Aqueous

Reactions

TITRATIONS :Experiment 3

Measuring devices

1

Aqueous

Reactions

THE STOICHIOMETRY OF

REACTIONS IN AQUEOUS

SOLUTION: TITRATIONS

2

Aqueous

Reactions

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

• From the balanced equation the reacting

ratio of HCl to NaOH is 1 : 1

• At the endpoint:

• no. of moles of HCl = no. of moles of NaOH

• i.e MHCl x VHCl = MNaOH x VNaOH

• MNaOH = MHCl x VHCl / VNaOH

MHCl= given value VHCl = 0.025 dm3

VNaOH ( average vol. from titration)

MNaOH = ----A--- mol dm-3

Conc. of NaOH in g dm-3 = A mol dm-3 × molar mass of

NaOH( in g mol-1)3

Aqueous

Reactions

Aqueous Reactions and

Solution Stoichiometry

Chemistry, The Central Science, 2nd edition

Brown; LeMay; Bursten; Murphy; Langford

and Sagatys

4

Aqueous

Reactions

Solutions:

• Homogeneous

mixtures of two or

more pure

substances.

• The solvent is

present in greatest

abundance.

• All other substances

are solutes.

5

Aqueous

Reactions

Dissociation• Ionic compounds are formed between

metals and non-metals.

• Ionic compound separates into ions

e.g. NaCl Na+ + Cl-

6

Aqueous

Reactions

Electrolytes

• Substances that

dissociate into ions

when dissolved in

water are called

electrolytes.

• A nonelectrolyte may

dissolve in water, but

it does not dissociate

into ions when it does

so.7

Aqueous

Reactions

Electrolytes

• A strong electrolyte

dissociates completely

when dissolved in

water.

• A weak electrolyte

only dissociates

partially when

dissolved in water.

8

Aqueous

Reactions

• Strong electrolytes completely ionise in

solution

– Strong acids

– Strong bases

– Soluble salts

Non-electrolyte weak electrolyte strong

electrolyte

9

Aqueous

Reactions

Strong Electrolytes Are…

• Strong acids

10

Aqueous

Reactions

Strong Electrolytes Are…

• Strong bases

11

Aqueous

Reactions

Strong Electrolytes Are…

• Soluble ionic salts

12

Aqueous

Reactions

Acids:

• Substances that

increase the

concentration of H+

when dissolved in

water (Arrhenius).

• Proton donors

(Brønsted–Lowry).

13

Aqueous

Reactions

Acids

There are only seven

strong acids:

• Hydrochloric (HCl)

• Hydrobromic (HBr)

• Hydroiodic (HI)

• Nitric (HNO3)

• Sulfuric (H2SO4)

• Chloric (HClO3)

• Perchloric (HClO4)

14

Aqueous

Reactions

Bases:

• Substances that

increase the

concentration of

OH− when dissolved

in water (Arrhenius).

• Proton acceptors

(Brønsted–Lowry).

15

Aqueous

Reactions

Bases

The strong bases

are the soluble salts

of hydroxide ion:

• Alkali metals

• Calcium

• Strontium

• Barium

16

Aqueous

Reactions

Rules for Recognizing Acids

(1)Compounds that do not contain C and have

an H written in front of their formulas. e.g.

H2S. (*****Refer to page 43 in textbook)

(2) Compounds that COOH in their formulas

e.g. CH3COOH is acetic acid.

(3) A positive ion that contains a N atom and

has a H bound to the N is frequently an acid.

NH4Cl is actually two ions, NH4+ and Cl–. The

ammonium ion is an acid.L.Pillay, 2009

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Aqueous

Reactions

(1) An ionic compound that contains a

hydroxide (OH–) ion. e.g. NaOH, Mg(OH)2

(2) A neutral compound that contains a N atom.

An H+ can bond to the lone pair of electrons,

forming , so the lone pair on the N allows

the molecule to accept an H+ (which is the

definition of base). **e.g. NH3

(3) An ionic compound that does not contain

the following: Cl–, Br–, I–, NO3–, SO4

2–,ClO4–,

or H+. **e.g. Na2CO3

Rules for Recognizing a Base

L.Pillay, 2009

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Aqueous

Reactions

Solubility Rules

1. All common compounds of Group I, Group 2

and ammonium ions are soluble.

2. All nitrates, acetates, and perchlorates are

soluble.

3. Salts of Ag, Hg(I), and Pb are insoluble. (Pb

halides are soluble in hot water.)

4. Chlorides, bromides and iodides are soluble

5. All sulfates are soluble, except those of

barium, strontium.

6. Except for rule 1, carbonates, hydroxides,

sulphides, oxides, silicates, and phosphates

are insoluble. 19

Aqueous

Reactions

Solubility - Example

Ag+ Cu2+ Mg2+

S2-

Cl-

OH-

ppt

ppt

ppt

ppt

sol

ppt

sol

sol

ppt

Salts of Ag+ are insoluble! Rule 3

Sulphides

insoluble

Rule 6Chlorides

soluble *

Rule 4

OH insoluble *

Rule 6

Group II

L.Pillay, 2009

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Aqueous

Reactions

Solubility Rules

Will a precipitate form when aqueous

solutions of Ca(NO3)2 and NaCl are

mixed?

– Ca(NO3)2 and NaCl are both soluble

– Combinations could be CaCl2 and NaNO3

– CaCl2 would be soluble (Rule **4--)

– NaNO3 would be soluble (Rule **2--)

L.Pillay, 2009

21

Aqueous

Reactions

SAMPLE EXERCISE 1 Using Solubility

Rules

• Classify the following ionic compounds

as soluble or insoluble in water:

• (a) sodium carbonate (Na2CO3),

• (b) lead sulfate (PbSO4).

22

Aqueous

Reactions

• (a)Most carbonates are insoluble, but

carbonates of the alkali metal cations

(such as sodium ion) are an exception

to this rule and are soluble. Thus,

Na2CO3 is soluble in water.

• (b) Although most sulfates are water

soluble, the sulfate of Pb2+ is an

exception. Thus, PbSO4 is insoluble in

water.

23

Aqueous

Reactions

SAMPLE EXERCISE 2

• a) Predict the identity of the precipitate

that forms when solutions of BaCl2 and

K2SO4 are mixed.

• (b) Write the balanced chemical

equation for the reaction.

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Aqueous

Reactions

(a) Most compounds of SO42– are soluble

but those of Ba2+ are not.

Thus, BaSO4 is insoluble and will

precipitate from solution.

KCl, on the other hand, is soluble.

25

Aqueous

Reactions

• b) From part (a) we know the chemical

formulas of the products, BaSO4 and

KCl.

• The balanced equation with phase

labels shown is

26

Aqueous

Reactions

Types of reactions

• Acid – base reactions

(neutralisation reaction)

• Precipitation reactions

• Redox reactions

(Oxidation –reduction reactions)

• Disproportionation reactions

L.Pillay, 2009

27

Aqueous

Reactions

Solution Chemistry

• It is helpful to pay attention to exactly

what species are present in a reaction

mixture (i.e., solid, liquid, gas, aqueous

solution).

• If we are to understand reactivity, we

must be aware of just what is changing

during the course of a reaction.

28

Aqueous

Reactions

Writing Net Ionic Equations

1.Write a balanced molecular equation.

2.Dissociate all strong electrolytes.

3.Cross out anything that remains

unchanged from the left side to the right

side of the equation, i.e. identify and

cancel spectator ions.

4.Write the net ionic equation with the

species that remain.29

Aqueous

Reactions

Molecular Equation

The molecular equation lists the reactants

and products in molecular form.

Unbalanced equation

Pb(NO3)2 (aq) + KI (aq) PbI2 (s) + 2KNO3 (aq)

Balanced equation

Pb(NO3)2 (aq) + 2KI (aq) PbI2 (s) + 2KNO3 (aq)

30

Aqueous

Reactions

Ionic Equation

• In the ionic equation all strong electrolytes

(strong acids, strong bases, and soluble

ionic salts) are dissociated into their ions.

• This more accurately reflects the species

that are found in the reaction mixture.

Balanced Molecular equation:

Pb(NO3)2 (aq) + 2KI (aq) PbI2 (s) + 2KNO3 (aq)

Balanced Ionic equation:

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

-(aq)

PbI2 (s) + 2K+ (aq) + 2NO3-(aq) 31

Aqueous

Reactions

Net Ionic Equation

• To form the net ionic equation, cross out

anything that does not change from the

left side of the equation to the right: i.e.

cross out the spectator ions

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

-(aq)

PbI2 (s) + 2K+ (aq) + 2NO3

-(aq)

32

Aqueous

Reactions

Net Ionic Equation

• The only things left in the equation are

those things that change (i.e. react)

during the course of the reaction.

Pb2+(aq) + 2I-(aq) PbI2 (s)

33

Aqueous

Reactions

Net ionic equations

• Total molecular equation:

2AgNO3(aq) + Cu(s) 2Ag(s) + Cu(NO3)2 (aq)

• Total ionic equation:

2Ag+(aq) + 2NO3

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

2+(aq) +

2NO3-(aq)

• Net Ionic equation: (remove spectator ions)

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

2+

L.Pillay, 2009

34

Aqueous

Reactions

Net ionic equations -

exampleCaCl2 + Na2CO3 ?

1.Determine products

CaCl2 + Na2CO3 CaCO3 + NaCl

2.Balance equations and predict solubility

CaCl2(aq) + Na2CO3(aq) CaCO3(s) + 2NaCl(aq)

L.Pillay, 2009

35

Aqueous

Reactions

3. Write total ionic equations

Ca2+(aq) + 2Cl-(aq) + 2Na+(aq) + CO32-(aq)

CaCO3(s) + 2Na+(aq) + 2Cl-(aq)

4. Write net ionic equation

Ca2+ (aq) + CO32-(aq) CaCO3(s)

36

Aqueous

Reactions

Acid-Base (neutralisation)

reactions

• acid + base → salt + water

• Driving force is the reaction of H+ and OH- to

produce H2O

HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)L.Pillay, 2009

37

Aqueous

Reactions

Neutralization Reactions

Consider the reaction between a strong acid, HCl and a strong

base, NaOH:

Total molecular equation:

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

Total ionic equation:

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

Na+ (aq) + Cl- (aq) + H2O (l)

38

Aqueous

Reactions

Neutralization Reactions

Cancel common reactants and products

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

Na+ (aq) + Cl- (aq) + H2O (l)

Net Ionic equation:

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

39

Aqueous

Reactions

Gas-Forming ReactionsThe reaction below does not give the expected product,

H2CO3:

NaHCO3(aq) + HCl(aq) NaCl(aq) + H2CO3(aq)

Instead the expected product decomposes to give a

gaseous product (CO2):

**H2CO3 (aq) CO2 (g) + H2O (l)

Overall:

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

Write the total ionic equation here!!

And the net ionic equation is:

HCO3- (aq) + H+ (aq) CO2 (g) + H2O (l)

40

Aqueous

Reactions

Gas-Forming Reactions

• This reaction gives the predicted product, but you had better carry it out in the fumehood, or you will be very unpopular!

• Just as in the previous examples, a gas is formed as a product of this reaction:

Molecular equation is:

**Na2S (aq) + 2HCl (aq) 2NaCl (aq) + H2S (g)

Write the total ionic equation here!!

The net ionic equation is:

S2- (aq) + 2H+ (aq) H2S (g)

41

Aqueous

Reactions

Displacement reactions

• Activity Series:

• Any free metal higher on list will displace

another metal lower on series.

e.g. Cu(s) + FeSO4(aq)

Zn(s) + 2HCl(aq)

Li

K

Ca

Na

Mg

Al

Mn

Zn

Cr

Fe

Cd

Co

Ni

Sn

Pb

H

Cu42

Aqueous

Reactions

Precipitation Reactions

When one mixes ions

that form compounds

that are insoluble (as

could be predicted by

the solubility

guidelines), a

precipitate is formed.

43

Aqueous

Reactions

Precipitation reactions

• Step #1: Determine the possible products using

the general double displacement equation.

AB + CD AD + CB

• Step #2: Predict whether either of the possible

products is water insoluble.

• Step #3: Write the complete balanced equation.

• Step #4: Write the net ionic equation.

L.Pillay, 2009

44

Aqueous

Reactions

AgNO3(aq) + Na2S(aq) ?

1. The possible products from the mixture of

AgNO3(aq) and Na2S(aq) are Ag2S and NaNO3.

(Remember to consider charge for the possible

products.)

AgNO3(aq) + Na2S(aq) Ag2S + NaNO3

2. According to solubility rules, most sulfides are

insoluble, therefore, Ag2S would be insoluble.

Compounds containing Na+ and NO3- are soluble,

NaNO3 is soluble.

AgNO3(aq) + Na2S(aq) Ag2S(s) + NaNO3(aq)L.Pillay, 2009

45

Aqueous

Reactions

3. Balance the equation:

2AgNO3(aq) + Na2S(aq) Ag2S(s) + 2NaNO3(aq)

4.Write the complete ionic equation, describing the aqueous ionic compounds, AgNO3(aq), Na2S(aq) and NaNO3(aq), as ions.

Describe the solid with a complete formula.

2Ag+(aq) + 2NO3-(aq) + 2Na+(aq) + S2-(aq)

Ag2S(s) + 2Na+(aq) + 2NO3-(aq)

L.Pillay, 2009

46

Aqueous

Reactions

• The nitrate and sodium ions have the same

form on each side of the equation, so they are

eliminated as spectator ions.

2Ag+(aq) + S2-(aq) Ag2S(s)

Ex. Pb(NO3)2 (aq) + K2CrO4(aq) ?

L.Pillay, 2009

47

Aqueous

Reactions

Metathesis (Exchange)

Reactions

• Metathesis comes from a Greek word

that means “to transpose”

• It appears the ions in the reactant

compounds exchange, or transpose,

ions

AgNO3 (aq) + KCl (aq) AgCl (s) + KNO3 (aq)

48

Aqueous

Reactions

Oxidation-Reduction Reactions

• An oxidation occurs

when an atom or ion

loses electrons.

• A reduction occurs

when an atom or ion

gains electrons.

49

Aqueous

Reactions

Oxidation Numbers

To determine if an oxidation-reduction

reaction has occurred, we assign an

oxidation number to each element in a

neutral compound or charged entity.

50

Aqueous

Reactions

Oxidation Numbers

• Elements in their elemental form have

an oxidation number of 0.

• The oxidation number of a monatomic

ion is the same as its charge.

51

Aqueous

Reactions

Oxidation Numbers

In order to keep

track of what loses

electrons and what

gains them, we

assign oxidation

numbers.

52

Aqueous

Reactions

Oxidation and Reduction

• Oxidation ( loss of electrons)

• A species is oxidized when it loses electrons.

– Here, zinc loses two electrons to go from neutral

zinc metal to the Zn2+ ion.

53

Aqueous

Reactions

Oxidation and Reduction

• Reduction (gain in electrons)

• A species is reduced when it gains electrons.

– Here, each of the H+ gains an electron and they

combine to form H2.

54

Aqueous

Reactions

Oxidation and Reduction

• What is reduced is the oxidizing agent.

– H+ oxidizes Zn by taking electrons from it.

• What is oxidized is the reducing agent.

– Zn reduces H+ by giving it electrons.

55

Aqueous

Reactions

Assigning Oxidation Numbers

1. Elements in their elemental form have

an oxidation number of 0.

2. The oxidation number of a monatomic

ion is the same as its charge.

56

Aqueous

Reactions

Assigning Oxidation Numbers

3. Nonmetals tend to have negative oxidation numbers, although some are positive in certain compounds or ions.

– Oxygen has an oxidation number of −2, except in the peroxide ion in which it has an oxidation number of −1.

**(Refer to page 39 in textbook)

– Hydrogen is −1 when bonded to a metal, +1 when bonded to a nonmetal,e.g. NaH*

57

Aqueous

Reactions

Assigning Oxidation Numbers

– Fluorine always has an oxidation number of −1.

– The other halogens have an oxidation number of −1 when they are negative; they can have positive oxidation numbers, however, most notably in oxyanions.****e.g. ClO4

-

58

Aqueous

Reactions

Assigning Oxidation Numbers

4. The sum of the oxidation numbers in a

neutral compound is 0.

5. The sum of the oxidation numbers in a

polyatomic ion is the charge on the

ion.

59

Aqueous

Reactions

Group1

Group18

H+1-1

Group 2Group

13Group

14Group

15Group

16Group

17

He

Li+1

Be+2

B+3

C+4, +2

-1-4

NAll from +5 to -3

O-1-2

F-1

Ne

Na+1

Mg+2

Group3

Group4

Group5

Group6

Group7

Group8

Group9

Group10

Group11

Group12

Al+3

Si+4-4

P+5+3-3

S+6+4

+2, -2

Cl+7, +5

+3+1, -1

Ar

K+1

Ca+2

Sc+3

Ti+4+3+2

V+5+4

+3, +2

Cr+6+3+2

Mn+7, +6

+4+3, +2

Fe+3+2

Co+3+2

Ni+2

Cu+2+1

Zn+2

Ga+3

Ge+4+2-4

As+5+3-3

Se+6+4-2

Br+7, +5

+3+1, -1

Kr

Rb+1

Sr+2

Y+3

Zr+4+3

Nb+5+4+2

Mo+6+5

+4, +3

Tc+7+5+4

Ru+8, +5

+4+3

Rh+4+3

Pd+4+2

Ag+1

Cd+2

In+3

Sn+4+2-4

Sb+5+3-3

Te+6+4-2

I+7, +5

+3+1, -1

Xe

Cs+1

Ba+2

Lu+3

Hf+4+3

Ta+5+4+3

W+6+5+4

Re+7+5+4

Os+8, +6

+4+3, +2

Ir+4+2

Pt+4+2

Au+3+1

Hg+2+1

Tl+3+1

Pb+4+2

Bi+5+3

+6+4-2

At+7, +5

+3+1, -1

Rn

Fr+1

Ra+2

Lr+3

The Common Oxidation Numbers of the Elements

60

Aqueous

Reactions

SAMPLE EXERCISE 4.8 Determining

Oxidation Numbers

• Determine the oxidation number of

sulfur in each of the following:

• (a) H2S,

• (b) S8 ,

• (c) SCl2,

• (d) Na2SO3,

• (e) SO42–.

61

Aqueous

Reactions

Oxidation Numbers

• Nonmetals tend to have negative

oxidation numbers, although some are

positive in certain compounds or ions.

Oxygen has an oxidation number of −2,

except in the peroxide ion in which it has

an oxidation number of −1.

Hydrogen is −1 when bonded to a metal,

+1 when bonded to a nonmetal.

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Aqueous

Reactions

• a) When bonded to a nonmetal,

hydrogen has an oxidation number of

+1 . Because the H2S molecule is

neutral, the sum of the oxidation

numbers must equal zero.

• Letting x equal the oxidation number of

S, we have 2(+1) + x = 0.

• Thus, S has an oxidation number of –2.

63

Aqueous

Reactions

Answers

• (b) 0

• (c) +2

• (d) +4

• (e) +6

64

Aqueous

Reactions65

Aqueous

Reactions

Oxidation-reduction (redox)

reactionsBreathalyser Test

3CH3CH2OH + 2K2Cr2O7 + 8H2SO4 →

3CH3COOH + 2Cr2(SO4)3 + K2SO4 + 11 H2O

• Ethanol → acetic acid (oxidised)

• Dichromate → chromic ion

Orange → Green

66

Aqueous

Reactions

Balancing Oxidation-Reduction

Equations

Perhaps the easiest way to balance the

equation of an oxidation-reduction

reaction is via the half-reaction method.

67

Aqueous

Reactions

Balancing Oxidation-Reduction

Equations

This involves treating (on paper only) the

oxidation and reduction as two separate

processes, balancing these half reactions,

and then combining them to attain the

balanced equation for the overall reaction.

68

Aqueous

Reactions

Half-Reaction Method- acid

medium

1. Assign oxidation numbers to

determine what is oxidized and what is

reduced.

2. Write the oxidation and reduction half-

reactions.

69

Aqueous

Reactions

Half-Reaction Method

3. Balance each half-reaction.

a. Balance elements other than H and O.

b. Balance O by adding H2O.

c. Balance H by adding H+.

d. Balance charge by adding electrons.

4. Multiply the half-reactions by integers so that the electrons gained and lost are the same.

70

Aqueous

Reactions

Half-Reaction Method

5. Add the half-reactions. **Cancel what is common to both sides.

6. Make sure the equation is balanced according to mass.

7. Make sure the equation is balanced according to charge.

71

Aqueous

Reactions

Half-Reaction Method

Consider the reaction between MnO4− and C2O4

2− :

MnO4−(aq) + C2O4

2−(aq) Mn2+(aq) + CO2(aq)

72

Aqueous

Reactions

Half-Reaction Method

First, we assign oxidation numbers.

MnO4− + C2O4

2- Mn2+ + CO2

+7 +3 +4+2

Since the manganese goes from +7 to +2, it is reduced

( decrease in oxidation number).Since the carbon goes from +3 to +4, it is oxidized

(increase in oxidation number).73

Aqueous

Reactions

Oxidation Half-Reaction

C2O42− CO2

To balance the carbon, we add a

coefficient of 2:

C2O42− 2 CO2

74

Aqueous

Reactions

Oxidation Half-Reaction

C2O42− 2 CO2

The oxygen is now balanced as well.

To balance the charge, we must add 2

electrons to the right side.

C2O42− 2 CO2 + 2 e−

75

Aqueous

Reactions

Reduction Half-Reaction

MnO4− Mn2+

The manganese is balanced; to balance

the oxygen, we must add 4 waters to

the right side.

MnO4− Mn2+ + 4 H2O

76

Aqueous

Reactions

Reduction Half-Reaction

MnO4− Mn2+ + 4 H2O

To balance the hydrogen, we add 8 H+

to the left side.

8 H+ + MnO4− Mn2+ + 4 H2O

77

Aqueous

Reactions

Reduction Half-Reaction

8 H+ + MnO4− Mn2+ + 4 H2O

To balance the charge, we add 5 e− to

the left side.

5 e− + 8 H+ + MnO4− Mn2+ + 4 H2O

78

Aqueous

Reactions

Combining the Half-Reactions

Now we evaluate the two half-reactions

together:

C2O42− 2 CO2 + 2 e−

5 e− + 8 H+ + MnO4− Mn2+ + 4 H2O

To attain the same number of electrons

on each side, we will multiply the first

reaction by 5 and the second by 2.79

Aqueous

Reactions

Combining the Half-Reactions

5 C2O42− 10 CO2 + 10 e−

10 e− + 16 H+ + 2 MnO4− 2 Mn2+ + 8 H2O

When we add these together, we get:

10 e− + 16 H+ + 2 MnO4− + 5 C2O4

2−

2 Mn2+ + 8 H2O + 10 CO2 +10 e−

80

Aqueous

Reactions

Combining the Half-Reactions

10 e− + 16 H+ + 2 MnO4− + 5 C2O4

2−

2 Mn2+ + 8 H2O + 10 CO2 +10 e−

The only thing that appears on both sides are the electrons. Subtracting them, we are left with:

16 H+ + 2 MnO4− + 5 C2O4

2−

2 Mn2+ + 8 H2O + 10 CO2

81

Aqueous

Reactions

Balancing in Basic Solution

• If a reaction occurs in basic solution, one

can balance it as if it occurred in acid.

• Once the equation is balanced, add OH−

to each side to “neutralize” the H+ in the

equation and create water in its place.

• If this produces water on both sides, you

might have to subtract water from each

side.82

Aqueous

Reactions

eg. MnO4- + H2C2O4 +... → Mn2+ + CO2 +...

1.Oxidation Numbers

2.Half-reactions

MnO4- → Mn2+

H2C2O4 → CO2

3. Balance Equations

Balance all elements except hydrogen and

oxygen by proper multiplication of

compounds or ions.

MnO4- → Mn2+

H2C2O4 → 2CO2

L.Pillay, 2009

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Aqueous

Reactions

Balance oxygen by adding water, H2O, as

necessary.

MnO4- → Mn2+ + 4H2O

H2C2O4 → 2CO2

Balance hydrogen by adding hydrogen ions,

H+, as necessary.

8H+ + MnO4- → Mn2+ + 4H2O

H2C2O4 → 2CO2 + 2H+

L.Pillay, 2009

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Aqueous

Reactions

4 & 5. Balance charge by adding electrons &

combine

(8H+ + MnO4- + 5 e-) x 2 → Mn2+ + 4H2O

(H2C2O4 → 2CO2 + 2H+ + 2e-) x 5

Multiplying and combining the two equations

16H+ + 2MnO4- + 5H2C2O4 + 10e- → 2Mn2+ +

8H2O + 10CO2 + 10H+ + 10e-

After cancellation

6H+ + 2MnO4- + 5H2C2O4 → 2Mn2+ + 8H2O +

10CO285

Aqueous

Reactions

6H+ + 2MnO4- + 5H2C2O4 → 2Mn2+ + 8H2O +

10CO2

6. Add number of OH- ion equal to number of H+

ions on both sides of overall reaction

6H+ + 2MnO4- + 5H2C2O4 + 6OH- →

2Mn2+ + 8H2O + 10CO2 + 6OH-

7 & 8. Combine hydrogen ions (protons) and

hydroxide ions to form water when they appear on

the same side of the equation

2MnO4- + 5H2C2O4 --> 2Mn2+ + 2H2O + 10CO2 + 6OH-

L.Pillay, 2009

86

Aqueous

Reactions

THE STOICHIOMETRY OF

REACTIONS IN AQUEOUS

SOLUTION: TITRATIONS

87

Aqueous

Reactions

Concentration of

Solutions

The concentration of a solution is the amount

of solute present in a given quantity of

solvent or solution.

Concentration can be expressed in various

ways; most commonly used is molarity.

Other ways of expressing units of

concentration are: parts per million and mass

(weight) percent. 88

Aqueous

Reactions

Molarity• Two solutions can contain the same compounds but

be quite different because the proportions of those

compounds are different eg. 3M and 6M HCl.

• Molarity is one way to measure the concentration of a

solution.

• Molarity (M) or molar concentration is the

number of moles of solute in 1 litre (L) of

solution.moles of solute (n)

volume of solution in liters (V)Molarity (M) =

89

Aqueous

Reactions90

Aqueous

Reactions

• Percent Concentration

• Can be expressed in several ways. Three

common methods are:

1. mass percent (%m/m) =mass solute(g) x 100%

mass solution(g)

2. volume percent (%v/v) =volume solute x 100%

volume solution

3. mass/volume percent (%m/v)= mass solute, g x 100

volume solution, mL

91

Aqueous

Reactions

Mixing a Solution

• To create a solution of a

known molarity, one

weighs out a known mass

(and, therefore, number of

moles) of the solute.

• The solute is added to a

volumetric flask, and

solvent is added to the line

on the neck of the flask.

92

Aqueous

Reactions

Example

How many grams of potassium dichromate(K2Cr2O7) are required to prepare a 250 mLsolution whose concentration is 2.16 M?

n = M x v= 2.16 mol /1L x 0.25 L

= 0.540 mol

mass = n x molar mass= 0.540 mol x 294.2 g /1mol

= 159 g

93

Aqueous

Reactions

Dilution

• One can also dilute a more concentrated solution by– Using a pipet to deliver a volume of the solution to

a new volumetric flask, and

– Adding solvent to the line on the neck of the new flask.

94

Aqueous

Reactions95

Aqueous

Reactions

Dilution

The molarity of the new solution can be determined from the equation

Mconc Vconc = Mdil Vdil, (for dilution only!)

where Mconc and Mdil are the molarity of the concentrated and dilute solutions, respectively, and Vconc and Vdil are the volumes of the two solutions.

96

Aqueous

Reactions97

Aqueous

Reactions98

Aqueous

Reactions

What is the new concentration of the

solution?

MconcVconc = MdilVdil

(0.020 M)(0.500 dm3) = (Md)(0.1000 dm3)

Md = 0.010 M

99

Aqueous

Reactions

Dilution of Solutions

Dilution is a procedure for preparing a less concentrated

solution from a more concentrated one.

moles of solute before dilution = moles of solute after dilution

Example

Describe how you would prepare 5.00 x 102 mL ofa 1.75 M H2SO4 solution starting with an 8.61 Mstock solution of H2SO4.

**Mconc = 8.61 M , Vconc = ?Mdil = 1.75 M , Vdil = 5.00 x 102 mL

100

Aqueous

Reactions

Vconc =Mdil Vdil

Mconc=

1.75 M x 500 mL8.61 M

= 102 mL

We must dilute 102 mL of 8.61 M H2SO4 withsufficient water to give a final volume of 5.00 x102 mL in a 500 mL volumetric flask to obtainthe desired concentration.

101

Aqueous

Reactions

Titration

Titration is an

analytical

technique in

which one can

calculate the

concentration

of a solute in

a solution.

102

Aqueous

Reactions103

Aqueous

Reactions

THE STOICHIOMETRY OF

REACTIONS IN AQUEOUS

SOLUTION: TITRATIONS

• The reaction of MnO4- with Sn2+ in

acidic solution follows the equation:2MnO4

- + 5Sn2+ + 16H+ 2Mn2+ + 5Sn4+ + 8H2O

•

• (a) How many mL of 0.02000 mol dm-3

KMnO4 solution would be needed to

react with all the tin in 25.00 mL of

0.05000 mol dm-3 SnCl2 solution?

• 104

Aqueous

Reactions

• From the balanced equation the reacting

ratio of MnO4- to Sn2+ is 2 : 5

• i.e moles of MnO4- / moles of Sn2+ = 2/5

• i.e MMnO4- x VMnO4

- = 2/5 MSn2+ x VSn2+

• i.e. VMnO4- = 2/5 MSn2+ x V Sn2+ / MMnO4

- +

= 2/5 x 0.05000 mol dm-3 x 0.02500 dm3 / 0.02000

mol dm-3

= 0.02500 dm3

= 25.00 mL

105

Aqueous

Reactions

• (b) A 0.5500 g sample of solder, is an alloy

containing lead and tin, was dissolved in

acid and all the tin was converted to Sn2+.

The solution was then titrated with 0.02000

mol dm-3 KMnO4 solution. The titration

required 27.73 mL of the KMnO4 solution.

Calculate the percent by mass of tin in the

solder.

106

Aqueous

Reactions

• From the balanced equation2MnO4

- + 5Sn2+ + 16H+ 2Mn2+ + 5Sn4+ + 8H2O

• the reacting ratio: mol of Sn2+/ mol

of MnO4- = 5 / 2

• mol of Sn2+= 5/2 x mol of MnO4-

•

= 5/2 x MMnO4- x VMnO4-

• = 5/2 x 0.02000 mol dm-3 x 0.02773 dm3

•

• = 0.001387 mol of Sn

•

107

Aqueous

Reactions

• Mass of Sn = mol of Sn x molar mass

• = 0.001387 mol x 118.7 g mol-1

• = 0.1646 g

•% Sn in Solder = mass of Sn / mass of solder x 100

• = 0.1646 g / 0.5500 g x 100

• = 29.93 %

108

Aqueous

Reactions

• Example

• What mass of Ag2CO3 (275.7g/mol) is formed

when 25.00mL of 0.200M AgNO3 are mixed

with 50.00mL of 0.0800M Na2CO3?

• Solution:

• Na2CO3(aq) + 2AgNO3(aq) Ag2CO3(s) + 2NaNO3(aq)

•

• Mixing these 2 solutions will result in one of three

possible outcomes:

• (1) an excess of AgNO3 will remain after the reaction

is complete.

• (2) an excess of Na2CO3 will remain after reaction is

complete.

• (3) neither reagent will be in excess.

•

109

Aqueous

Reactions

• Determine amount of Ag2CO3 produced from each

reactant:

• From the stoichiometric relationship

• n Na2CO3 = n Ag2CO3

• 1 1

• n Ag2CO3 = 1 x nNa2CO3

• i.e. n Ag2CO3 = 1 x 0.0800 mol dm-3 x 0.050 dm3

•

• = 0.004 mol

•110

Aqueous

Reactions

• n AgNO3 = n Ag2CO3

2 1

• n Ag2CO3 = 0.200 mol dm-3 x 0.025dm3

2

= 0.005 mol AgNO3 x 1

• 2

= 0.0025 mol 111

Aqueous

Reactions

• Compare the amounts of Ag2CO3 produced form

each reactant:

• From: Na2CO3 0.004 mol

• AgNO3 0.0025 mol

• AgNO3 - the limiting reagent

• The amount of Ag2CO3 produced will be limited

by the amount of AgNO3 available.

• mass Ag2CO3 = 0.0025 mol x 275.7 g/mol

• = 0.689g Ag2CO3112

Aqueous

Reactions

• Initial amounts are

• Amount AgNO3, nAgNO3 = 0.200 M x 0.025dm3

• = 0.005mol AgNO3

• amount Na2CO3, nNa2CO3 = 0.0800 M x 0.050 dm3

• = 0.004 mol Na2CO3

•

• From the stoichiometric relationship

• n Na2CO3 = nAgNO3

• 1 2

• n AgNO3 = 2 x nNa2CO3

• i.e. n AgNO3 = 2 x 0.004 mol

• = 0.008 mol

= the amount of AgNO3 that would

be required to react with Na2CO3.

•

113

Aqueous

Reactions

• From the stoichiometric relationship

• n Na2CO3 = nAgNO3

• 1 2

• n AgNO3 = 2 x nNa2CO3

• i.e.n AgNO3 = 2 x 0.004 mol

• = 0.008 mol

= the amount of AgNO3

that would be required to react with

Na2CO3.114

Aqueous

Reactions

• Go through Sample Exercises 3.9 to

3.15. and the Practice Exercises.

115

Aqueous

Reactions

Summary of Concepts

Aqueous Solutions

Ionic equations

Acids and Bases

Precipitation reactions

Redox reactions

Disproportionation

The stoichiometry of

reactions in aqueous

solution

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116