of 0.175 M Pb(NO3)2 in the

reaction 2 KCl(aq) + Pb(NO3)2(aq) PbCl2(s) + 2 KNO3(aq)

since need 2x moles of KCl as Pb(NO3)2,and

the molarity of Pb(NO3)2> KCl, the volume of

KCl should be more than 2x volume Pb(NO3)2

1 L Pb(NO3)2= 0.175 mol, 1 L KCl = 0.150 mol,

1 mol Pb(NO3)2= 2 mol KCl

0.150 M KCl, 0.150 L of 0.175 M Pb(NO3)2

L KCl

Check: Check

Solution: Follow theConcept

Plan to

Solvetheproblem

Concept Plan:

Relationships:

Strategize

Given:

Find:

SortInformation

23)Pb(NOL1

mol0.175

KClL.3500

mol.1500

KClL1

)Pb(NOmol1

KClmol2

)Pb(NOL1

mol0.175)Pb(NOL.1500

232323

23)Pb(NOmol1

KClmol2

L Pb(NO3)2 mol KCl L KClmol Pb(NO3)2

mol0.150

KClL1


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38

What Happens When a Solute Dissolves? there are attractive forces between the solute particles

holding them together there are also attractive forces between the solvent

molecules

when we mix the solute with the solvent, there are

attractive forces between the solute particles and thesolvent molecules

if the attractions between solute and solvent are strongenough, the solute will dissolve


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Table Salt Dissolving in WaterEach ion is attracted

to the surroundingwater molecules and

pulled off and away

from the crystal

When it enters thesolution, the ion is

surrounded by water

molecules, insulating

it from other ionsThe result is a solution

with free moving

charged particles able

to conduct electricity


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Electrolytes and Nonelectrolytes

materials that dissolvein water to form asolution that willconduct electricity are

called electrolytes

materials that dissolvein water to form a

solution that will notconduct electricity arecalled nonelectrolytes


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Molecular View of

Electrolytes and Nonelectrolytes

in order to conduct electricity, a material must havecharged particles that are able to flow

electrolyte solutions all contain ions dissolved in the

water ionic compounds are electrolytes because they all dissociate

into their ions when they dissolve

nonelectrolyte solutions contain whole molecules

dissolved in the watergenerally, molecular compounds do not ionize when they

dissolve in water

the notable exception being molecular acids


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Salt vs. Sugar Dissolved in Water

ionic compounds dissociate

into ions when they dissolve

molecular compounds do not

dissociate when they dissolve


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Acids

acids are molecular compounds that ionizewhen theydissolve in water the molecules are pulled apart by their attraction for the water

when acids ionize, they form H+cations and anions

the percentage of molecules that ionize varies from oneacid to another acids that ionize virtually 100% are called strong acids

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

acids that only ionize a small percentage are calledweak acids

HF(aq) H+(aq) + F-(aq)


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Strong and Weak Electrolytes strong electrolytesare materials that dissolve

completely as ions ionic compounds and strong acids

their solutions conduct electricity well

weak electrolytesare materials that dissolve mostly as

molecules, but partially as ionsweak acids

their solutions conduct electricity, but not well

when compounds containing a polyatomic ion dissolve,

the polyatomic ion stays togetherNa2SO4(aq) 2 Na+(aq) + SO4

2-(aq)

HC2H3O2(aq) H+(aq) + C2H3O2

-(aq)


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Classes of Dissolved Materials


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Solubility of Ionic Compounds

some ionic compounds, like NaCl, dissolve very well inwater at room temperature

other ionic compounds, like AgCl, dissolve hardly at allin water at room temperature

compounds that dissolve in a solvent are said to besoluble, while those that do not are said to be insoluble

NaCl is soluble in water, AgCl is insoluble in water

the degree of solubility depends on the temperature even insoluble compounds dissolve, just not enough to be

meaningful


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When Will a Salt Dissolve?

Predicting whether a compound will dissolve inwater is not easy

The best way to do it is to do some experimentsto test whether a compound will dissolve in

water, then develop some rules based on those

experimental resultswe call this method the empirical method


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Compounds Containing the

Following Ions are Generally

Soluble

Exceptions

(when combined with ions on the

left the compound is insoluble)

Li+, Na+, K+, NH4+ none

NO3, C2H3O2

none

Cl

, Br

, I

Ag+

, Hg22+

, Pb2+

SO4

2 Ag+, Ca2+, Sr2+, Ba2+, Pb2+

Solubility Rules

Compounds that Are Generally Soluble in Water


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Compounds Containing the

Following Ions are Generally

Insoluble

Exceptions

(when combined with ions on the

left the compound is soluble or

slightly soluble)

OH Li+, Na+, K+, NH4+,

Ca2+, Sr2+, Ba2+

S2 Li+, Na+, K+, NH4+,

Ca2+, Sr2+, Ba2+

CO32, PO4

3 Li+, Na+, K+, NH4+

Solubility Rules

Compounds that Are Generally Insoluble


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Precipitation Reactions

reactions between aqueous solutions of ioniccompounds that produce an ionic compound

that is insoluble in water are called

precipitation reactionsand the insoluble

product is called a precipitate

2 KI(aq) + Pb(NO ) (aq) PbI (s) + 2


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51

2 KI(aq)+ Pb(NO3)2(aq)PbI2(s) + 2

KNO3(aq)


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No Precipitate Formation =

No ReactionKI(aq) + NaCl(aq) KCl(aq) + NaI(aq)

all ions still present, no reaction

Process for Predicting the Products of


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a Precipitation Reaction

1. Determine what ions each aqueous reactant has2. Determine formulas of possible products Exchange ions

(+) ion from one reactant with (-) ion from other

Balance charges of combined ions to get formula of eachproduct

3. Determine Solubility of Each Product in Water Use the solubility rules

If product is insoluble or slightly soluble, it will precipitate

4. If neither product will precipitate, write no reactionafter the arrow



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a Precipitation Reaction

5. If either product is insoluble, write the formulasfor the products after the arrowwriting (s)after the product that is insoluble and will

precipitate, and (aq) after products that are

soluble and will not precipitate

6. Balance the equation

Example 4.10Write the equation for the


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p q

precipitation reaction between an aqueous solution

of potassium carbonate and an aqueous solution of

nickel(II) chloride1. Write the formulas of the reactants

K2CO3(aq) + NiCl2(aq)

2. Determine the possible productsa) Determine the ions present

(K++ CO32-) + (Ni2++ Cl-)

b) Exchange the Ions

(K++ CO32-) + (Ni2++ Cl-) (K++ Cl-) + (Ni2++ CO32-)c) Write the formulas of the products

cross charges and reduce

K2CO3(aq) + NiCl2(aq) KCl + NiCO3

Example 4 10 Write the equation for the


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3. Determine the solubility of each productKCl is soluble

NiCO3is insoluble

4. If both products soluble, write no reactiondoes not apply since NiCO3is insoluble

Example 4.10Write the equation for the



nickel(II) chloride

Example 4 10 Write the equation for the


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5. Write (aq) next to soluble products and (s) nextto insoluble products

K2CO3(aq) + NiCl2(aq) KCl(aq) + NiCO3(s)

6. Balance the Equation

K2CO3(aq) + NiCl2(aq) 2 KCl(aq) + NiCO3(s)

Example 4.10 Write the equation for the



nickel(II) chloride

I i E ti


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Ionic Equations equations which describe the chemicals put into the water

and the product molecules are called molecular equations2 KOH(aq) + Mg(NO3)2(aq) 2 KNO3(aq) + Mg(OH)2(s)

equations which describe the actual dissolved species are

called completeionic equations aqueous strong electrolytes are written as ions

soluble salts, strong acids, strong bases

insoluble substances, weak electrolytes, and nonelectrolytes

written in molecule formsolids, liquids, and gases are not dissolved, therefore molecule form

2K+1(aq) + 2OH-1

(aq)+ Mg+2

(aq)+ 2NO3-1

(aq) 2K+1

(aq)+ 2NO3-1

(aq)+ Mg(OH)2(s)


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Ionic Equations

ions that are both reactants and products are calledspectator ions

2K+1(aq) + 2OH-1

(aq)+ Mg+2

(aq)+ 2NO3-1

(aq) 2K+1

(aq)+ 2NO3-1

(aq)+ Mg(OH)2(s)

an ionic equation in which the spectator ions are

removed is called a net ionic equation

2OH-1(aq)+ Mg+2(aq) Mg(OH)2(s)


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Acid-Base Reactions

also called neutralization reactionsbecause theacid and base neutralize each others properties

2 HNO3

(aq) + Ca(OH)2

(aq) Ca(NO3

)2

(aq) + 2 H2

O(l)

the net ionic equation for an acid-base reaction is

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

as long as the salt that forms is soluble in water


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Acids and Bases in Solution acids ionize in water to form H+ionsmore precisely, the H from the acid molecule is donated to a

water molecule to form hydronium ion, H3O+

most chemists use H+and H3O+interchangeably

bases dissociate in water to form OHions

bases, like NH3, that do not contain OH

ions, produce OH

bypulling H off water molecules

in the reaction of an acid with a base, the H+from theacid combines with the OHfrom the base to make water

the cation from the base combines with the anion fromthe acid to make the salt

acid + base salt + water

Common Acids


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Chemical Name Formula Uses Strength

Perchloric Acid HClO4 explosives, catalyst Strong

Nitric Acid HNO3 explosives, fertilizer, dye, glue Strong

Sulfuric Acid H2SO4explosives, fertilizer, dye, glue,

batteriesStrong

Hydrochloric Acid HClmetal cleaning, food prep, ore

refining, stomach acidStrong

Phosphoric Acid H3PO4 fertilizer, plastics & rubber,food preservationModerate

Chloric Acid HClO3 explosives Moderate

Acetic Acid HC2H3O2plastics & rubber, food

preservation, vinegarWeak

Hydrofluoric Acid HF metal cleaning, glass etching Weak

Carbonic Acid H2CO3 soda water Weak

Hypochlorous Acid HClO sanitizer Weak

Boric Acid H3BO3 eye wash Weak

Common Bases


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Common BasesChemical

NameFormula

Common

NameUses Strength

sodium

hydroxideNaOH

lye,

caustic soda

soap, plastic,

petrol refiningStrong

potassium

hydroxideKOH caustic potash

soap, cotton,

electroplatingStrong

calciumhydroxide

Ca(OH)2 slaked lime cement Strong

sodium

bicarbonateNaHCO3 baking soda cooking, antacid Weak

magnesium

hydroxide Mg(OH)2

milk of

magnesia antacid Weak

ammonium

hydroxide

NH4OH,

{NH3(aq)}

ammonia

water

detergent,

fertilizer,

explosives, fibers

Weak

HCl( ) + NaOH( ) NaCl( ) + H O(l)


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


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Example - Write the molecular, ionic, and net-

ionic equation for the reaction of aqueous nitric

acid with aqueous calcium hydroxide1. Write the formulas of the reactants

HNO3(aq) + Ca(OH)2(aq)

2. Determine the possible productsa) Determine the ions present when each reactant dissociates(H++ NO3

-) + (Ca+2+ OH-)

b) Exchange the ions, H+1 combines with OH-1to make H2O(l)

(H

+

+ NO3-

) + (Ca

+2

+ OH

-

)

(Ca

+2

+ NO3-

) + H2O(l)c) Write the formula of the salt

cross the charges

(H++ NO3-) + (Ca+2+ OH-) Ca(NO3)2+ H2O(l)


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3. Determine the solubility of the saltCa(NO3)2is soluble

4. Write an (s) after the insoluble products and a(aq) after the soluble products

HNO3(aq) + Ca(OH)2(aq) Ca(NO3)2(aq) + H2O(l)

5. Balance the equation2 HNO3(aq) + Ca(OH)2(aq) Ca(NO3)2(aq) + 2 H2O(l)



acid with aqueous calcium hydroxide


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acid with aqueous calcium hydroxide6. Dissociate all aqueous strong electrolytes to

get complete ionic equation

not H2O

2 H+(aq) + 2 NO3-(aq) + Ca+2(aq) + 2 OH-(aq)

Ca+2(aq) + 2 NO3-(aq) + H2O(l)

7. Eliminate spectator ions to get net-ionic

equation2 H+1(aq) + 2 OH-1(aq) 2 H2O(l)

H+1(aq) + OH-1(aq) H2O(l)


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Titration often in the lab, a solutions concentration is

determined by reacting it with another materialand using stoichiometrythis process is calledtitration

in the titration, the unknown solution is addedto a known amount of another reactant untilthe reaction is just completed, at this point,called the endpoint, the reactants are in theirstoichiometric ratio

the unknown solution is added slowly from aninstrument called a burette

a long glass tube with precise volume markings thatallows small additions of solution


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Acid-Base Titrations the difficulty is determining when there has been just

enough titrant added to complete the reaction

the titrant is the solution in the burette

in acid-base titrations, because both the reactant and

product solutions are colorless, a chemical is added thatchanges color when the solution undergoes large

changes in acidity/alkalinity

the chemical is called an indicator

at the endpoint of an acid-base titration, the number ofmoles of H+equals the number of moles of OH

aka the equivalence point
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Titration

Ti i
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TitrationThe base solution is the

titrant in the burette.As the base is added to

the acid, the H+ reacts with

the OHto form water.

But there is still excessacid present so the color

does not change.

At the titrations endpoint,

just enough base has beenadded to neutralize all the

acid. At this point the

indicator changes color.

Example 4.14:

Th i i f 10 00 L f HCl


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The titration of 10.00 mL of HCl

solution of unknown concentration

requires 12.54 mL of 0.100 M

NaOH solution to reach the endpoint. What is the concentration of

the unknown HCl solution?

Write down the given quantity and its units.Given: 10.00 mL HCl

12.54 mL of 0.100 M NaOH

Information

Gi 10 00 L HCl

Example 4.14:

Th i i f 10 00 L f HCl


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Write down the quantity to find, and/or its units.

Find: concentration HCl, M

Given: 10.00 mL HCl

12.54 mL of 0.100 M NaOH






Information

Gi 10 00 L HCl

Example 4.14:

Th tit ti f 10 00 L f HCl


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Collect Needed Equations and Conversion Factors:

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

1 mole HCl = 1 mole NaOH

0.100 M NaOH 0.100 mol NaOH 1 L soln

Given: 10.00 mL HCl

12.54 mL of 0.100 M NaOH

Find: M HCl






solutionliters

solutemolesMolarity

Information

Gi 10 00 L HCl

Example 4.14:



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Write a Concept Plan:

mL

NaOH

L

NaOH

mol

NaOH

NaOHL1

NaOHmol1000.

mL1

L0010.

mol

HCl

NaOHmol1

HClmol1

Given: 10.00 mL HCl

12.54 mL of 0.100 M NaOH

Find: M HCl

CF: 1 mol HCl = 1 mol NaOH0.100 mol NaOH = 1 L

M = mol/L






mL

HCl

L

HCl

mL1

L0010. HClliters

HClmolesMolarity

Information

Given: 10 00 mL HClExample:



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NaOHmole1

HClmol1

L1

NaOHmol1000

mL1

L0.001NaOHmL2.541

.

Apply the Solution Map:

= 1.25 x 10-3mol HCl

Given: 10.00 mL HCl

12.54 mL of 0.100 M NaOH

Find: M HCl

CF: 1 mol HCl = 1 mol NaOH

0.100 mol NaOH = 1 L

M = mol/L

CP: mL NaOH L NaOH

mol NaOH mol HCl;

mL HCl L HCl & mol M






Information




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HClL010000mL1

L0.001NaOHmL0.001 .


Given: 10.00 mL HCl

12.54 mL NaOH

Find: M HCl



M = mol/L

CP: mL NaOH L NaOH

mol NaOH mol HCl;







M1250HClL0.01000

HClmoles10x1.25Molarity

-3

.

Information




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Check the Solution:HCl solution = 0.125 M

The units of the answer, M, are correct.The magnitude of the answer makes sense since

the neutralization takes less HCl solution than

NaOH solution, so the HCl should be more concentrated.

Given: 10.00 mL HCl

12.54 mL NaOH

Find: M HCl



M = mol/L

CP: mL NaOH L NaOH

mol NaOH mol HCl;







G E l i R ti


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Gas Evolving Reactions Some reactions form a gas directly from the ion

exchangeK2S(aq) + H2SO4(aq) K2SO4(aq) + H2S(g)

Other reactions form a gas by the decomposition of one

of the ion exchange products into a gas and waterK2SO3(aq) + H2SO4(aq) K2SO4(aq) + H2SO3(aq)

H2SO3 H2O(l) + SO2(g)

N HCO ( ) HCl( ) N Cl( ) CO ( ) H O(l)


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NaHCO3(aq) + HCl(aq) NaCl(aq) + CO2(g) + H2O(l)

Compounds that Undergo


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Compounds that Undergo

Gas Evolving Reactions

Reactant

Type

Reacting

With

Ion

Exchange

Product

Decom-

pose?

Gas

Formed

Example

metalnS,metal HS

acid H2S no H2S K2S(aq) + 2HCl(aq) 2KCl(aq) + H2S(g)

metalnCO3,

metal HCO3

acid H2CO3 yes CO2 K2CO3(aq) + 2HCl(aq)

2KCl(aq) + CO2(g) + H2O(l)

metalnSO3

metalHSO3

acid H2SO3 yes SO2 K2SO3(aq) + 2HCl(aq)

2KCl(aq) + SO2(g) + H2O(l)

(NH4)nanion base NH4OH yes NH3 KOH(aq) + NH4Cl(aq)

KCl(aq) + NH3(g) + H2O(l)

Example 4 15 When an aqueous solution of


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Example 4.15 - When an aqueous solution of

sodium carbonate is added to an aqueous solution

of nitric acid, a gas evolves1. Write the formulas of the reactants

Na2CO3(aq) + HNO3(aq)

2. Determine the possible productsa) Determine the ions present when each reactant dissociates

(Na+1+ CO3-2) + (H+1+ NO3

-1)

b) Exchange the anions

(Na+1+ CO3-2) + (H+1+ NO3

-1) (Na+1+ NO3-1) + (H+1+ CO3

-2)

c) Write the formula of compounds

cross the charges

Na2CO3(aq) + HNO3(aq) NaNO3 + H2CO3



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3. Check to see either product H2S -No

4. Check to see if either product decomposes

Yes H2CO3decomposes into CO2(g) + H2O(l)

Na2CO3(aq) + HNO3(aq) NaNO3 + CO2(g) + H2O(l)



of nitric acid, a gas evolves



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5. Determine the solubility of other productNaNO3is soluble

6. Write an (s) after the insoluble products and a(aq) after the soluble products

Na2CO3(aq) + 2 HNO3(aq) 2NaNO3(aq) + CO2(g) + H2O(l)

7. Balance the equationNa2CO3(aq) + 2 HNO3(aq) 2NaNO3 + CO2(g) + H2O(l)



of nitric acid, a gas evolves


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Other Patterns in Reactions

the precipitation, acid-base, and gas evolvingreactions all involved exchanging the ions inthe solution

other kinds of reactions involve transferringelectrons from one atom to anotherthese arecalled oxidation-reduction reactions

also known as redox reactions

many involve the reaction of a substance with O2(g)

4 Fe(s) + 3 O2(g) 2 Fe2O3(s)

Combustion as Redox


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2 H2(g) + O2(g)2 H2O(g)

Redox without Combustion


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

2 Na2 Na++ 2 e

Cl2+ 2 e2 Cl

R ti f M t l ith N t l


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Reactions of Metals with Nonmetals

consider the following reactions:4 Na(s) + O2(g) 2 Na2O(s)

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

the reaction involves a metal reacting with a nonmetal in addition, both reactions involve the conversion of

free elements into ions

4 Na(s) + O2(g) 2 Na+

2O(s)

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

O id i d R d i


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Oxidation and Reduction

in order to convert a free element into an ion, theatoms must gain or lose electrons

of course, if one atom loses electrons, another mustaccept them

reactions where electrons are transferred from oneatom to another are redox reactions atoms that lose electrons are being oxidized, atoms

that gain electrons are being reduced

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

Na Na++ 1 e oxidation

Cl2+ 2 e 2 Clreduction

Leo

Ger

El t B kk i


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Electron Bookkeeping for reactions that are not metal + nonmetal, or do

not involve O2, we need a method for determining

how the electrons are transferred

chemists assign a number to each element in a

reaction called an oxidation statethat allows themto determine the electron flow in the reaction

even though they look like them, oxidation states are

not ion charges!

oxidation states are imaginary charges assigned based on aset of rules

ion charges are real, measurable charges

Rules for Assigning Oxidation States


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rules are in order of priority1. free elements have an oxidation state = 0 Na = 0 and Cl2= 0 in 2 Na(s) + Cl2(g)

2. monatomic ions have an oxidation state equalto their charge Na = +1 and Cl = -1 in NaCl

3. (a) the sum of the oxidation states of all theatoms in a compound is 0 Na = +1 and Cl = -1 in NaCl, (+1) + (-1) = 0



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3. (b) the sum of the oxidation states of all the atoms ina polyatomic ion equals the charge on the ion N = +5 and O = -2 in NO3

, (+5) + 3(-2) = -1

4.(a) Group I metals have an oxidation state of +1 in alltheir compounds

Na = +1 in NaCl

4. (b) Group II metals have an oxidation state of +2 inall their compounds Mg = +2 in MgCl2



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5. in their compounds, nonmetals have oxidationstates according to the table below

nonmetals higher on the table take priority

Nonmetal Oxidation State Example

F -1 CF4

H +1 CH4

O -2 CO2

Group 7A -1 CCl4Group 6A -2 CS2

Group 5A -3 NH3

Practice Assign an Oxidation State to


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Each Element in the following

Br2

K+

LiF

CO2

SO42-

Na2O2



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Each Element in the following

Br2 Br = 0, (Rule 1)

K+ K = +1, (Rule 2)

LiF Li = +1, (Rule 4a) & F = -1, (Rule 5)

CO2 O = -2, (Rule 5) & C = +4, (Rule 3a)

SO42- O = -2, (Rule 5) & S = +6, (Rule 3b)

Na2O2 Na = +1, (Rule 4a) & O = -1, (Rule 3a)

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Another Definition oxidation occurs when an atoms oxidation stateincreases during a reaction

reduction occurs when an atoms oxidation statedecreases during a reaction

CH4 + 2 O2 CO2+ 2 H2O-4+1 0 +42 +1 -2

oxidation

reduction

OxidationReduction
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oxidation and reduction must occur simultaneously

if an atom loses electrons another atom must take them the reactant that reduces an element in another reactant

is called the reducing agent

the reducing agent contains the element that is oxidized

the reactant that oxidizes an element in another reactantis called the oxidizing agent

the oxidizing agent contains the element that is reduced

2 Na(s) + Cl2(g) 2 Na+Cl(s)Na is oxidized, Cl is reduced

Na is the reducing agent, Cl2is the oxidizing agent

Id tif th O idi i d R d i A t


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Identify the Oxidizing and Reducing Agents

in Each of the Following

3 H2S + 2 NO3 + 2 H+ 3S + 2 NO + 4 H2O

MnO2+ 4 HBr MnBr2+ Br2+ 2 H2O

Id tif th O idi i d R d i A t


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Identify the Oxidizing and Reducing Agents

in Each of the Following

3 H2S + 2 NO3 + 2 H+ 3S + 2 NO + 4 H2O

MnO2+ 4 HBr MnBr2+ Br2+ 2 H2O

+1 -2 +5 -2 +1 0 +2 -2 +1 -2

ox agred ag

+4 -2 +1 -1 +2 -1 0 +1 -2

oxidationreduction

oxidation

reduction

red agox ag

Combustion Reactions


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Combustion Reactions

Reactions in which O2(g) is areactant are calledcombustion reactions

Combustion reactions release

lots of energy Combustion reactions are a

subclass of oxidation-reduction reactions

2 C8H18(g) + 25 O2(g) 16 CO2(g) + 18 H2O(g)

Combustion Products


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Combustion Products

to predict the products of a combustionreaction, combine each element in the other

reactant with oxygen

Reactant Combustion Product

contains C CO2(g)

contains H H2O(g)

contains S SO2(g)

contains N NO(g) or NO2(g)

contains metal M2On(s)


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PracticeComplete the Reactions

combustion of C3H7OH(l)

combustion of CH3NH2(g)


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PracticeComplete the Reactions

C3H7OH(l) + 5 O2(g) 3 CO2(g) + 4 H2O(g)

CH3NH2(g) + 3 O2(g) CO2(g) + 2 H2O(g) + NO2(g)

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Chem chapter 04 LEC