CHAPTER 5 - Armstrongchemistry.armstrong.edu/nivens/GeneralChemistry/Chapter5kotz.pdf · CHAPTER 5 • Chemical Reactions in Aqueous Solution 2 ... • ionize in water essentially

1

CHAPTER 5

• Chemical Reactions in Aqueous Solution

2

Aqueous Solutions: An Introduction

1. Electrolytes and Extent of Ionization

• Aqueous solutions consist of a solute dissolved in water.

• Classification of solutes:– Nonelectrolytes – solutes that do not conduct

electricity in water – do not ionize

• Examples:

• C2H5OH – ethanol

• Sugars – glucose, etc.

3

Strong and Weak Electrolytes

• Electrolytes produce ions in solution and conduct electricity

• Some are Strong electrolytes,

• Strong electrolytes ionize or dissociate completely.– Strong electrolytes approach 100% dissociation in aqueous solutions.

– NaCl(s)�Na+(aq) + Cl-(aq)

• Weak electrolytes ionize or dissociate partially, much less than 100%.– HF(l) H+(aq) + F-(aq)

4


– strong electrolytes - conduct electricity extremely well in dilute aqueous solutions

• Examples of strong electrolytes

1. HCl, HNO3, etc.• strong soluble acids

2. NaOH, KOH, etc.• strong soluble bases

3. NaCl, KBr, etc.• soluble ionic salts

• ionize in water essentially 100%

5


• Classification of solutes

– weak electrolytes - conduct electricity poorly in dilute aqueous solutions

1. CH3COOH, (COOH)2• weak acids

2. NH3, Fe(OH)3• weak bases

6


Nonelectrolytes are covalent compounds that dissolve in water, but do not conduct electricity

7


8

Strong Electrolytes

Strong Water Soluble Acids

−+≈

−+≈

+ →

+ →+

3(aq)(aq)

%100

)3(

3(aq)(aq)3

100%

)(2)3(

NOHHNO

or

NOOH OHHNO

l

ll

9

Strong Electrolytes

Strong Water Soluble Bases

-

(aq)

2

(aq)

100% OH

2(s)

-

(aq)(aq)

100% OH

(s)

OH 2Sr Sr(OH)

OHKKOH

2

2

+ →

+ →

+≈

+≈

10

Strong Electrolytes

Most Water Soluble Salts

( )−+≈

+≈

+ →

+ →

3(aq)

2

(aq)

100% OH

s23

-

(aq)(aq)

100% OH

(s)

NO 2Ca )Ca(NO

ClNaNaCl

2

2

11

Solubility

• Solubility is the maximum amount of solute that can dissolve in a given amount of solvent– Usually expressed as the

amount that dissolves in 100 g solvent

– Unsaturated Solution—contains less than the maximum amount

– Saturated –contains the maximum amount

– Increasing the temperature makes more dissolve

• Supersaturated solution

12

Solubility

• There are rules to determine if something is soluble (dissolves) or insoluble (does not dissolve)

13

Solubility

14


5. Solubility Guidelines for Compounds in Aqueous Solutions

1) All common compounds of the Group 1 metal ions and the ammonium ion are soluble. – Alkalis are soluble

–– LiLi++, NaNa++, KK++, RbRb++, CsCs++, and NHNH44++

15


3) Common nitrates, acetates, chlorates, and perchlorates are soluble.–– NONO33

--, CHCH33COOCOO--, ClOClO33--, and ClOClO44

--

4) Common chlorides. bromides and iodides are soluble.– Exceptions – AgClAgCl, HgHg22ClCl22, & PbClPbCl22

– Common fluorides are water soluble.• Exceptions – MgFMgF22, CaFCaF22, SrFSrF22, BaFBaF22, and

PbFPbF22

16


5) Sulfates are soluble.– Exceptions – PbSOPbSO44, BaSOBaSO44, & HgSOHgSO44

6) Common metal hydroxides are insolubleinsoluble.– Exceptions – LiOHLiOH, , NaOHNaOH, KOH, , KOH, RbOHRbOH &

CsOHCsOH

17


7) Carbonates, phosphates, and arsenates are insolubleinsoluble.–– COCO33

22--, POPO4433--, & AsOAsO44

33--

– Exceptions- IA metals IA metals and NHNH44++ plus

Ca to Ca to BaBa

– Moderately soluble – MgCOMgCO33

8) Sulfides are insolubleinsoluble.– Exceptions – IA metals IA metals and NHNH44

++ plus

IIA metalsIIA metals

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Reactions in Aqueous Solutions

• There are three ways to write reactions in aqueous solutions.

1. Molecular equation – Show all reactants & products in molecular or ionic

form

2. Total ionic equation

– Show the ions and molecules as they exist in solution

( ) ( ) ( ) ( )aq3saqaq3 NaNOAgClNaCl AgNO +→+

Ag+(aq) + NO3-(aq) + Na+ (aq) + Cl-(aq)

� AgCl(s) + Na+ (aq) + NO3-(aq)

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Reactions in Aqueous Solutions

3. Net ionic equation

– Shows ions that participate in reaction and removes spectator ions.

• Spectator ions do not participate in the reaction.

Ag+(aq) + Cl-(aq) � AgCl(s)

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Metathesis Reactions

•• Metathesis reactionsMetathesis reactions occur when two ionic aqueous solutions are mixed and the ions switch partners.

AX + BY → AY + BX

• Metathesis reactions remove ions from solution in 3 ways:1. form H2O - neutralization

2. form an insoluble solid

3. Form a gas

• Ion removal is the driving force of metathesis reactions.

21


1. Acid-Base (neutralization) Reactions

– Formation of the nonelectrolyte H2O

– acid + base → salt + water

22


• Molecular equation

)(2 (aq)(aq)(aq) OH + KBr KOH + HBrl

→

�Total ionic equation

( ) ( ) ( ) ( ) ( ) ( ) )(2

-

aqaq

-

aqaq

-

aqaq OH + Br+KOH+K+Br+Hl

+++ →

�Net ionic equation

( ) ( ) )(2

-

aqaq OH OH +Hl

→+

23



)(2aq)(23(aq)3(aq)2 OH 2 + )Ca(NOHNO 2 + Ca(OH)l

→

�Total ionic equation

( ) ( ) ( ) ( ) ( ) ( ) )(2

-

aq3

2

aq

-

aq3aq

-

aq

2

aq OH 2 +NO 2+ CaNO 2+ H 2+OH 2+Cal

+++ →

�Net ionic equation

( ) ( )

( ) ( ) )(2aq

-

aq

)(2aq

-

aq

OH H+OH

betteror

OH 2 H 2+OH 2

l

l

→

→

+

+

24


Precipitation reactionsPrecipitation reactions are metathesis reactions in which an insoluble compound is formed.– The solid precipitates out of the solution

much like rain or snow precipitates out of the air.

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• Precipitation Reactions


(s)3)aq(3aq)(32(aq)23 CaCO +KNO 2 COK + )Ca(NO →

�Total ionic reaction

( ) ( ) ( ) ( )

( ) ( ) ( )s3

-

aq3aq

-2

aq3aq

-

aq3

2

aq

CaCO NO 2K 2

COK 2 NO 2 Ca

++

→+++

+

++

( ) ( ) ( )s3

-2

aq3

2

aq CaCO CO Ca →++

�Net ionic reaction

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Gas forming reactionsGas forming reactions are metathesis in which a gas is formed and removed from solution

H2CO3 � H2O(l) + CO2 (g)

H2SO3 � H2O(l) + SO2 (g)

HCN(g)

NH4OH � NH3(g) + H2O(l)

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PONa 2 + CaCl 3aq)(43(aq)2 →

→aq)(32(aq) SONa + HCl 2

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2. Strong and Weak Acids

• Acids are substances that generate H+

in aqueous solutions.

• Strong acids ionize 100% in water.

( ) ( ) ( )-

aqaq

%100

g Cl H HCl + → +≈

( ) ( ) ( )

( ) ( )-

aq3aq

OH

3

-

aq3aq3

100%

2 3

NO + H HNO

or

NO + OH OH HNO

2 +

+≈

→

→+l

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• Strong Acids

• Formula Name

1. HCl hydrochloric acid

2. HBr hydrobromic acid

3. HI hydroiodic acid

4. HNO3 nitric acid

5. H2SO4 sulfuric acid

6. HClO3 chloric acid

7. HClO4 perchloric acid

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• Weak acids ionize significantly less than 100% in water.

•Common Weak Acids

•Formula Name

1.HF hydrofluoric acid

2.CH3COOH acetic acid (vinegar)

3.HCN hydrocyanic acid

4.HNO2 nitrous acid

5.H2CO3 carbonic acid (soda water)

6.H3PO4 phosphoric acid

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3. Reversible Reactions

• CH3COOH acetic acid

( ) ( )←→++

≈

aq3-

aq3

7%

23 OH + COOCH OH COOHCH

( ) ( )←→ +≈

aq

-

aq3

7%

3 H + COOCH COOHCH

32


4. Strong Bases, Insoluble Bases, and Weak Bases

• Characteristic of common inorganic bases is that they produce OH- ions in solution.

• Examples of Common Strong Bases

1. LiOH, NaOH, KOH, RbOH, CsOH, Ca(OH)2 Sr(OH)22. Ba(OH)2

3. Notice that they are all hydroxides of IA and IIA metals

33


• Similarly to strong acids, strong bases ionize 100% in water.

(aq)OH 2 + (aq)Ba Ba(OH)

(aq)OH + (aq)K KOH

-+2

2

-+

→

→

34


• Weak bases are covalent compounds that ionize slightly in water.

• Ammonia is most common weak base– NH3

( ) ( ) ( )←→ + -

(aq)aq42g3 OH + NH OH + NHl

35

The Arrhenius Theory

• Neutralization reactions are the combination of H+ (or H3O

+) with OH- to form H2O.

• Strong acids are acidic substances that ionize 100% in water.– List of aqueous strong acids:

– HCl, HBr, HI, H2SO4, HNO3, HClO4, HClO3

• Strong bases are basic substances that ionize 100% in water.– List of aqueous strong bases:

– LiOH, NaOH, KOH, RbOH, CsOH,

– Ca(OH)2, Sr(OH)2, Ba(OH)2, Na 2O, K 2O, etc.

36

The Arrhenius Theory

• Strong acid-strong base reaction The formula unit equation is:

( ) ( ) ( ) ( ) ( ) ( ) )(lOH Cl Na OH Na Cl H 2

-

aqaq

-

aqaqaqaq ++→+++ ++−+

� The total ionic equation is:

( ) ( ) ( ) )(lOH NaCl NaOH HCl 2aqaqaq +→+

�The net ionic equation is:

( ) ( ) )(lOH OH H 2

-

aqaq →++

37

Naming Some Inorganic Compounds

• Salts are formed by the reaction of the acid with a strong base.

• Acid Salt

• HNO2 NaNO2

nitrous acid sodium nitrite

• HNO3 NaNO3

nitric acid sodium nitrate

• H2SO3 Na2SO3

sulfurous acid sodium sulfite

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• There are two other possible acid and salt combinations.

• Acids that have 1 more O atom than the “ic” acid are given the prefix “per”.

• Acids that have one less O atom than the “ous” acid are given the prefix “hypo”.

39


• Illustrate this series of acids and salts with the Cl ternary acids and salts.

• Acid Na Salt

• HClO NaClOhypochlorous acid sodium hypochlorite

• HClO2 NaClO2

chlorous acid sodium chlorite

• HClO3 NaClO3

chloric acid sodium chlorate

• HClO4 NaClO4

perchloric acid sodium perchlorate

40

Acid-Base Reactions in Aqueous Solutions

• There are four acid-base reaction combinations that are possible:

1. Strong acids – strong bases

2. Weak acids – strong bases

3. Strong acids – weak bases

4. Weak acids – weak bases

• Let us look at one example of each acid-base reaction.

41


1. Strong acids - strong bases

– forming soluble salts

• This is one example of several possibilities

hydrobromic acid + calcium hydroxide

• The molecular equation is:

2 HBr(aq) + Ca(OH)2(aq) → CaBr2(aq) + 2 H2O(l)

Net:

H+ (aq) + OH

-( aq)→ H2O(l)

42


1. Strong acids-strong bases

– forming insoluble salts

• There is only one reaction of this type:

sulfuric acid + barium hydroxide


H2SO4(aq) + Ba(OH)2(aq) → BaSO4(s)+ 2H2O(l)

43

Acid-Base Reactions

2. Weak acids - strong bases


nitrous acid + sodium hydroxide


HNO2(aq) + NaOH(aq) → NaNO2(aq) + H2O(l)

•The total ionic equation is:

HNO2(aq) + Na+(aq) + OH

-(aq)→ Na+(aq) + NO2

-(aq)+ H2O(l)

•The net ionic equation is:

HNO2(aq) + OH-(aq) → NO2

-(aq) + H2O(l)

44

Acid-Base Reactions3. Strong acids - weak bases- forming soluble salts

nitric acid + ammonia


HNO3(aq) + NH3(aq) → NH4NO3(aq)

• The net equation is:

H+(aq) + NH3(aq) → NH4

+(aq)

45

Acid-Base Reactions4. Weak acids - weak bases


acetic acid + ammonia


CH3COOH(aq) + NH3(aq) → NH4CH3COO(aq)

• The total ionic equation is:

CH3COOH(aq) + NH3(aq) → NH4+(aq) + CH3COO

-(aq)

• The net ionic equation is:

CH3COOH(aq) + NH3(aq) → NH4+(aq) + CH3COO

-(aq)

46

Acidic Salts and Basic Salts

• Polyprotic acids with less than the stoichiometric amount of base.

• 1:1 ratio.

H2SO4(aq) + NaOH(aq) → NaHSO4(aq) + H2O(l)

The acidic salt sodium hydrogen sulfate is formed.

• 1:2 ratio.

H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l)

The normal salt sodium sulfate is formed.

47

Soultions• A solution is a mixture of two or more substances

dissolved in another.

– Solute is the substance present in the smaller amount.

– Solvent is the substance present in the larger amount.

– In aqueous solutions, the solvent is water.

48

Types of Solutions

49

Concentrations of Solutions

• Common unit of concentration:

L

moles

solution of liters ofnumber

solute of moles ofnumber molarity

=

=

M

50


51


• Example 9-5: Calculate the molarity of a solution that contains 12.5 g of sulfuric acid in 1.75 L of solution.

42

424242

SOH g 98.1

SOH mol 1

nsol' L 75.1

SOH g 12.5

nsol' L

SOH mol ?×=

52


• Example 9-6: Calculate the molarity of a solution that contains 12.5 g of sulfuric acid in 1.75 L of solution.

42

42

42

424242

SOH 0728.0

L

SOH mol 0728.0

SOH g 98.1

SOH mol 1

nsol' L 75.1

SOH g 12.5

nsol' L

SOH mol ?

M=

=

×=

53


• Example 9-7: Determine the mass of calcium nitrate required to prepare 3.50 L of 0.800 M Ca(NO3)2 .

You do it!You do it!

54


• Example 9-7: Determine the mass of calcium nitrate required to prepare 3.50 L of 0.800 M Ca(NO3)2 .

? g Ca(NO L 0.800 mol Ca(NO

L

164 g Ca(NO

mol Ca(NO g Ca(NO

33

3

3

3

) .)

)

))

22

2

2

2

3 50

1459

= × ×

=

55

Calculations Involving Molarity

• Example: If 100.0 mL of 1.00 M NaOHand 100.0 mL of 0.500 M H2SO4

solutions are mixed, what will the concentration of the resulting solution be?

56


• Example: If 130.0 mL of 1.00 M KOH and 100.0 mL of 0.500 M H2SO4

solutions are mixed, what will be the concentration of KOH and K2SO4 in the resulting solution?

57


• Example: What volume of 0.750 M NaOH solution would be required to completely neutralize 100 mL of 0.250 M H3PO4?

58

Dilution of Solutions

• To dilute a solution, add solvent to a concentrated solution.

• The number of moles of solute in the two solutions remains constant.

• The relationship M1V1 = M2V2 is appropriate for dilutions, but not for chemical reactions.

59


• Example: If 10.0 mL of 12.0 M HCl is added to enough water to give 100. mLof solution, what is the concentration of the solution?

M

MM

MM

MM

20.1

mL 100.0

mL 0.100.12

mL 100.0mL 0.10 0.12

VV

2

2

2211

=

×=

×=×

=

60


• Example: What volume of 18.0 M sulfuric acid is required to make 2.50 L of a 2.40 M sulfuric acid solution?

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• Example: What volume of 18.0 M sulfuric acid is required to make 2.50 L of a 2.40 M sulfuric acid solution?

mL 333or L 0.333

18.0

2.40 L 2.50V

V V

V V

1

1

221

2211

=

×=

×=

=

M

M

M

M

MM

62

Solutions in Chemical Reactions

• Combine the concepts of molarity and stoichiometry to determine the amounts of reactants and products involved in reactions in solution.

63

Using Solutions in Chemical Reactions

• Example: What volume of 0.500 M BaCl2 is required to completely react with 4.32 g of Na2SO4?

NaCl 2 + BaSO BaCl + SONa 4242 →

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• Example: What volume of 0.500 M BaCl2 is required to completely react with 4.32 g of Na2SO4?

L 0.0608 BaCl mol 0.500

BaCl L 1

SONa mol 1

BaCl mol 1

SONa g 142

SONa mol 1 SOgNa 4.32 BaCl L ?

NaCl 2 + BaSO BaCl + SONa

2

2

42

2

42

42422

4242

=×

××=

→

65


• Example: (a)What volume of 0.200 M NaOH will react with 50.0 mL 0f 0.200 M aluminum nitrate, Al(NO3)3?

( ) ( )

You do it!

3333 NaNO 3OHAlNaOH 3NOAl +→+

66


• Example: (a)What volume of 0.200 M NaOH will react with 50.0 mL 0f 0.200 M aluminum nitrate?

( )

nsol' NaOH mL 150or L 0.150 NaOH mol 0.200

NaOH L 1

)Al(NO mol 1

NaOH mol 3

nsol' )Al(NO L 1

n sol' )Al(NO mol 0.200

mL 1000

L 1nsol' )Al(NO mL 50.0 = NaOH mL ?

NaNO 3Al(OH)NaOH 3NOAl

3333

33

33

3333

=

××

×

+→+

67


• (b)What mass of Al(OH)3 precipitates in (a)?

68


• (b) What mass of Al(OH)3 precipitates in (a)?

3

3

3

33

3

33

33

333

Al(OH) g 780.0

Al(OH) mol 1

Al(OH) g 0.78

)Al(NO mol 1

Al(OH) mol 1

nsol' )Al(NO L1

)Al(NO mol 0.200

mL 1000

L1nsol' )Al(NO mL 50.0 Al(OH) g ?

=

××

×=

69


• Titrations are a method of determining the concentration of an unknown solutions from the known concentration of a solution and solution reaction stoichiometry.

70

Titrations

• Acid-base Titration Terminology1. **Titration – A method of determining the

concentration of one solution by reacting it with a solution of known concentration.

2. **Primary standard – A chemical compound which can be used to accurately determine the concentration of another solution. Examples include KHP and sodium carbonate.

71

Titrations

• Acid-base Titration Terminology1. **Standard solution – A solution whose

concentration has been determined using a primary standard.

2. **Standardization – The process in which the concentration of a solution is determined by accurately measuring the volume of the solution required to react with a known amount of a primary standard.

72

Titrations

**Indicator – A substance that exists in different forms with different colors depending on the concentration of the H+ in solution. Examples are phenolphthalein and bromothymol blue.

5. **Equivalence point – The point at which stoichiometrically equivalent amounts of the acid and base have reacted.

6.**End point – The point at which the indicator changes color and the titration is stopped.

73

The Mole Method and Molarity

• Potassium hydrogen phthalate is a very good primary standard.– It is often given the acronym, KHP.

– KHP has a molar mass of 204.2 g/mol.

• A very common mistake is for students to see the acronym KHP and think that this compound is made of potassium, hydrogen, and phosphorous.

CH

CH

CH

C

C

CH C

OH

O

C

OH

O

CH

CH

CH

C

C

CH C

O

O

C

OH

O

+ KOH

-K+

+ H2O

KHP

acidic H

74


• Example: Calculate the molarity of a NaOHsolution if 27.3 mL of it reacts with 0.4084 g of KHP.

NaOH + KHP NaKP + H O2→

NaOH 0.0733NaOH L 0.0273

NaOH mol 0.00200 = NaOH ?

NaOH mol 0.00200KHP mol 1

NaOH mol 1

KHP g 204.2

KHP mol 1KHP g 0.4084 = NaOH mol ?

MM =

=

××

NaOH mol 0.00200KHP mol 1

NaOH mol 1

KHP g 204.2

KHP mol 1KHP g 0.4084 = NaOH mol ?

=

××

75


• Example: What is the molarity of a KOH solution if 38.7 mL of the KOH solution is required to react with 43.2 mL of 0.223 M HCl?

OH + KCl HCl + KOH 2→

76


• Example: What is the molarity of a KOH solution if 38.7 mL of the KOH solution is required to react with 43.2 mL of 0.223 M HCl?

KOH 249.0KOH mL 38.7

KOH mmol 9.63

KOH mmol 63.9HCl mmol 1

KOH mmol 1HCl mmol 9.63

HCl mmol 9.63 = HCl 0.223 mL 43.2

OH + KCl HCl + KOH 2

M

M

=

=×

×

→

77


• Example: What is the molarity of a barium hydroxide solution if 44.1 mL of 0.103 M HClis required to react with 38.3 mL of the Ba(OH)2 solution?

78


• Example: What is the molarity of a barium hydroxide solution if 44.1 mL of 0.103 M HClis required to react with 38.3 mL of the Ba(OH)2 solution?

2

2

2

2

2

222

Ba(OH) 0593.0Ba(OH) L 0383.0

Ba(OH) mol 00227.0

Ba(OH) mol 0.00227

HCl mol 2

Ba(OH) mol 1HCl mol 00454.0

HCl mol 0.00454 = HCl) HCl)(0.103 L (0.0441

OH 2 + BaCl HCl 2 + Ba(OH)

M

M

=

=

×

→

79


• Example: Calculate the molarity of a sulfuric acid solution if 23.2 mL of it reacts with 0.212 g of Na2CO3.

Na CO + H SO Na SO + CO + H O2 3 2 4 2 4 2 2→

80


• Example: An impure sample of potassium hydrogen phthalate, KHP, had a mass of 0.884 g. It was dissolved in water and titrated with 31.5 mL of 0.100 M NaOH solution. Calculate the percent purity of the KHP sample.

NaOH + KHP → NaKP + H2O

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CHAPTER 5 - Armstrongchemistry.armstrong.edu/nivens/GeneralChemistry/Chapter5kotz.pdf · CHAPTER 5 • Chemical Reactions in Aqueous Solution 2 ... • ionize in water essentially