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1

CHAPTER 5

Read Chapter 5Read Chapter 5

Study examples and exercises.Study examples and exercises.

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Chapter 5 Outline

• Electrolytes

• NIE’s

• 5 General Chemical Reactions

• Redox

• Acid/Bases

• Molarity

• Titration

IONIC COMPOUNDSCompounds in Aqueous Solution

• Many reactions involve ionic compounds, especially reactions in water — aqueous solutions.aqueous solutions.

KMnOKMnO44 in water in water KK++(aq) + MnO(aq) + MnO44--(aq)(aq)

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Aqueous SolutionsHow do we know ions are present in aqueous

solutions?

The solutions conduct electricity!

They are called ELECTROLYTESELECTROLYTES

HCl, K2CrO4, MgCl2, and NaCl are strong strong electrolytes. electrolytes. They dissociate completely (or nearly so) into ions.

K2CrO4(aq) ---> 2 K+(aq) + CrO42-(aq)

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NaCl dissolving in waterNaCl dissolving in water

Cl-

Na+ Negative O atom

Positive H atom

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Aqueous SolutionsHow do we know ions are present in aqueous

solutions?

The solutions conduct electricity!

They are called ELECTROLYTESELECTROLYTES

HCl, MgCl2, and NaCl are strong strong

electrolytes.electrolytes. They dissociate completely (or nearly so) into ions.

7Figure 5.2

Strong Electrolyte

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Aqueous SolutionsAcetic acid ionizes only to a small extent, so it is a

weak electrolyte.weak electrolyte.

HC2H3O2(aq) C2H3O2-(aq) + H+

(aq)

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Figure 5.3

Weak Electrolyte

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Ionized acetic acid

H+

Acetic acid — Weak Electrolyte

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Aqueous SolutionsAcetic acid ionizes only to a small extent, so it is a

weak electrolyte.weak electrolyte.CH3CO2H(aq) -CH3CO2

-(aq) + H+(aq)

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Aqueous SolutionsSome compounds dissolve in water but do not

conduct electricity. They are called nonelectrolytes.nonelectrolytes.

Examples include:

sugar

ethanol

ethylene glycol (in antifreeze)

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Figure 5.3 b

Nonelectrolyte

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Nonelectrolyte— Ethanol, C2H5OH

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WATER SOLUBILITY OF IONIC COMPOUNDS

Not all ionic compounds dissolve in water.

Some are INSOLUBLEINSOLUBLE..

See Figure 5.4Figure 5.4

As long as one ion from the list As long as one ion from the list is present in a compound, the is present in a compound, the compound is water soluble.compound is water soluble.

16

See yourSolubilityTable

17WATER SOLUBILITY OF IONIC COMPOUNDS

Common minerals are often formed with anions that lead to insolubility:

sulfide fluoride

carbonate oxide

Azurite, a copper carbonate

Iron pyrite, a sulfide Orpiment, arsenic sulfide

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An acid An acid HH++ in water in water

ACIDS

Some strongstrong acids are

HCl hydrochloric

HNO3 nitric

HClO4 perchloric

H2SO4 sulfuric

HNOHNO33

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An acidAn acid H H++ in water in water

ACIDS

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

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The Nature of Acids

The Nature of Acids

HCl

H2O H3O+

Cl-

hydronium ion

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Weak AcidsWEAK ACIDS = WEAK ELECTROLYTESWEAK ELECTROLYTES

HC2H3O2 acetic acid

H2CO3 carbonic acid

H3PO4 phosphoric acid

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ACIDSNonmetal oxides can be acids

CO2(aq) + H2O(l)

H2CO3(aq)

SO3(aq) + H2O(l) H2SO4(aq)

and can come from burning

coal and oil.

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Base Base OH OH-- in water in water

BASES

NaOH(aq) Na+(aq) + OH-(aq)

NaOH is NaOH is a strong a strong base.base.

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Ammonia, NH3

An Important BaseNH3(aq) + H2O(l) NH4

+(aq) + OH-(aq)

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Figure 5.11

Ammonia Ammonia is a weak is a weak basebase

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BASES

Metal oxides are bases

CaO(s)+ H2O(l)

Ca(OH)2(aq)

CaO in water. Indicator shows solution is basic.

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Net Ionic Equations

Mg(s) + 2 HCl(aq) H2(g) + MgCl2(aq)

We really should write

Mg(s) + 2 H+(aq) + 2 Cl-(aq) H2(g) + Mg2+(aq)

+ 2 Cl-(aq)

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Ionic Equations

The two Cl- ions are SPECTATOR IONSSPECTATOR IONS — they do not participate.


Mg(s) + 2 H+(aq) + 2 Cl-(aq) H2(g) + Mg2+(aq) +

2 Cl-(aq)

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Net Ionic Equations


Mg(s) + 2 H+(aq) + 2 Cl-(aq) H2(g) + Mg2+(aq)

+ 2 Cl-(aq)

We leave the spectator ions out in writing theWe leave the spectator ions out in writing the

NET IONIC NET IONIC

EQUATION (NIE)EQUATION (NIE)Mg(s) + 2 H+(aq) H2(g) + Mg2+(aq)

CHEMICAL REACTIONS IN WATER

We will look at We will look at EXCHANGE EXCHANGE REACTIONSREACTIONS

The anions exchange The anions exchange places between cations.places between cations.

Pb(NOPb(NO33) ) 22(aq)(aq) + 2 KI + 2 KI(aq)(aq) ----> PbI ----> PbI22(s)(s) + 2 KNO + 2 KNO33 (aq)(aq)

AX + B

Y AY + BX

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Precipitation Reactions

The “driving force” is the formation of an insoluble The “driving force” is the formation of an insoluble compound — a precipitate. compound — a precipitate.

Pb(NOPb(NO33))22(aq) + 2 KI(aq) (aq) + 2 KI(aq) 2 KNO2 KNO33(aq) + (aq) + PbIPbI22(s)(s)

Net ionic equationNet ionic equation

PbPb2+2+(aq) + 2 I(aq) + 2 I--(aq) (aq) PbIPbI22(s)(s)

32Acid-Base Reactions• The “The “driving forcedriving force” is the formation of water.” is the formation of water.

NaOH(aq) + HCl(aq) NaOH(aq) + HCl(aq) NaCl(aq) + HNaCl(aq) + H22O(l)O(l)

Net ionic equationNet ionic equation

OHOH--(aq) + H(aq) + H++(aq) (aq) HH22O(l)O(l)

• This applies to ALL reactions of This applies to ALL reactions of STRONGSTRONG acids and bases.acids and bases.

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Acid-Base Reactions• A-B reactions are sometimes called A-B reactions are sometimes called

NEUTRALIZATIONSNEUTRALIZATIONS because the solution is because the solution is neither acidic nor basic at the end.neither acidic nor basic at the end.

• The other product of the A-B reaction is a The other product of the A-B reaction is a SALTSALT, MX., MX.

HHXX + + MMOH ---> OH ---> MMXX + H + H22OO

MMn+n+ comes from the comes from the base base && XXn-n- comes comes from the from the acid.acid.

This is one way to make compounds!This is one way to make compounds!

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Gas-Forming Reactions

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Gas-Forming Reactions

CaCOCaCO33(s) + H(s) + H22SOSO44(aq) (aq) CaSOCaSO44(s) + (s) + HH22COCO33(aq) (aq)

Carbonic acid is unstable and forms COCarbonic acid is unstable and forms CO22 & H & H22OO

HH22COCO33(aq) (aq) CO CO22 (g) + water (g) + water

(Antacid tablet has citric acid + NaHCO(Antacid tablet has citric acid + NaHCO33))

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Oxidation-Reduction Reactions

FeFe22OO33(s) + 2 Al(s) (s) + 2 Al(s) 2 Fe(s) + Al2 Fe(s) + Al22OO33(s)(s)

Thermite Thermite reactionreaction

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REDOX REDOX REACTIONSREACTIONS

EXCHANGEEXCHANGEAcid-BaseAcid-BaseReactionsReactions

EXCHANGEEXCHANGEGas-FormingGas-FormingReactionsReactions

EXCHANGE: Precipitation ReactionsEXCHANGE: Precipitation Reactions

REACTIONSREACTIONS

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REDOX REACTIONSOxidation — Oxidation — ( H, Mg, and Al )( H, Mg, and Al )

2 H2 H22(g) + O(g) + O22(g) (g) 2 H 2 H22O(l)O(l)

Mg(s) + 2 HCl(aq) Mg(s) + 2 HCl(aq) MgClMgCl22(aq) + H(aq) + H22(g)(g)

All corrosion reactions are oxidations.All corrosion reactions are oxidations.

2 Al(s) + 3 Cu2 Al(s) + 3 Cu2+2+(aq) (aq) 2 Al2 Al3+3+(aq) + 3 Cu(s)(aq) + 3 Cu(s)

Reduction — Reduction — ( Fe( Fe+3 +3 ))FeFe22OO33(s) + 2 Al(s) (s) + 2 Al(s) 2 Fe(s) + Al2 Fe(s) + Al22OO33(s)(s)

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But notice that in all But notice that in all reactions if something has reactions if something has been oxidized then been oxidized then something has also been something has also been reduced.reduced.

Cu(s) + 2 AgCu(s) + 2 Ag++(aq) (aq)

CuCu2+2+(aq) + 2 Ag(s)(aq) + 2 Ag(s)

REDOX REDOX REACTIONSREACTIONS

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Why Study Redox Reactions

Manufacturing metalsManufacturing metals

FuelsFuels

CorrosionCorrosion

BatteriesBatteries

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Redox reactions are characterized by Redox reactions are characterized by ELECTRON TRANSFER between an electron ELECTRON TRANSFER between an electron donor and electron acceptor.donor and electron acceptor.

Transfer leads to — Transfer leads to —

1. Increase in oxidation number of some 1. Increase in oxidation number of some element = OXIDATIONelement = OXIDATION

2. Decrease in oxidation number of some 2. Decrease in oxidation number of some element = REDUCTIONelement = REDUCTION

REDOX REACTIONSREDOX REACTIONS

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OXIDATION NUMBERS

The electric charge an element APPEARS to have The electric charge an element APPEARS to have when electrons are counted by some arbitrary when electrons are counted by some arbitrary rules:rules:

1. Each atom in free element has ox. no. = 01. Each atom in free element has ox. no. = 0

Zn OZn O22 I I22 S S88

2. In simple ions, ox. no. = charge on ion2. In simple ions, ox. no. = charge on ion

-1 for Cl-1 for Cl-- +2 for Mg +2 for Mg2+2+

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OXIDATION NUMBERS

3. O has ox. no. = -23. O has ox. no. = -2

(except in peroxides: in H(except in peroxides: in H22OO22, O = -1), O = -1)

4. Ox. no. of H = +14. Ox. no. of H = +1

(except when H is associated with a (except when H is associated with a metal as in NaH where it is -1)metal as in NaH where it is -1)

5. Algebraic sum of oxidation numbers 5. Algebraic sum of oxidation numbers

= 0 for a compound = 0 for a compound

= overall charge for an ion= overall charge for an ion

44

OXIDATION NUMBERS

NHNH33 N = N =

ClOClO-- Cl = Cl =

HH33POPO44 P = P =

MnOMnO44- - Mn = Mn =

CrCr22OO772-2- Cr = Cr =

CC33HH88 C = C =

Oxidation Oxidation number of F in number of F in HF?HF?

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Recognizing a Redox Reaction

Corrosion of aluminumCorrosion of aluminum

2 Al2 Al(s)(s) + 3 Cu + 3 Cu22++(aq)(aq) ---> 2 Al ---> 2 Al33++

(aq)(aq) + 3 Cu + 3 Cu(s)(s)

Al(s) --> AlAl(s) --> Al3+3+(aq) + 3 e(aq) + 3 e--

• Ox. no. of Al increases as eOx. no. of Al increases as e-- are donated by the are donated by the metal. metal.

• Therefore, Therefore, Al is OXIDIZED Al is OXIDIZED and is the and is the REDUCING REDUCING AGENTAGENT in this balanced in this balanced half-reaction.half-reaction.

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Corrosion of aluminumCorrosion of aluminum

2Al2Al(s)(s)+ 3Cu+ 3Cu2+2+(aq)(aq)--> 2Al--> 2Al3+3+

(aq)(aq)+ 3Cu+ 3Cu(s)(s)

CuCu2+2+(aq) + 2 e(aq) + 2 e- - --> Cu(s) --> Cu(s)

• Ox. no. of Cu decreases as eOx. no. of Cu decreases as e-- are accepted by are accepted by the ion. the ion.

• Therefore, Therefore, Cu is REDUCED Cu is REDUCED and is the and is the OXIDIZING AGENTOXIDIZING AGENT in this balanced in this balanced half-half-reaction.reaction.

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Notice that the 2 half-reactions add up to give Notice that the 2 half-reactions add up to give the overall reaction if we use 2 mol of Al and 3 the overall reaction if we use 2 mol of Al and 3 mol of Cumol of Cu2+2+..

2 Al(s) --> 2 Al2 Al(s) --> 2 Al3+3+(aq) + 6 e(aq) + 6 e--

3 Cu3 Cu2+2+(aq) + 6 e(aq) + 6 e- - --> 3 Cu(s)--> 3 Cu(s)

----------------------------------------------------------------------------------------------------------------------

2 Al(s) + 3 Cu2 Al(s) + 3 Cu2+2+(aq) ---> 2 Al(aq) ---> 2 Al3+3+(aq) + 3 Cu(s)(aq) + 3 Cu(s)

Final equation is balanced for mass and charge.Final equation is balanced for mass and charge.


48

Examples of Redox ReactionsExamples of Redox Reactions

Metal + acidMetal + acidMg + HClMg + HClMg = reducing agentMg = reducing agentHH++ = oxidizing agent = oxidizing agent

Metal + acidMetal + acidCu + HNOCu + HNO33

Cu = reducing agentCu = reducing agentHNOHNO3 3 = oxidizing agent= oxidizing agent

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Examples of Redox ReactionsExamples of Redox Reactions

Metal + HalogenMetal + Halogen

2 Al + 3 Br2 Al + 3 Br22 ---> Al ---> Al22BrBr66

50


See Table 5.5

In terms of oxygenIn terms of oxygen gain gain lossloss

In terms Ox. No.In terms Ox. No. increase increase decreasedecrease

In terms of electronsIn terms of electrons lossloss gaingain

Reaction TypeReaction Type OxidationOxidation ReductionReduction

51

Common Oxidizing and Reducing Agents

See Table 5.4

Metals Metals (Cu) are (Cu) are reducing reducing agentsagents

HNOHNO33 is an is an

oxidizing oxidizing agentagent

2 K + 2 H2 K + 2 H22O --> O -->

2 KOH + H 2 KOH + H22

Metals Metals (Na, K, (Na, K, Mg, Fe) Mg, Fe) are are reducing reducing agentsagents

Cu + 4 HNOCu + 4 HNO33 --> Cu(NO --> Cu(NO33))22

+ 2 NO+ 2 NO2 2 + 2 H+ 2 H22OO

52

Learn to recognize common oxidizing and reducing agents.

See Table 5.4.

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Table 5-4

54

BALANCING REDOX EQUATIONSSection 21.1

BALANCING REDOX EQUATIONSSection 21.1

The Half-Reaction MethodThe Half-Reaction Method

• Separate the equation into half-reactions.Separate the equation into half-reactions.

• Balance the half-reactions.Balance the half-reactions.

• Combine the half-reactions to form a Combine the half-reactions to form a

balanced equation containing no electrons.balanced equation containing no electrons.

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Balancing Half-ReactionsBalancing Half-Reactions

• First balance the element changing oxidation state.First balance the element changing oxidation state.

• Balance the oxygen atoms with water.Balance the oxygen atoms with water.

• Balance the hydrogen atoms with HBalance the hydrogen atoms with H++..

• Balance theBalance the charge with electrons.charge with electrons.

After combining the half-reactions, After combining the half-reactions,

check for mass and charge check for mass and charge

balance.balance.

56

Practice ProblemsPractice Problems

Balance the following equations:

MnO4- + H2SO3 ----> Mn+2 + SO4

-2

Al + NO3- ---> Al(OH)4

- + NH3

REACTIONS IN SOLUTION

Section 5.8

58

Terminology

In solution we need to define theIn solution we need to define the

• SOLVENTSOLVENT

the component whose physical state is the component whose physical state is preserved when solution preserved when solution formsforms

• SOLUTESOLUTE

the other solution componentthe other solution component

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Concentration of Solute

The amount of solute in a solution The amount of solute in a solution is given by its is given by its concentrationconcentration.

Molarity (M) = moles solute

liters of solution

60

PROBLEM: Dissolve 5.00 g of NiCl2•6 H2O in enough water to make 250.0 mL of solution. Calculate molarity.

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PROBLEM: Dissolve 5.00 g of NiCl2•6 H2O in enough water to make 250.0 mL of solution. Calculate molarity.

= 0.0841 mole/L

= 0.0841 M

5.00 g mole 5.00 g mole 0.25000.2500 L 237.7 gL 237.7 g

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The Nature of the KMnO4 Solution

KMnOKMnO44(aq) --> K(aq) --> K++(aq) + MnO(aq) + MnO44--(aq)(aq)

If you make a solution that is 0.30 M in KMnOIf you make a solution that is 0.30 M in KMnO44, , this means that this means that __

[K[K++] = [MnO] = [MnO44--] = 0.30 M] = 0.30 M

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The Nature of a Na2CO3 Solution

This water-soluble compound is ionicThis water-soluble compound is ionic

NaNa22COCO33(aq) --> 2 Na(aq) --> 2 Na++(aq) + CO(aq) + CO332-2-(aq)(aq)

If [NaIf [Na22COCO33] = 0.100 M, then] = 0.100 M, then

[Na[Na++] = 0.200 M] = 0.200 M

[CO[CO332-2-] = 0.100 M] = 0.100 M

NaNa22COCO33

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USING MOLARITY

What mass of oxalic acid, HWhat mass of oxalic acid, H22CC22OO44, , is required to make 250. mL of a is required to make 250. mL of a 0.0500 M solution?0.0500 M solution?

Conc (M) = moles/volume = mol/VConc (M) = moles/volume = mol/V

This means thatThis means that

moles = M • Vmoles = M • V

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Preparing Solutions From Solids

What mass of oxalic acid, HWhat mass of oxalic acid, H22CC22OO44, , is required to make 250. mL of a is required to make 250. mL of a 0.0500 M solution?0.0500 M solution?

0.250 L 0.0500 mole 90.0 g L mole

= 1.12 g

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Preparing Solutions• Weigh out a solid solute Weigh out a solid solute

and dissolve in a given and dissolve in a given quantity of solvent.quantity of solvent.

• Dilute a concentrated Dilute a concentrated solution to give one that is solution to give one that is less concentrated.less concentrated.

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Preparing Solutions by Dilution

Preparing aPreparing a

1.64 1.64 ×× 10 10-4-4M or M or

1.64 1.64 ×× 10 10-4-4mol/Lmol/L

solution.solution.

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PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do?

3.0 M NaOH 0.50 M NaOH

H2O

Concentrated Dilute

But how much water do we add?

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PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do?

How much water is added?

The important point is that

moles of NaOH in ORIGINAL solution = moles of NaOH in FINAL solution

70PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do?

Moles of NaOH in original solution =

M • VM • V =

(3.0 mol/L)(0.050 L) = 0.15 mol NaOH

Therefore, moles of NaOH in final solution must also = 0.15 mol NaOH

(0.15 mol NaOH)(1 L/0.50 mol) = 0.30 L

or 300 mL = volume of final solution.300 mL = volume of final solution.

71PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do?

Conclusion:

add enough add enough

waterwater to 50.0 mL of 3.0 M NaOH to make 300 mL of 0.50 M NaOH.3.0 M NaOH 0.50 M NaOH

H2O

Concentrated Dilute

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A shortcutA shortcut

MMinitialinitial • V • Vinitialinitial = M = Mfinalfinal • V • Vfinalfinal



Zinc reacts with acids to produce H2 gas.

If you have 10.0 g of Zn, what volume of 2.50 M HCl is needed to convert the Zn completely?

SOLUTION SOLUTION STOICHIOMETRYSTOICHIOMETRY

Section 5.9

74

Step 1:Step 1: Write the balanced equationWrite the balanced equation

Zn(s) + 2 HCl(aq) --> ZnClZn(s) + 2 HCl(aq) --> ZnCl22(aq) + H(aq) + H22(g)(g)

Zinc reacts with acids to produce H2 gas. If you have 10.0 g of Zn, what volume of 2.50 M HCl is needed to convert the Zn completely?

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Step 2: Step 2: Write the given and requested information below the equation.


10.0 g 2.50 M10.0 g 2.50 M ?mL ?mL


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Step 3: Step 3: Calculate using the information.


10.0 g 2.50 M10.0 g 2.50 M

?mL?mL

= 0.122 L HCl

10.0 gZn moleZn 2 moleHCl L HCl

65.4 gZn moleZn 2.50 moleHCl


77

ACID-BASE REACTIONSTitrations

ACID-BASE REACTIONSTitrations

H2C2O4(aq) + 2 NaOH(aq) --->Na2C2O4(aq) + 2 H2O(liq)

acidacid basebase

Carry out this reaction using a Carry out this reaction using a TITRATIONTITRATION..

Oxalic acid,

H2C2O4

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Titration Titration setupsetup Buret contains a

solution whoseconcentration isknown exactly.

Solution ofunknownconcentration

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TitrationTitration1. Add solution from the buret.

2. Reagent (base) reacts with compound (acid) in solution in the flask.

3. Indicator shows when exact stoichiometric reaction has occurred.

4. At equivalence point moles H+ = moles OH-

80

1.065 g of H2C2O4 (oxalic acid)

requires 35.62 mL of NaOH for

titration to an equivalence point.

What is the concentration of the

NaOH?

LAB PROBLEM #1: Standardize a solution of NaOH — i.e., accurately determine its concentration.

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Step 1: Step 1: Write the balanced equationWrite the balanced equation

2 NaOH(aq) + H2 NaOH(aq) + H22CC22OO44(aq) --> Na(aq) --> Na22CC22OO44(aq) + 2 HOH(l)(aq) + 2 HOH(l)

1.065 g of H2C2O4 (oxalic acid) requires 35.62 mL of NaOH for titration to an equivalence point. What is the concentration of the NaOH?

82


2NaOH(aq) + H2NaOH(aq) + H22CC22OO44(aq) --> Na(aq) --> Na22CC22OO44(aq) +2 HOH(l) (aq) +2 HOH(l)

35.62 mL 1.065 g35.62 mL 1.065 g ? M ? M


83

= 0.664 M NaOH

Step 3: Step 3: Calculate using the information. 2NaOH(aq) + H2NaOH(aq) + H22CC22OO44(aq) --> Na(aq) --> Na22CC22OO44(aq) +2 HOH(l) (aq) +2 HOH(l)

35.62 mL 1.065 g35.62 mL 1.065 g ? M ? M

1.065g A mole A 2 mole B

0.03562 L B 90.0 g A mole A = 0.664 mole B/L B


84LAB PROBLEM #2: Use standardized NaOH to determine the amount of an acid in an unknown.

Apples contain malic acid, C4H6O5.

C4H6O5(aq) + 2 NaOH(aq) --->

Na2C4H4O5(aq) + 2 H2O(liq)

76.80 g of apple requires 34.56 mL of 0.663 M

NaOH for titration.

What is mass % of malic acid?

8576.80 g of apple requires 34.56 mL of 0.663 M NaOH for titration. What is weight % of malic acid?

Step 1: Step 1: Write the balanced equationWrite the balanced equation

C4H6O5(aq) + 2 NaOH(aq) ---> Na2C4H4O5(aq) + 2 H2O(l)

86


C4H6O5(aq) + 2 NaOH(aq) ---> Na2C4H4O5(aq) + 2 H2O(l)? g 34.56 mL

(? % A) 0.663 M


87

Step 3: Step 3: Calculate using the information.C4H6O5(aq) + 2 NaOH(aq) ---> Na2C4H4O5(aq) + 2 H2O(l)

? g 34.56 mL(? % A) 0.663 M

.03456 L B .663 mole B mole A 134.0 g A

L B 2 mole B mole A= 1.54 g A

1.54 g A 76.80 g apple

% A = x 100 = 2.00 % A


88Sample Problems

1) What volume of 0.50 M sulfuric acid is 1) What volume of 0.50 M sulfuric acid is needed to react completely with 10.0 mL of 2.0 needed to react completely with 10.0 mL of 2.0 M potassium hydroxide?M potassium hydroxide?

HH22SOSO44 + 2 KOH --> 2 HOH + K + 2 KOH --> 2 HOH + K22SOSO44

0.50 M 2.0 M0.50 M 2.0 M

? mL 10.0 mL? mL 10.0 mL

.0100 L KOH 2.0 mole KOH mole H.0100 L KOH 2.0 mole KOH mole H22SOSO44 L H L H22SOSO44

L KOH 2 mole KOH .50 mole HL KOH 2 mole KOH .50 mole H22SOSO44

= 0.020 L H= 0.020 L H22SOSO44

89Sample Problems

2) 16 mL of 2.0 M NaOH neutralizes 25 mL of 2) 16 mL of 2.0 M NaOH neutralizes 25 mL of HCl. What is the molarity of the acid?HCl. What is the molarity of the acid?

0.016 L NaOH 2.0 mole NaOH mole HCl 0.016 L NaOH 2.0 mole NaOH mole HCl

0.025 L HCl L NaOH mole NaOH 0.025 L HCl L NaOH mole NaOH

= 1.3 M HCl= 1.3 M HCl

HCl + NaOH --> HOH + NaClHCl + NaOH --> HOH + NaCl

? M 2.0 M? M 2.0 M

25 mL 16 mL25 mL 16 mL

90Sample Problems

3) A 0.15 M solution of calcium chloride is 3) A 0.15 M solution of calcium chloride is added to a solution of ammonium carbonate added to a solution of ammonium carbonate and 2.010 g of calcium carbonate is and 2.010 g of calcium carbonate is precipitated. What volume of calcium chloride precipitated. What volume of calcium chloride solution was added?solution was added?

CaClCaCl22 + (NH + (NH44))22COCO33 --> CaCO --> CaCO33 + 2 NH + 2 NH44ClCl

0.15 M 0.15 M 2.010 g 2.010 g

? mL ? mL

= 0. 13 L CaCl= 0. 13 L CaCl22

2.010 g CaCO2.010 g CaCO33 mole CaCO mole CaCO33 mole CaCl mole CaCl22 L CaCl L CaCl22

100.1 gCaCO100.1 gCaCO33 moleCaCO moleCaCO3 3 .15 moleCaCl.15 moleCaCl22

91Sample Problems

4) What volume(mL) of 0.100 M HCl is needed 4) What volume(mL) of 0.100 M HCl is needed to react completely with 15.7 g of barium to react completely with 15.7 g of barium hydroxide?hydroxide?

2 HCl + Ba(OH)2 HCl + Ba(OH)22 --> 2 HOH + BaCl --> 2 HOH + BaCl22

0.100 M 15.7 g0.100 M 15.7 g

? mL ? mL

= 1.83 L HCl = 1830 mL HCl= 1.83 L HCl = 1830 mL HCl

15.7 g Ba(OH)15.7 g Ba(OH)22 mole Ba(OH) mole Ba(OH)22 2 mole HCl L HCl 2 mole HCl L HCl

171.3 g Ba(OH)171.3 g Ba(OH)22 moleBa(OH) moleBa(OH)2 2 .100moleHCl.100moleHCl

92Practice Problems• Write the balanced formula, ionic, and net ionic

equations for:

• Mixture of solutions of barium chloride and sodium phosphate.

• Mixtures of solutions of silver nitrate and sodium carbonate.

• Mixtures of solutions of nitric acid and barium hydroxide.

• Mixtures of solutions of ammonia and acetic acid. • Silver carbonate solid reacts with nitric acid

93

Practice Problems

1. Write the NIE for each of the following:1. Write the NIE for each of the following:

lead(II) nitrate + potassium iodidelead(II) nitrate + potassium iodide

perchloric acid + potassium hydroxideperchloric acid + potassium hydroxide

sodium sulfite + hydroiodic acidsodium sulfite + hydroiodic acid

2. Identify the substance being reduced, the 2. Identify the substance being reduced, the substance being substance being

oxidized, the oxidizing agent, and the reducing oxidized, the oxidizing agent, and the reducing agent:agent:

4 Fe + 3 O4 Fe + 3 O22 --> 2 Fe --> 2 Fe22OO33

94

Practice Problems

3. a) Balance the following equation in acidic 3. a) Balance the following equation in acidic solution.solution.

KMnOKMnO44 + NaCl --> Cl + NaCl --> Cl22 + MnCl + MnCl2 2

b) Balance the following equation in basic b) Balance the following equation in basic solution.solution.

II22 + NO + NO33- - -----> IO-----> IO22

- - + N+ N22OO

4. Calculate the molarity (M) of 32.7 g H4. Calculate the molarity (M) of 32.7 g H33POPO44 in 250 in 250 mL of solution.mL of solution.

5. Explain how to prepare 500. mL of 0.10 M 5. Explain how to prepare 500. mL of 0.10 M (NH(NH44))22COCO33

6. Explain how to prepare 75 mL of 0.025 M NaOH 6. Explain how to prepare 75 mL of 0.025 M NaOH from 4.0 M NaOH.from 4.0 M NaOH.

95

Practice Problems

7. How many grams of solute is there in 75.0 mL of 7. How many grams of solute is there in 75.0 mL of 0.25 M FeCl0.25 M FeCl33??

8. Explain how to prepare 75.0 mL of 0.105 M 8. Explain how to prepare 75.0 mL of 0.105 M ammonium phosphate.ammonium phosphate.

9. 50.0 mL of a HCl solution is required to react 9. 50.0 mL of a HCl solution is required to react completely with 1.204 g of CaCOcompletely with 1.204 g of CaCO33. What is the M?. What is the M?

10. How many grams of BaSO10. How many grams of BaSO44 can be formed from can be formed from a barium nitrate solution by adding 42.6 mL of a barium nitrate solution by adding 42.6 mL of 0.15 M sulfuric acid?0.15 M sulfuric acid?

11. How many moles of sulfur dioxide are produced 11. How many moles of sulfur dioxide are produced when 25.0 mL of 0.15 M ammonium sulfite reacts when 25.0 mL of 0.15 M ammonium sulfite reacts with 25.0 mL of 0.25 M chloric acid?with 25.0 mL of 0.25 M chloric acid?

96

Practice Problems Answers

1. Pb1. Pb2+2+ (aq) + 2 I (aq) + 2 I1- 1- (aq) --> PbI(aq) --> PbI22 (s) (s)

HH++ (aq) + OH (aq) + OH-- (aq) --> HOH (l) (aq) --> HOH (l)

SOSO332-2- (aq) + 2 H (aq) + 2 H++ (aq) --> SO (aq) --> SO22 (g) + H (g) + H22O (l)O (l)

2. Reduced, OA = oxygen2. Reduced, OA = oxygen

Oxidized, RA = ironOxidized, RA = iron

3. a) 16 H3. a) 16 H++ (aq) + 2 MnO (aq) + 2 MnO44-- (aq) + 10 Cl (aq) + 10 Cl-- (aq) (aq) ----

> 2 Mn> 2 Mn2+2+ (aq) + 8 H (aq) + 8 H22O (l) + 5 ClO (l) + 5 Cl22 (g) (g)

b) 2 OHb) 2 OH-- (aq) + 4 I (aq) + 4 I22 (aq) + 6 NO (aq) + 6 NO33-- (aq) --> (aq) -->

8 IO8 IO22-- (aq) + 3 N (aq) + 3 N22O (g) + HO (g) + H22O (l)O (l)

4. 1.3 M4. 1.3 M

97

Practice Problems Answers

5. Dissolve 4.8 g ammonium carbonate in 5. Dissolve 4.8 g ammonium carbonate in some water, then dilute to a total some water, then dilute to a total volume of 500. mL.volume of 500. mL.

6. Take 0.47 mL of 4.0 M NaOH and dilute 6. Take 0.47 mL of 4.0 M NaOH and dilute to a total volume of 75 mL.to a total volume of 75 mL.

7. 3.0 g7. 3.0 g

8. Dissolve 1.17 g ammonium phosphate 8. Dissolve 1.17 g ammonium phosphate in some water, then dilute to a total in some water, then dilute to a total volume of 75.0 mL.volume of 75.0 mL.

9. 0.481 M9. 0.481 M 10. 1.5 g10. 1.5 g

11. .0031 mole11. .0031 moleEnd of Chapter 5

98

Sample Problems

Write the NIEWrite the NIE1. ferric nitrate + sodium sulfide1. ferric nitrate + sodium sulfide

Fe(NOFe(NO33))33 NaNa22SS FeFe22SS33 NaNONaNO33++ -->--> ++

-->-->2 Fe2 Fe3+3+ ++ 6 NO6 NO33--

2 Fe(NO2 Fe(NO33))33 3 Na3 Na22SS FeFe22SS33 6 NaNO6 NaNO33++ -->--> ++

++ 6 Na6 Na++ 3 S3 S2-2-++

FeFe22SS33 6 NO6 NO33--++6 Na6 Na++++

-->-->2 Fe2 Fe3+3+ 3 S3 S2-2-++ FeFe22SS33

99

Sample Problems

Write the NIEWrite the NIE2. barium hydroxide + acetic acid2. barium hydroxide + acetic acid

Ba(OH)Ba(OH)22 HCHC22HH33OO22 Ba(CBa(C22HH33OO22))22 HOHHOH++ -->--> ++

-->--> BaBa2+2+ ++ 2 OH2 OH-- ++ 2 HC2 HC22HH33OO22

2 HOH2 HOH++2 C2 C22HH33OO22--++BaBa2+2+

Ba(OH)Ba(OH)22 2 HC2 HC22HH33OO22 Ba(CBa(C22HH33OO22))22 2 HOH2 HOH++ -->--> ++

-->-->2 OH2 OH-- 2 HC2 HC22HH33OO22++ 2 HOH2 HOH 2 C2 C22HH33OO22--++

-->--> OHOH-- HCHC22HH33OO22++ HOH HOH CC22HH33OO22--++

100

Sample Problems

Write the NIEWrite the NIE3. hydrochloric acid + calcium carbonate3. hydrochloric acid + calcium carbonate

HClHCl CaCOCaCO33 CaClCaCl22 HH22COCO33++ -->--> ++

-->--> 2 H2 H++ ++ 2 Cl2 Cl-- ++ CaCOCaCO33

CaCa2+2+ ++ 2 Cl2 Cl--

2 HCl2 HCl CaCOCaCO33 CaClCaCl22 HH22COCO33++ -->--> ++

HOHHOH++COCO22++

-->-->2 H2 H++ CaCOCaCO33++ COCO22 HOHHOH++++CaCa2+2+

101

Sample Problems

Write the NIEWrite the NIE4. sodium nitrate + potassium chloride4. sodium nitrate + potassium chloride

NaNONaNO33 KClKCl NaClNaCl KNOKNO33++ -->--> ++

-->-->NaNa++ ++ NONO33-- ++ KK++ ClCl--++

NONO33--++KK++++ClCl--NaNa++ ++

All spectators, no reaction (N/R)All spectators, no reaction (N/R)

102

Sample Problems

Write the NIEWrite the NIE5. sodium sulfite + sulfuric acid5. sodium sulfite + sulfuric acid

NaNa22SOSO33 HH22SOSO44 NaNa22SOSO44 HH22SOSO33++ -->--> ++

-->-->2 Na2 Na++ ++ SOSO332-2- ++ HH++ HSOHSO44

--++

SOSO332- 2- + H + H++ + HSO + HSO44

- - --> SO--> SO442-2- + SO + SO22 + H + H22OO

HH22OO++SOSO22++ SOSO442-2-2 Na2 Na++ ++

103

Sample Problems

Write the NIEWrite the NIE6. barium + sodium chloride6. barium + sodium chloride

BaBa NaClNaCl BaClBaCl22 NaNa++ -->--> ++

-->-->BaBa ++ 2 Na2 Na++ 2 Cl2 Cl--++

2 Na2 Na++ BaBa2+2+ 2 Cl2 Cl--++

-->--> BaBa 2 Na2 Na++++ BaBa2+2+ + 2 Na + 2 Na

BaBa 2 NaCl2 NaCl BaClBaCl22 2 Na2 Na++ -->--> ++

104


Balance the following equation:Balance the following equation:

MnOMnO44-- + H + H22SOSO3 3 ----> Mn ----> Mn+2+2 + SO + SO44

-2-2

MnOMnO44-- ----> Mn ----> Mn+2+2

HH22SOSO3 3 ----> SO ----> SO44-2-2

2. Balance atoms:2. Balance atoms:

8 H8 H++ + MnO + MnO44-- ----> Mn ----> Mn+2+2 + 4 H + 4 H22O O

HH22O + HO + H22SOSO3 3 ----> SO ----> SO44-2-2 + 4 H + 4 H++

1. Separate into half reactions:1. Separate into half reactions:

105


3. Balance charges:3. Balance charges:

5 e5 e-- + 8 H + 8 H++ + MnO + MnO44-- ----> Mn ----> Mn+2+2 + 4 H + 4 H22O O

HH22O + HO + H22SOSO3 3 ----> SO ----> SO44-2-2 + 4 H + 4 H++ + 2 e + 2 e--

4. Equal electrons gained and lost:4. Equal electrons gained and lost:

2(2(5 e5 e-- + 8 H + 8 H++ + MnO + MnO44-- ----> Mn ----> Mn+2+2 + 4 H + 4 H22OO))

5(5(HH22O + HO + H22SOSO3 3 ----> SO ----> SO44-2-2 + 4 H + 4 H++ + 2 e + 2 e--))

10 16 2 2 810 16 2 2 8

5 5 5 20 105 5 5 20 10

106


5. Simplify and Add:5. Simplify and Add:

2(2(5 e5 e-- + 8 H + 8 H++ + MnO + MnO44-- ----> Mn ----> Mn+2+2 + 4 H + 4 H22OO))

5(5(HH22O + HO + H22SOSO3 3 ----> SO ----> SO44-2-2 + 4 H + 4 H++ + 2 e + 2 e--))

10 16 2 2 810 16 2 2 8

5 5 5 20 105 5 5 20 10

2MnO2MnO44-- + 5 H + 5 H22SOSO3 3 ----> ---->

2 Mn2 Mn+2 +2 + 5 SO+ 5 SO44-2-2 + 4 H + 4 H++ + + 33 HH22O O

44

33

107


Balance the following equation:Balance the following equation:

Al + NOAl + NO33-- ---> Al(OH) ---> Al(OH)44

-- + NH + NH33

Al ----> Al(OH)Al ----> Al(OH)44--

NONO33-- ----> NH ----> NH33

2. Balance atoms:2. Balance atoms:

4 H4 H22O + Al ----> Al(OH)O + Al ----> Al(OH)44-- + 4 H + 4 H++

9 H9 H+ + + NO+ NO33-- ----> NH ----> NH3 3 + 3 H+ 3 H22OO

1. Separate into half reactions:1. Separate into half reactions:

108


3. Balance charges:3. Balance charges:

4 H4 H22O + Al ----> Al(OH)O + Al ----> Al(OH)44-- + 4 H + 4 H++ + 3 e + 3 e--

8 e8 e-- + 9 H + 9 H+ + + NO+ NO33-- ----> NH ----> NH3 3 + 3 H+ 3 H22OO

4. Equal electrons gained and lost:4. Equal electrons gained and lost:

8(8(4 H4 H22O + Al ----> Al(OH)O + Al ----> Al(OH)44-- + 4 H + 4 H++ + 3 e + 3 e--))

3(3(8 e8 e-- + 9 H + 9 H+ + + NO+ NO33-- ----> NH ----> NH3 3 + 3 H+ 3 H22OO))

32 8 8 32 2432 8 8 32 24

24 27 3 3 924 27 3 3 9

109


5. Simplify and Add:5. Simplify and Add:

8(8(4 H4 H22O + Al ----> Al(OH)O + Al ----> Al(OH)44-- + 4 H + 4 H++ + 3 e + 3 e--))

3(3(8 e8 e-- + 9 H + 9 H+ + + NO+ NO33-- ----> NH ----> NH3 3 + 3 H+ 3 H22OO))

32 8 8 32 2432 8 8 32 24

24 27 3 3 924 27 3 3 9

23 H23 H22O + 8 Al + 3 NOO + 8 Al + 3 NO33-- ----> ---->

8 Al(OH)8 Al(OH)44-- + 5 H + 5 H++ + 3 + 3

NHNH3 3

552323

110


6. Change to basic solution:6. Change to basic solution:

23 H23 H22O + 8 Al + 3 NOO + 8 Al + 3 NO33-- ----> ----> 8 Al(OH)8 Al(OH)44

-- + 5 H + 5 H++ + 3 + 3

NHNH3 3 + 5 OH+ 5 OH-- + 5 OH + 5 OH --

5 H5 H22OO

1818

18 H18 H22O + 5 OHO + 5 OH-- + 8 Al + 3 NO + 8 Al + 3 NO33-- ---->8 Al(OH) ---->8 Al(OH)44

-- + 3 NH + 3 NH3 3

111


1. 1. MnOMnO22 + HBr --> Br + HBr --> Br22 + MnBr + MnBr22

2e2e-- + 4 H + 4 H++ + MnO + MnO22 ----> Mn ----> Mn2+2+ + 2 H + 2 H22OO

2 Br2 Br-- ----> Br ----> Br2 2 + 2e+ 2e--

4 H4 H++ + MnO + MnO22 + 2 Br + 2 Br-- ----> Mn ----> Mn2+2+ + 2 H + 2 H22O + BrO + Br22

112


2. 2. ClCl22 + NaBr --> NaCl + Br + NaBr --> NaCl + Br22

2e2e-- + Cl + Cl22 --> 2 Cl --> 2 Cl--

2 Br2 Br-- --> Br --> Br22 + 2e + 2e--

ClCl22 + 2 Br + 2 Br-- ----> 2 Cl ----> 2 Cl-- + Br + Br2 2

113


3. 3. HH22S + HNOS + HNO33 --> S + NO --> S + NO

3(3(HH22S ----> S + 2 HS ----> S + 2 H++ + 2 e + 2 e--))

2(2(3 e3 e-- + 4 H + 4 H+ + + NO+ NO33-- ----> NO ----> NO + 2 H+ 2 H22OO))

3 3 6 63 3 6 6

6 8 2 2 46 8 2 2 422

3 H3 H22S + 2 HS + 2 H++ + 2 NO + 2 NO33-- ----> 3 S + 2 NO + 4 H ----> 3 S + 2 NO + 4 H22OO

114


4. 4. PbOPbO22 + Sb --> PbO + NaSbO + Sb --> PbO + NaSbO2 2

(base)(base)

3(3(2 e- +2 e- + 2H2H++ + PbO + PbO22 ----> PbO + H ----> PbO + H22OO))

2(2(2 H2 H22O + Sb ----> SbOO + Sb ----> SbO22-- + 4 H + 4 H++ + 3 e + 3 e--))

6 6 3 3 36 6 3 3 3

4 2 2 8 64 2 2 8 6

3 PbO3 PbO22 + H + H22O + 2 Sb --->3 PbO + 2 SbOO + 2 Sb --->3 PbO + 2 SbO22-- + 2 H + 2 H++

2211

+ 2 OH+ 2 OH-- + 2 + 2 OHOH--

2 H2 H22OO

3PbO3PbO22 + 2 OH- + 2Sb --->3PbO + 2SbO + 2 OH- + 2Sb --->3PbO + 2SbO22-- + H + H22OO

11

115Sample Problems

1) Calculate the molarity(M) of 40.0 g NaOH in 1) Calculate the molarity(M) of 40.0 g NaOH in 700. mL of solution.700. mL of solution.

40.0 g40.0 g

0.700 L0.700 L

molemole

40.0 g40.0 g= 1.43 M= 1.43 M

116Sample Problems

2) Explain how to prepare 2.0 L of 1.5 M LiBr.2) Explain how to prepare 2.0 L of 1.5 M LiBr.

2.0 L2.0 L 1.5 mole1.5 mole

LL= 260 g= 260 g

86.8 g86.8 g

molemole

Dissolve 260 g LiBr in Dissolve 260 g LiBr in some water, then dilute some water, then dilute to a to a totaltotal volume of 2.0 L. volume of 2.0 L.

117Sample Problems

3) Explain how to prepare 100. mL of 3) Explain how to prepare 100. mL of 0.10 M HCl from 6.0 M HCl0.10 M HCl from 6.0 M HCl

MMDDVVDD = M = MCCVVCC

mole HCl = mole HClmole HCl = mole HCl

(0.10M)(100. mL) = (6.0 M)V(0.10M)(100. mL) = (6.0 M)VCC

VVCC = 1.7 mL = 1.7 mL

Take 1.7 mL of 6.0 M HCl Take 1.7 mL of 6.0 M HCl and dilute to a and dilute to a totaltotal volume of 100. mL.volume of 100. mL.

118Sample Problems

4) How many grams of solute is there in 50.0 mL of 0.15 M NaOH?

0.0500L0.0500L 0.15 mole0.15 mole

LL= 0.30 g= 0.30 g

40.0 g40.0 g

molemole