Many chemical reactions occur in water. Water is a polar solvent that can dissolve ionic compounds via hydration. When an ionic compound dissolves in water, it separates into its constituent ions which are surrounded by water molecules. The concentration of a solution is expressed as molarity, which is the number of moles of solute per liter of solution. Solutions can be prepared by accurately weighing out and dissolving the solute in a volumetric flask and diluting the solution as needed.

1. 1
Chapter 4

2. 2
Many chemical reactions and virtually all biological processes take
place in water.
e.g.: all chemistry that makes life possible occur in aqueous solution,
various medical tests involve aqueous reactions like sugar
tests..cholesterol .. ect
To understand the chemistry that occur in such diverse places as
human body, the atmosphere, the ground water, the oceans, and so
on, we must understand how substances dissolved in water react with
each other.

3. 3
Before we can understand solution reactions, we need to
discuss the nature of solutions in which water is the
dissolving medium or solvent.
These solutions- (homogeneous mixture) are called aqueous
solutions.
aqueous solutions
Solvent Solute what gets
Soluble- Can be dissolved.
Miscible- liquids dissolve in each other.
dissolved
what does the
dissolving

4. 4
Water, the common solvent
One of the most valuable properties of water
is ability to dissolve many different
substances.
Water is a good solvent for ionic compounds
because its molecules are polar. The oxygen
atoms have a partial negative charge. The
hydrogen atoms have a partial positive
charge.
The angle is 104.5ºC.
It is the polarity that gives water its great
ability to dissolve compounds
+ δ
+ δ
-δ

5. 5
A schematic of an ionic solid dissolving in water is
shown in the following figure.
The “positive ends” of water molecules are attracted to
the negatively charged anions and the “negative ends” are
attracted to the positively charged cations
This process is called hydration
The hydration process is breaking the ionic substances
(salts) apart into individual cations and anions and
surrounded by water molecules
When NaCl dissolves in water, the resulting solution
contains Na+ and Cl- ions moving around independently.
NaCl(s) Na+
(aq) + Cl-
H (aq) 2O(ℓ)

6. How Ionic solids dissolve
H H
O
6
O
H H
H H
O
O
H H
H H
O
H H
O
O
H H
H H
O
H
O
H

7. 7
The solubility is how much of a substance will dissolve
in a given amount of water, usually g/100 mL
The solubility of ionic substances in water varies greatly,
e.g: NaCl soluble in water, whereas AgCl insoluble in
water
If the ionic substance do dissolve the ions are separated,
and they can move around.
Water can also dissolve non-ionic compounds if they
have polar bonds like ethanol (C2H5OH).
“Like dissolve like” is a useful rule for predicting
solubility.

8. 8
Types of solutions
Solutions are classified three ways
Strong electrolytes- completely dissociate (fall apart into
ions). - Many ions - Conduct electricity well.
e.g.: NaCl give Na+ , Cl- when dissolve in water
Weak electrolytes- Partially fall apart into ions.
– Few ions -Conduct electricity slightly.
e.g.: acetic acid dissociate partially in water.
Non-electrolytes- Don’t fall apart. - No ions- Don’t
conduct electricity.
e.g.: sugar ( C12H22O12) , ethanol (C2H5OH)

9. 9
The composition of solution
Chemical reactions take place when two solution are mixed.
To perform stoichiometric calculations in such cases, two
things must be known:
1.The nature of the reaction
2.The amount of chemicals present in the solution, expressed
as concentrations.
Concentration: is how much solute is dissolved in a given
amount of solvent.
The concentration of solution can be expressed in many
different ways. We will consider only the most commonly
used expression of concentration, molarity

10. Molarity (M): moles of solute per volume of solution in liters
1 M = 1 mol solute / 1 liter solution
1 M = 1 mol solute dissolved water until the volume of solute and
solvent (solution) become 1 liter
The concentration is an intensive property, so its value does not
depend on how much of the solution is present.
Example (1):
Calculate the molarity of a solution with 34.6 g of NaCl (molar mass =
58.5g/mol)dissolved in 125 mL of solution.
10
M = molarity = moles of solutes
volume of solution (L)
mol NaCl = 34.6 g NaCl x 1 mol NaCl
58.5 g NaCl
= 0.59 mol NaCl
M = 0.59 mol NaCl
125 x 10-3 L = 4.73 mol/L

11. Example (2):
How many grams of HCl (molar mass = 36.5 g/mol) would
be required to make 50.0 mL of a 2.7 M solution?
11
M =
moles of HCl
volume of solution (L)
2.7 (mol/L) = moles of HCl
50.0 x10-3 L
moles of HCl = 2.7 (mol/L) x 50.0 x10-3 L = 0.135 mol HCl
36.5 g HCl = 4.93 g HCl
Mass of HCl = 0.135 mol HCl x 1 mol HCl

12. mol of NaCl =1.0 x 10-3g NaCl x 1 mol NaCl
12
Example(3):
Typical blood serum is about 0.14 M. what volume of
blood contain 1.0 mg NaCl (molar mass = 58.5 g/mol).
0.14 mol/L =
58.5 g NaCl= 1.71 x 10-5 mol NaCl
1.71 x 10-5 mol NaCl
volume of blood serum
volume of blood serum (L) = 1.71 x 10-5 mol NaCl
0.14 mol/L
= 1.2 x 10-4 L solution = 0.12 ml solution
Thus: 0.12 ml of blood contain 1.71x10-5 mol NaCl or 1.0mg NaCl

13. 13
Solution concentration is always given in term of the form of the
solute before it dissolves and does not take into account any
subsequent processes, such as the dissociation of salts.
e.g.: 1.0 M NaCl, this mean that 1.0 mol (58.5g) solid NaCl dissolved
in enough water to make 1.0 L solution; it does not mean that the
solution contain 1.0 mol of NaCl units.
Actually, the solution contains 1.0 mol Na+ ions and 1.0 mol Cl- ions.
Example (4):
What is the concentration of each ions in 0.5M CaCl2 solution:
CaCl2(s) Ca+
(aq) + 2 Cl-
(aq)
H2O
1 mol CaCl2 that dissolved, the solution contain 1 mol Ca+2 and 2 mol
Cl- ions. Thus 0.5 M CaCl2 contain 0.5 M Ca+2 and 2 x 0.5 M Cl- or 1
M Cl-

14. 1 mol CaCl2 = 0.24 mol CaCl2 a)
14
Example (5):
A 27 g of CaCl2 (molar mass = 111 g/mol) dissolved in
500 ml solution. Calculate: a) concentration of CaCl2
b) concentration of each ion
mol of CaCl2 = 27 g CaCl2 x
111 g CaCl2
M (CaCl2) = 0.24 mol CaCl2
b) 1 mol CaCl2
500 x 10-3 L = 0.48 M CaCl2
1 mol Ca+2
2 mol Cl-
0.48 mol CaCl2
0.48 mol Ca+2
2 x 0.48 mol Cl- = 0.96 mol Cl-

15. Example (6):
A 45.6 g of Fe2(SO4)3 (molar mass = 400 g/mol) dissolved in
475 ml solution. Calculate: a) concentration of Fe2(SO4)3
15
b) concentration of
eaMch (iFoen 2(SO4)3) = (mass/molar mass) Fe2(SO4)3
V (L) solution
M (Fe2(SO4)3) = 45.6 g/(400g/mol)
475 x 10-3 L
= 0.24 M or mol/L
0.24 (mol/L) Fe2(SO4)3
2 x 0.24 M Fe+3 = 0.48 M Fe+3
3 x 0.24 M SO4
-2 = 0.72 M SO4
-2
a)
b)

16. 16
The number of moles of solute present in a given volume
of solution of known molarity can be calculated as follows:
Molarity (mol/L)= mol of solute
volume of solution (L)
mol of solute = molarity (mol/L) x volume of solution (L)
Example (7):
Calculate the number of mol of Cl- ions in 1.75L of 1x10-3 M ZnCl2
1x10-3 M ZnCl2
1x10-3 M Zn+2
2 x(1x10-3)M Cl- = 2 x 10-3 M Cl-mol
of Cl- = 2 x 10-3 mol Cl-
L
x 1.75 L = 3.5 x10-3 mol Cl-

17. Preparing a Solution of Known Concentration
A standard solution: is a solution whose concentration is accurately
known. It can be prepared as follows. First, the solute is accurately
weighed and transferred to volumetric flask as shown in the following
figure. Next water is added to the flask, to dissove the solid. After all
the solid has dissolved, more
water is added to bring the
level of solution exactly to
the volume mark.
17 17

18. Example (8): Describe how to make 100.0 mL of a 1.0 M K2Cr2O4
(molar mass =246 g/mol) solution.
mol of K2Cr2O4 = 1.0 mol K2Cr2O4
18
L
x 100 x 10-3 L = 0.1 mol K2Cr2O4
mass of K2Cr2O4 = 0.1 mol K2Cr2O4 x
246g K2Cr2O4 =24.6 g K2Cr2O4
1 mol K2Cr2O4
Thus, to prepare 100 ml of 1.0 M K2Cr2O4, weigh out 24.6g K2Cr2O4 ,
add distilled water until the volume of solution = 100ml
Example (9): Describe how to make 250. mL of an 2.0 M copper (II)
sulfate dihydrate (CuSO4.2H2O, molar mass = 200g/mol) solution.
mol of CuSO4 = 2.0 mol CuSO4 x 250 x 10-3 L = 0.5 mol CuSO4 L
mass of CuSO4 = 0.5 mol CuSO4 200 g CuSO4
1 mol CuSO4
= 100 g CuSO4
Thus, to prepare 250.0 ml of 2.0 M CuSO4.2H2O, weigh out 100 g
CuSO4.2H2O , add distilled water until the volume of solution = 250 ml

19. 19
Dilution
In the lab. a concentrated solutions are present, these solutions are
called Stock solution.
Water is added to these solutions to achieve the molarity desired
for a particular solution.
This process is called dilution: the procedure for preparing a less
concentrated solution from more concentrated one
The moles of solute stay the same.
Dilution
Add Solvent
Moles of solute
before dilution (i)
Moles of solute
after dilution (f)
=
MiVi = MfVf

20. 20
Example:
You have a 4.0 M stock solution. Describe how to make 1.0L of a
0.75 M solution from the stock solution.
(M V)conc. = (M V)dil.
0.4 M x V = 0.75 M x 1.0 L
Vconc. = 0.75 M x 1.0 L/ 4 M = 0.1875 L = 187.5 ml
Take 187.5 ml of the stock solution, add water until the volume of
the solution (solute and solute) become 1.0 L
Example:
18.5 mL of 2.3 M HCl is diluted to 250 mL with water. What is the
concentration of the solution?
M1 V1 = M2 V2 2.3 M x 18.5 ml = M2 x 250 ml
M2 = 2.3 M x 18.5 ml / 250 ml
M2 = 0.17 M

21. Example:
A solution is prepared by dissolving 10.8 g of (NH4)2SO4
(molar mass = 132 g/mol) in enough water to make 100 ml
stock solution. A 10.0 ml of this stock solution is added to
water to make 50.0 ml solution. Calculated the
concentration of NH4
21
+ and SO4
-2 ions in the final solution.
moles of (NH4)2SO4 = 10.8 g
132 g /mol = 0.0818 moles of (NH4)2SO4
molarity of (NH4)2SO4 = 0.818 moles
100 x 10-3 L
= 0.818 M (NH4)2SO4
(M V) conc. = (M V) dil.
0.818 M x 10.0 ml = M x 50 ml
M dil. = 0.16 M of (NH4)2SO4
0.16 M of (NH4)2SO4
0.16M SO4
-2
2 x 0.16M NH4
+

22. 22
Classes of Reactions
CChheemmiiccaall rreeaaccttiioonnss
Precipitation
reactions
Acid-Base
Reactions
Oxidation-Reduction
Reactions

23. 23
1) Precipitation reactions:
Occur in aqueous solutions of ionic compounds which results in
the formation of an insoluble product, or a precipitate.
A precipitate is an insoluble solid that separates from the solution.
For example: Pb(NO3)2(aq) + 2 KI(aq) → PbI2(s) + KNO3(aq)
This reaction is an example of a metathesis reaction (also called
double-displacement reaction): a reaction that involves the
exchange of parts between the two compounds.

24. 24
Pb(NO3)2(s) Pb(NO3)2(aq)
Pb2+
Ions in Aqueous Solution
NO3
1–
NO3
1–
Na1+ I1–
Pb2+(aq) + 2 NO3
1–(aq)
add
water
NaI(s)
+ H2O(ℓ)
dissociation:
+ H2O(ℓ)
Pb2+
NaI(aq)
NO3
1–
NO3
1–
Na1+(aq) + I1–(aq)
Mix them and get…
Balance to get overall ionic equation…
Cancel spectator ions to get net ionic equation…
in solution
Na1+ I1–

25. 25
Mix them and get…
Pb(NO3)2(aq) + 2 NaI(aq) PbI2(s) + 2 NO3
NO3
Balance to get overall ionic equation…
Pb2+(aq) + 2 NO3
1–(aq) + 2 Na1+(aq)
1–(aq) + 2 Na1+(aq) + 2 I1–(aq) PbI2(s) + 2 NO3
1–(aq) + 2 Na1+(aq)
Cancel spectator ions to get net ionic equation…
Pb2+(aq) + 2 I1–(aq) PbI2(s)
solid
in solution
Pb2+
NO3
1–
Na1+ I1–
1–
Na1+ I1–
Pb2+
NO3
1–
Na1+
I1–
NO3
1–
Na1+
I1–

26. 26
Mix together Zn(NO3)2(aq) and Ba(OH)2(aq):
Ba(NO3)2 and (aq) Zn(OH)2 (ppt)
Zn(NO3)2(aq) + Ba(OH)2(aq) Zn(OH)2(s) + 2 NO3
Ba(OH)2(aq)
Ba2+
OH1–
OH1–
NO3
1–
NO3
1–
Zn2+
(aq) + 2 NO3
1–
(aq) Ba2+
(aq) + 2 OH1–
(aq)
Zn(NO3)2(aq)
Zn2+
Balance to get overall ionic equation…
1–
(aq) + Ba2+
(aq)
Zn2+
(aq) + 2 NO3
1–
(aq) + Ba2+
(aq) + 2OH1–
(aq) Zn(OH)2(s) + 2 NO3
1–
(aq) + Ba2+
(aq)
Mix them and get…
Cancel spectator ions to get net ionic equation…
Zn2+
(aq) + 2 OH1–
(aq) Zn(OH)2(s)

27. 27
Three Types of Equations
Molecular Equation- written as whole formulas, not the ions.
K2CrO4(aq) + Ba(NO3)2(aq) → BaCrO4(s) + 2KNO3(aq)
Complete Ionic equation show dissolved electrolytes as the ions.
2K+
(aq) + CrO4
-2
(aq) + Ba+2
(aq) + 2 NO3
-
(aq) →
BaCrO4(s) + 2K+
(aq) + 2 NO3
-
(aq)
Spectator ions are those that don’t react.
Net Ionic equations show only those ions that react, not the
spectator ions
Ba+2
(aq)
+ CrO4
-2
(aq) → BaCrO4(s)

28. How can we predict whether a precipitate will form or not when
two aqueous solutions are mixed? It depends on the solubility of
the solute.
The solubility is the maximum amount of solute that will dissolve
in a given quantity of solvent at specific temperature.
Substances can be classified as soluble, slightly soluble, or
insoluble
All ionic compounds are strong electrolytes, but they are not
equally soluble
the solubility rules for common ionic compounds in water are as
follows:
28

29. 29
Solubility Rules
1. Most nitrates (NO3
-) are soluble.
2. Most salts containing Group (I) ion and ammonium ion,
NH4
+, are soluble.
3. Most chloride (Cl-), bromide (Br-), and iodide (I-) salts are
soluble, except Ag+, Pb2+ and Hg2
2+.
4. Most sulfate (SO4
-2) salts are soluble, except BaSO4, PbSO4,
Hg2SO4, Ag2SO4 and CaSO4.
5. Most hydroxides (OH-) except Group (1) and Ba(OH)2.
6. Most sulfides (S-2), carbonates (CO3
-2), chromates (CrO4
-2), and
phosphates (PO4
-3)are only slightly soluble.

30. Examples:
Predict if a precipitate is formed when the following
solutions are mixed. Write the three types of equations for
the reactions that lead to the formation of a precipitate.
iron (III) sulfate and potassium sulfide
Lead (II) nitrate and sulfuric acid.
NH4Cl(aq) + NaNO3(aq) →
AgNO3(aq) + KCl(aq) →
Zn(NO3)2(aq) + BaCr2O7(aq) →
CdCl2(aq) + Na2S(aq) →
NaOH(aq) + FeCl3(aq) →
30

31. 31
Stoichiometry of Precipitation Reactions
The procedure for doing stoichiometric calculations for
solutions reactions are very similar to those for other
types of reactions.
The following steps are used:
Write the balanced net ionic equation for the reaction
Calculate the moles of the reactants.
Determine which reactant is limiting.
Calculate the moles of product or products, as
required.
Convert to grams, as required

32. 32
Example (1)
What mass of Na2CrO4 (molar mass = 162 g/mol) is
required to precipitate all Ag+ ions from 75.0ml of 0.1 M
solution AgNO3.
2 AgNO3(aq) + Na2CrO4(s) → Ag2CrO4(s) + NaNO3(aq)
Net ionic equation: 2Ag+
(aq)+ Na2CrO4(s)→Ag2CrO4(s) + 2Na+
(aq)
moles of Ag+ = M x V = 0.1 (mol/L) x 75.0 x 10-3 L = 7.5 x 10-3 mol
mol of Na2CrO4 = 7.5 x 10-3 mol Ag+ x 1 mol of Na2CrO4
2 mol Ag+
= 3.8 x 10-3 mol Na2CrO4
mass of Na2CrO4 = 3.8 x 10-3 mol Na2CrO4 x 162 g/mol
= 0.616 g Na2CrO4

33. Example (2):
Calculate the mass of Ba(OH)2 (molar mass = 171g/mol ) is
formed when 100.00 mL of 0.100 M BaCl2 is mixed with
100.00 mL of 0.100 M NaOH.
BaCl2(aq) + 2 NaOH(aq) → Ba(OH)2(s) + 2 NaCl(aq)
mol of of BaCl2 = M x V = 0.1(mol/L) x 100 x 10-3 L
33
= 0.01 mol
mol of NaOH = 0.1(mol/L) x 100 x 10-3 L = 0.01 mol NaOH
BaCl2 : 2 NaOH
0.01 0.01/2
NaOH is the limiting reactant
mol of Ba(OH)2 = 0.01 mol NaOH x 1 mol Ba(OH)2
2 mol NaOH
= 0.005 mol Ba(OH)2
mass Ba(OH)2 = 0.005 mol Ba(OH)2 x
171 g Ba(OH)2
1 mol Ba(OH)2
= 0.855 g Ba(OH)2

34. Example (3):
What volume of 0.204 M HCl is needed to precipitate all the
silver ions from 50.ml of 0.0500 M silver nitrate solution.
34
AgNO3(aq) + HCl(aq) → AgCl(s) +
mol of AgNO3 = M x V = 0.05 M x 50 x 10-3 L = 2.5 x 10-3 mol
1 mol HCl = 2.5 x 10-3 mol
mol of HCl = 2.5 x 10-3 mol AgNO3 x 1 mol AgNO3
Molarity of HCl = mol of HCl
Volume (L)
Volume (L) =mol of HCl
molarity of HCl
volume of HCl = 2.5 x 10-3 mol
0.204 mol/L = 0.01225 L = 12.23 ml

35. 35
2) Acid-Base reactions
Arrhenius’s concept of acids and bases:
Acid: is a substance that produce H+ ions when dissolved in
water.
Base: is a substance that produce OH- ions.
A more general definition of a base (which include
substances that do not contain OH-) was provided by
Brǿnsted-Lowry whose defined acids and bases as
follows:
an acid is a proton donor.
a base is a proton acceptor.

36. 36
When aqueous solution of HCl (strong electrolyte) is mixed with
aqueous solution of NaOH(strong electrolyte), the reaction can be
represented as follows:
Molecular equation: HCl(aq) + NaOH(aq)→ NaCl(aq) + H2O(ℓ)
Ionic equation :
H+
(aq) + Cl-
(aq) + Na+
(aq) + OH-
(aq) → Na+
(aq) + Cl-
(aq)+ H2O(ℓ)
Net ionic equation: H+
(aq) + OH-
(aq) → H2O(ℓ)
So, all acid-base reaction can be represented as :
Acid + Base → salt + water
Acid - Base Reaction is often called a neutralization reaction. When
enough acid is added to react exactly with the base, we say that the
acid has been neutralized.

37. When aqueous solution of weak acid such as HCN is mixed with
aqueous solution of NaOH, the reaction can be represented as
follows:
Molecular equation: HCN(aq) + NaOH(aq)→ NaCN(aq) + H2O(ℓ)
Ionic equation :
HCN
37
(aq) + Na+
(aq) + OH-
(aq) → Na+
(aq) + CN-
(aq)+ H2O(ℓ)
Net ionic equation: HCN
(aq) + OH-
(aq) → CN-
(aq) + H2O(ℓ)
Note that only Na+ is a spectator ion.
Example: write the ionic, and the net ionic equation for the following acid-base
reaction:
Na2CO3(aq) + 2 HCl → 2 NaCl(aq) + H2CO3(aq)
H2CO3(aq) decompose as follows:H2CO3(aq) → CO2(g) + H2O(ℓ)
Ionic equation:
2 Na+
(aq) + CO-3
(aq) + H+
(aq) + Cl-
(aq) → CO2(g) + H2O(ℓ) +2 Na+
(aq) + 2 Cl-
(aq)
Net ionic equation: CO-3
(aq) + H+
(aq) → CO2(g) + H2O(ℓ)

38. Acid-Base Titration:
Quantitative studies of acid-base neutralization reactions are carried
out using a technique known as titration.
In titration, a solution of accurately known concentration (standard
solution) is added gradually by using the buret to another solution
of unknown concentration, until the chemical reaction between the
two solutions is complete. The reaction is said to be complete when
we reach the Equivalence point.
Equivalence point – the point at which the reaction is complete
The equivalence point is usually
signaled by a sharp change in the
Color of an indicator
Indicator – substance that changes
color at (or near) the equivalence point
38

39. Example (1):
75 mL of 0.25M HCl is mixed with 225 mL of 0.055 M
Ba(OH)2 . calculate : a) the amount of H2O produce.
39
b) the concentration of the excess H+ or OH-Ba(
OH)2(aq) + 2HCl(aq) → BaCl2(aq) + 2 H2O(ℓ)
moles of HCl = 0.25 M x 75x10-3L = 0.01875 mol
mol of Ba(OH)= 0.055 M x 225x10-3L = 0.012375 mol
2 Ba(OH): 2 HCl
2 0.012375 : 0.01875/2 2 mol limiting HO reactant 2x 18 g His HCl
O
2Mass of HO = 0.01875 mol HCl x 2 mol HCl
1 mol HO
22= 0.3375 g H2O

40. 40
b) moles of Ba(OH)2 reacted = 0.01875 mol HCl x
1 mol Ba(OH)2
2 mol HCl
= 0.009375 mol Ba(OH)2
Total mol of Ba(OH)2 = mol Ba(OH)2reacted + mol Ba(OH)2 unreacted
0.012375 mol = 0.009375 mol + mol Ba(OH)2unreacted
mol Ba(OH)2 unreacted = 0.003 mol
mol of OH- = 2 x 0.003 mol = 0.006 mol
Molarity of OH- = 0.006 mol OH-total
volume (L)
= 0.006 mol
300 x 10-3 L
= 0.02 M OH-

41. Example (2):
A 50.00 mL sample of aqueous Ca(OH)2 requires 34.66 mL
of 0.0980 M HNO3 for neutralization. What is the
concentration of Ca(OH)2.
Ca(OH)2(aq) + 2HNO3(aq) → Ca(NO3)2(aq) + 2H2O(ℓ)
mol of HNO3 = 0.098 (mol/L) x 34.66x10-3 L = 3.4 x10-3 mol
mol of Ca(OH)2 = 3.4 x 10-3 mol HNO3 x 2 mol HNO3
41
1 mol Ca(OH)2
= 1.7 x 10-3 mol Ca(OH)2
molarity of Ca(OH)2= 1.7 x 10-3 mol
50 x 10-3 L
= 0.034 M

42. 3) Oxidation-Reduction (called Redox) Reactions:
For the following reaction:
2 Na(s) + Cl2(g) → 2 NaCl(s)
Na is a neutral atom react with Cl2 which is also a neutral
compound to form NaCl which contain Na+ and Cl- ions.
Reactions like this one, which involves the transfer of
electrons, are called oxidation-reduction (or redox)
reactions.
Many important redox reactions occur in our daily life, e.g.:
Photosynthesis, combustion of sugar in our body,..etc.
To explain how electrons are transfer, the concept of
oxidation states must be introduced.
42

43. 43
Oxidation States
The concept of oxidation states (or oxidation numbers)
provides a way of keeping track of the electrons in oxidation
reduction reactions.
need the rules for assigning oxidation states (memorize).
The sum of the oxidation states must be zero in
compounds or equal the charge of the ion.
The following table summarized the rules for assigning the
oxidation states

44. 44
…The oxidation state of summary examples
• elements in their standard
states is zero
Element: 0 Na(s), O2(g) , Hg(ℓ)
•monoatomic ions are the same
as their charge.
Monoatomic ions:
Charge of ion
Na+ , Cl-, Ca+2
•Oxygen is assigned an
oxidation state of -2 in its
covalent compounds except as a
peroxide (O2
-2, oxygen is -1, e.g.
H2O2)
Oxygen : -2
H2O , CO2 , CO
•hydrogen is assigned the
oxidation state +1, except when
it bonded to metals; LiH, NaH
…, its oxidation state is -1
Hydrogen: +1 HCN, HCl, NH3
•fluorine is always –1 in its
compounds.
Flourine: -1 HF, PF3

45. The following figure shows oxidation numbers of familiar elements,
arranged according to their position in the periodic table.
45

46. Example:
Assign the oxidation states to each element in the following
compounds.
CO2 : O is -2, so -2 x 2 + C = 0 C = +4
NO3
46
- : O is -2, so (3 x -2)+ N = -1 N = +5
H2SO4 : H is +1, O is -2, so [2x(+1)]+[(4x(-2)]+S =0
S = +6
FeO: O is -2, so [3x(-2)]+ 2xFe = 0 Fe = +3
23 CrO-2 : 2xCr +[7x(-2)]= -2 Cr = +6
27

47. Oxidation-reduction reactions are characterized by the
transfer electrons, so the oxidation states change.
For the following reactions:
1) 2 Na + Cl2 → 2NaCl
Oxidation
state 0 0 +1 -1
47
2) CH4 + 2O2 → CO2 + 2H2O
Oxidation
state -4 +1
(each H) 0
+4 -2
(each O) +1
each H
-2
Carbon undergoes a change in oxidation state from -4 in CH4
to +4 in CO2, 8 electrons (symbol e- stands for electrons) are
lost

48. 48
CH4 → CO2 + 8 e-
-4 +4
Each oxygen change from an oxidation state 0 in O2 to -2
in H2O and CO2 , each atom gain 2 electrons. Since four
oxygen atoms are involved , this is a gain of 8 electrons.
O2 + 8 e- → CO2 + 2 H2O
0 4 x (-2)= - 8
No change occurs in the oxidation state of hydrogen .

49. We can now define some important terms:
Oxidation is the loss of electrons or an increase in
oxidation state - lose electrons. .
Reduction is the gain of electrons or a decrease in
oxidation state - gain electrons.
The substance that is oxidized is called the reducing agent.
Reducing agent gets oxidized. Loses electrons.
More positive oxidation state.
The substance that is reduced is called the oxidizing agent.
Oxidizing agent gets reduced. Gains electrons. More
negative oxidation state.
49

50. 50
Example:
For the following reactions, identify
Substance oxidized
Substance reduced
Oxidizing agent
Reducing agent
1) Fe (s) + O2(g) → Fe2O3(s)
0 0 + 3 -2
The oxidation state for Fe increase from 0 to +3. thus Fe is oxidized.
The oxidation state for O decrease from 0 to -2. thus O is reduced.
The oxidizing agent is O2.
The reducing agent is Fe.

51. +3 -2 +2 -2 0 +4 -2
The oxidation state for C increase from +2 to +4. thus C is oxidized.
The oxidation state for Fe decrease from +3 to 0. thus O is reduced.
The oxidizing agent is Fe.
The reducing agent is CO.
51
2) Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g)
3) SO3
2- + H+ + MnO4
- → SO4
2- + H2O + Mn2+
+4 -2 +1 +7 -2 +6 -2 +1 -2 +2
The oxidation state for S increase from +4 to +6. thus S is oxidized.
The oxidation state for Mn decrease from +7 to +2. thus Mn is reduced.
The oxidizing agent is MnO-.
4
The reducing agent is SO3
2-.

52. 52
Balancing oxidation-reduction equations
It is difficult to balance redox equations by simple
inspection, the technique used for balancing redox reactions
called the half-reaction method.
All redox reactions can be thought of as happening in two
halves.
One produces electrons - Oxidation half.
The other requires electrons - Reduction half.
Write the half reactions for the following.
1) Na + Cl2 → Na+ + Cl-
2) Ce+4 + Sn+2 → Ce+3 + Sn+4

53. 53
1) Na + Cl2 → Na+ + Cl-
Na → Na+ oxidation
Cl2 → Cl- reduction
2) Ce+4 + Sn+2 → Ce+3 + Sn+4
Ce+4 → Ce+3 reduction
Sn+2 → Sn+4 oxidation
The method for balancing redox equations differ slightly
depending on whether the reaction takes place in acidic or
basic solution.

54. Balancing Redox Equations in acidic solution
In aqueous solutions the key is the number of electrons
produced must be the same as those required.
For reactions in acidic solution an 8 step procedure.
1.Write separate equations for the oxidation-reduction half reactions
2. For each half reaction balance all reactants except H and O
3. Balance O using H2O
4. Balance H using H+
5. Balance charge using e-
6. Multiply equations to make electrons equal
7. Add the half-reactions, and cancel identical species
8 Check that charges and elements are balanced.
54

55. 55
Examples:
Balance the following oxidation-reduction reactions occur
in acidic aqueous solution.
1) MnO- + Fe+2 → Mn+2 + Fe+3
4
1) Write the half reactions
MnO4
- → Mn+2 Fe+2 → Fe+3
2) Balance each half-reaction:
MnO4
- → Mn+2
-Mn is balanced
-balance oxygen by adding 4H2O to the write side of the equation.
- Balance hydrogen by adding 8H+ to the left side of equation.
8H+ + MnO4
- → Mn+2 + 4H2O

56. 56
-Balance the charge using electrons
8H+ + MnO4
- → Mn+2 + 4H2O
+8 -1 → +2 0
+7 +2
equalize the charges by adding five electrons to the left side.
5e- + 8H+ + MnO4
- → Mn+2 + 4H2O reduction
+2 +2
both the elements and charges are now balanced.
For the oxidation reaction: Fe+2 → Fe+3 + e-
+2 +3-1
+2 +2

57. Equalize the electron transfer in the two half-reactions
57
5e- + 8H+ + MnO4
- → Mn+2 + 4H2O
5 Fe+2 → 5 Fe+3 + 5e-
8H+ + MnO4
- + 5 Fe+2 → Mn+2 + 4H2O + 5 Fe+3
check that elements and charges are balanced.
elements balance: 5Fe, 1Mn, 4O, 8H→ 5Fe, 1Mn, 4O, 8H
charge balance: 8(+1)+(-1)+5(+2)→(+2)+0+5(+3)
+17 → +17
the equation is balanced.

58. 58
2) Cu + NO3
- → Cu+2 + NO(g)
3 / Cu → Cu+2 + 2e-
2 / 3e- + 4H+ + NO3
- → NO + 2H2O
3Cu + 8H+ + 2NO3
- → 3Cu+2 + 2NO + 4H2O
3) Mn+2 + BiO3
- → Bi+3 + MnO4
-
2 / 4 H2O + Mn+2 → MnO4
- + 8H+ +5e-
+2 +7
5/ 2e- + 6H+ +BiO3
- → Bi+3 + 3H2O
+5 +3
2Mn+2 + 5BiO3
- + 14H+ → 2MnO4
- + 5Bi+3 + 7H2O

59. Balancing Redox Equations in basic solution
Do everything you would with acid, but add one more
step.
Add enough OH- to both sides to neutralize the H+
Makes water
Examples:
Balance the following equation in basic solutions:
59
Cr(OH)3 + OCl- + OH- → CrO4
2- + Cl- + H2O

60. Cr(OH)3 + OCl- + OH- → CrO4
60
2- + Cl- + H2O
2/ Cr(OH)3 + H2O → CrO4
2- + 5H++3e-
0 +3
3/ 2e- + OCl- + 2H+ → Cl- + H2O
+1 -1
2- + 3Cl- +H2O +4H+ (acidic solution)
2Cr(OH)3 + 3OCl- → 2CrO4
4OH- +2Cr(OH)3 + 3OCl- → 2CrO4
2- + 3Cl- +H2O + 4H+ + 4OH-
4OH- +2Cr(OH)3 + 3OCl- → 2CrO4
2- + 3Cl- +5H2O (basic solution)

61. Example: 16.42 mL of 0.1327 M KMnO4 solution is needed to
oxidize 25.00 mL of an acidic FeSO4 solution. What is the molarity of
the iron solution:
5Fe2+ + MnO4
- + 8H+ Mn2+ + 5Fe3+ + 4H2O
0.01642 L 0.1327 mol KMnOx
4
1 L
x
5 mol Fe2+
1 mol KMnO4
x
1
0.02500 L Fe2+
= 0.4358 M Fe2+
61 61