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Reactions in Aqueous Solution
Chapter 4
Dr. Sa’ib Khouri
AUM- JORDAN
Chemistry
By Raymond Chang
2
Solution is a homogenous mixture of two or more
substances.
Solute is the substance present in a smaller amount.
(In a solution may be more than one solute)
Solvent is the substance present in a larger amount.
Solution Solvent Solute
Soft drink (l)
Air (g)
Soft Solder (s)
H2O
N2
Pb
Sugar, CO2
O2, Ar, CH4
Sn
Aqueous solutions:
the solute is a liquid
or a solid and the
solvent is water
General Properties of Aqueous Solutions
A solution may be gaseous (air), solid (alloy),
or liquid (seawater).
3
Electrolyte is a substance that, when dissolved in
water, results in a solution that can conduct electricity.
Nonelectrolyte does not conduct electricity when
dissolved in water.
nonelectrolyte weak electrolyte strong electrolyte
Electrolytic Properties
4
Examples of strong electrolytes, weak electrolytes, and nonelectrolytes.
Cations (+) and Anions (-) are responsible to conduct electricity in
solution.
Strong Electrolyte – 100% dissociation
NaCl(s) Na
+
(aq) + Cl
-
(aq)
H2O
5
Hydration is the process in which an ion is surrounded by
water molecules arranged in a specific manner.
d+
d-
H2O
Hydration helps to stabilize ions in solution and prevents cations
from combining with anions
The Na+
and Cl-
ions are separated from each other and undergo
hydration
6
A reversible reaction. The reaction can occur in both
directions (chemical equilibrium)
7
Precipitation Reactions
Precipitate – insoluble solid that separates from solution
Molecular equation
Ionic equation
Net ionic equation
Pb2+
+ 2NO3
-
+ 2K+
+ 2I-
PbI2 (s) + 2K+
+ 2NO3
-
Here K
+
and NO3
-
are spectator ions
Pb(NO3)2 (aq) + 2KI (aq) PbI2 (s) + 2KNO3 (aq)
Pb
2+
+ 2I
-
PbI2 (s)
e.g. Mixing of Pb(NO3)2 and KI solutions forms an insoluble
yellow product, PbI2 through a double-displacement reaction:
8
Precipitation of Lead Iodide
Pb
2+
+ 2I
-
PbI2 (s)
9
is the maximum amount of solute that will dissolve in
a given quantity of solvent at a specific temperature.
Solubility
10
Examples of Insoluble Compounds
CdS PbS Ni(OH)2 Al(OH)3
Practice Exercise Classify the following ionic compounds as
soluble or insoluble:
(a) CuS, (b) Ca(OH)2, (c) Zn(NO3)2, (d) Ag2SO4, (e) Na3PO4
11
Writing Net Ionic Equations
1. Write the balanced molecular equation.
2. Write the ionic equation showing the strong electrolytes
completely dissociated into cations and anions.
3. Cancel the spectator ions on both sides of the ionic equation
4. Check that charges and number of atoms are balanced in the
net ionic equation
AgNO3 (aq) + NaCl (aq) AgCl (s) + NaNO3 (aq)
Ag+ + NO3
- + Na+ + Cl- AgCl (s) + Na+ + NO3
-
Ag+ + Cl- AgCl (s)
e.g. Write the net ionic equation for the reaction between
silver nitrate and sodium chloride.
Write balance complete molecular equation, ionic equation, and
net ionic equations for the reaction that occurs when Pb(NO3)2
and Na2S solutions are mixed. Identify the spectator ions in the
reaction.
e.g.
Molecular equation
Ionic equation
Net ionic equation
General Properties of Acids and Bases
Acids
• Acids have a sour taste; e.g. Vinegar owes its sourness to
acetic acid, and lemons and other citrus fruits contain citric acid
• Acids cause color changes in plant
dyes, e.g. acids change the color of
litmus from blue to red.
• Acids react with certain metals to produce
hydrogen gas
2HCl (aq) + Mg (s) → MgCl2 (aq) + H2 (g)
Acid-Base Reactions
Acids react with carbonates and bicarbonates to produce
carbon dioxide gas
2HCl (aq) + CaCO3 (s) → CaCl2 (aq) + CO2 (g) + H2O (l)
HCl (aq) + NaHCO3 (s) → NaCl (aq) + CO2 (g) + H2O (l)
• Aqueous acid solutions conduct electricity
Bases
• Have a bitter taste
• Feel slippery, e.g. soaps, which contain bases, exhibit this
property.
• Cause color changes in plant dyes
• Aqueous base solutions conduct electricity
16
Arrhenius acid is a substance that produces H+ (H3O+) in water
Arrhenius base is a substance that produces OH- in water
Bronsted Acids and Bases
Arrhenius’s definitions of acids and bases apply only to
aqueous solutions
17
A Brønsted acid is a proton donor
A Brønsted base is a proton acceptor
Johannes Brønsted (Danish chemist) proposed broader definitions:
Note: Brønsted’s definition do not require acids and bases to be
in aqueous solution
NH3 (g) + HCl (g) → NH4Cl (s)
Here
HCl (g) is a Brønsted acid
NH3 (g) is a Brønsted base
18
Hydrochloric acid is a Brønsted acid because it donates a
proton in water:
HCl (aq) → H
+
(aq) + Cl
-
(aq)
H+
ion is a hydrogen atom that has lost its electron; that is, it is just a bare
proton and exists in the hydrated form:
The hydrated proton, H3O
+
, is called the hydronium ion
19
Monoprotic acids: yields one hydrogen ion upon ionization
HCl H+ + Cl-
HNO3 H+ + NO3
-
CH3COOH H+ + CH3COO-
Strong acid
Strong acid
Weak acid
Diprotic acids: gives up two H
+
ions, in two separate steps
H2SO4 H+ + HSO4
-
HSO4
- H+ + SO4
2-
Strong acid
Weak acid
Triprotic acids: yields three H+
ions, in three separate steps
H3PO4 H+ + H2PO4
-
H2PO4
- H+ + HPO4
2-
HPO4
2- H+ + PO4
3-
Weak acid
Weak acid
Weak acid
20
21
22
Acid-Base Neutralization
acid + base salt + water
HCl (aq) + NaOH (aq) NaCl (aq) + H2O
H+ + Cl
-
+ Na+
+ OH- Na
+
+ Cl
-
+ H2O
H
+
+ OH
-
H2O
A neutralization reaction is a reaction between an acid and a
base. Generally, aqueous acid-base reactions produce water
and a salt (ionic compound)
Acid-base reactions generally go to completion.
Here only if we had started the reaction with equal molar
amounts of the acid and the base.
23
Neutralization reaction involving a weak electrolyte
weak acid + base salt + water
HCN (aq) + NaOH (aq) NaCN (aq) + H2O
HCN + Na
+
+ OH
-
Na
+
+ CN
-
+ H2O
HCN (aq) + OH
-
(aq) CN
-
(aq) + H2O
The following are also examples of acid-base neutralization
reactions, represented by molecular equations:
HF(aq) + KOH(aq) ⟶ KF(aq) + H2O
H2SO4(aq) + 2NaOH(aq) ⟶ Na2SO4(aq) + 2H2O
HNO3(aq) + NH3(aq) ⟶ NH4NO3(aq)
HNO3(aq) + NH+
4 (aq) + OH−
(aq) ⟶ NH4NO3(aq) + H2O
24
Acid-Base Reactions Leading to Gas Formation
acid + base salt + water + CO2
2HCl (aq) + Na2CO3 (aq) 2NaCl (aq) + H2O +CO2
2H
+
+ 2Cl
-
+ 2Na
+
+ CO3
2- 2Na
+
+ 2Cl
-
+ H2O + CO2
2H+ (aq) + CO3
2- (aq) H2O + CO2
Certain salts like carbonates (containing the CO3
2- ion), bicarbonates
(containing the HCO3
1- ion), sulfites (containing the SO3
2- ion), and sulfides
(containing the S2- ion) react with acids to form gaseous products.
Similar reactions involving other mentioned salts are
25
26
Oxidation-Reduction, or Redox ,Reactions
(electron transfer reactions)
2Mg 2Mg2+ + 4e-
O2 + 4e- 2O2-
Oxidation half-reaction (lose e-)
Reduction half-reaction (gain e-)
2Mg + O2 + 4e- 2Mg2+ + 2O2- + 4e-
2Mg + O2 2MgO
Example (Two elements combine to form a compound)
The formation of magnesium oxide (MgO) from magnesium
and oxygen.
2Mg(s) + O2(g) ⟶2MgO(s)
27
Zn(s) + CuSO4 (aq) ZnSO4 (aq) + Cu (s)
(Zn is oxidized)Zn Zn2+ + 2e-
(Cu2+ is reduced)Cu2+ + 2e- Cu
Zn is the reducing agent
Cu2+ is the oxidizing agent
e.g. Copper wire reacts with silver nitrate to form silver metal.
What is the oxidizing agent in the reaction?
Cu (s) + 2AgNO3 (aq) Cu(NO3)2 (aq) + 2Ag (s)
Cu Cu2+ + 2e- Ag+ + 1e- Ag
(Ag+ is reduced) Ag+ is the oxidizing agent
e.g. When metallic zinc is added to a solution containing
copper(II) sulfate (CuSO4),
28
2
9
Oxidation number
(or oxidation state)
1. Free elements (uncombined state) have an oxidation
number of zero.
Na, Be, K, H2, O2, P4 = 0
2. In monatomic ions (composed of only one atom), the
oxidation number is equal to the charge on the ion.
Li+ = +1; Fe3+ = +3; O2- = -2
3. The oxidation number of oxygen is usually –2.
In H2O2 and O2
2- it is –1.
4.4
Signifies the number of charges the atom would have in a
molecule (or an ionic compound) if electrons were transferred
completely.
Rules to assign oxidation numbers
30
4. The oxidation number of hydrogen is +1 except when
it is bonded to metals in binary compounds. In these
cases (e.g. NaH), its oxidation number is –1.
6. The sum of the oxidation numbers of all the atoms in a
molecule or ion is equal to the charge on the
molecule or ion.
5. Group IA metals are +1, IIA metals are +2
HCO3
-
O = - 2
H = +1
For C: 3(- 2) + 1 + x = -1
x=+4 C = +4
e.g.
What are the oxidation numbers of all the elements in HCO3
- ?
7. Oxidation numbers do not have to be integers.
e.g. Oxidation number of oxygen in the superoxide ion,
O2
-, is –½.
31
The Oxidation Numbers of Elements in their Compounds
The more common oxidation numbers are in
color
32
NaIO3
Na = +1 O = -2
3(-2) + 1 + x = 0
x=+5, I = +5
IF7 F = -1
7(-1) + x = 0
x=+7, I = +7
K2Cr2O7
O = -2 K = +1
7(-2) + 2(+1) + 2x = 0
x=+6, Cr = +6
e.g. What are the oxidation numbers of all the elements in each
of these compounds?
NaIO3 IF7 K2Cr2O7
33
Types of Redox Reactions
Combination Reactions
Decomposition Reactions
A combination reaction is a reaction in which two or more
substances combine to form a single product.
Decomposition reactions are the opposite of combination
reactions, or the breakdown of a compound into two or
more components.
3
4
Combustion Reactions
A combustion reaction is a reaction in which a substance reacts with oxygen,
usually with the release of heat and light to produce a flame
Example: Burning of propane (C3H8), a component of natural gas that is used for
domestic heating and cooking:
Here, we focus only on the oxidation number of O atoms, which changes from 0 to -2.
Displacement Reactions
In a displacement reaction, an ion (or atom) in a compound is replaced by an ion
(or atom) of another element.
1. Hydrogen Displacement
All alkali metals and some alkaline earth metals (Ca, Sr, and Ba), which are
the most reactive of the metallic elements, will displace hydrogen from cold
water
C3H8(g) + 5O2(g) ⟶ 3CO2(g) + 4H2O(l)
Example next slide
35
Example for combustion rxn in the last slide
Zinc (Zn) and magnesium (Mg) do not react with cold water but do react with
hydrochloric acid, as follows:
2. Metal Displacement
A metal in a compound can be displaced by another metal in the elemental state
Example: zinc replacing copper ions and copper replacing silver ions
Metal displacement reactions find many applications, the goal of which is to
separate pure metals from their ores
e.g. vanadium is obtained by treating vanadium(V) oxide with metallic calcium
V2O5(s) + 5Ca(l) ⟶2V(l) + 5CaO(s)
37
The metals are arranged according to their ability to displace
hydrogen from an acid or water
The activity series for metals
According to this series, any
metal above hydrogen will
displace it from water or from
an acid, but metals below
hydrogen will not react with
either water or an acid
Any metal listed in the series
will react with any metal (in a
compound) below it. For
example, Zn is above Cu, so
zinc metal will displace copper
ions from copper sulfate.
38
3. Halogen Displacement. Another activity series summarizes
the halogens’ behavior in halogen displacement reactions:
F2 > Cl2 > Br2 > I2
The power of these elements as oxidizing agents decreases as
we move down Group 7A from fluorine to iodine, so molecular
fluorine can replace chloride, bromide, and iodide ions in
solution
Molecular bromine, in turn, can displace iodide ion in solution:
39
Disproportionation Reaction
an element in one oxidation state is simultaneously oxidized
and reduced.
Here the oxidation number of oxygen in the reactant (-1) both
increases to zero in O2 and decreases to -2 in H2O
This reaction describes the formation of household bleaching
agents
One reactant in a disproportionation reaction always contains an
element that can have at least three oxidation states. The
element itself is in an intermediate.
40
41
42
Concentration of solutions
The concentration of a solution is the amount of solute
present in a given quantity of solvent or solution.
Molarity (M) =
moles of solute (n)
liters of solution(V)
e.g. What mass of KI is required to make 500 mL of
a 2.80 M (mol/L) KI solution?
volume of KI solution (L) moles KI grams KI
500. mL = 232 g KI
166 g KI
1 mol KI
x
2.80 mol KI
1 L soln
x
1 L
1000 mL
x
Or directly: mass = V (L) X M (mol/L) X MM (g/mol)
43
Preparing a solution of known molarity
(a) known amount of a solid solute is transferred into the volumetric flask; then water
is added through a funnel.
(b) The solid is slowly dissolved by gently swirling the flask.
(c) After the solid has completely dissolved, more water is added to bring the
level of solution to the mark. Knowing the volume of the solution and the amount
of solute dissolved in it, we can calculate the molarity of the prepared solution.
44
45
Dilution of solutions: is the procedure for preparing a less
concentrated solution from a more concentrated solution.
Dilution
Add Solvent
Moles of solute
before dilution (i)
Moles of solute
after dilution (f)=
Mi Vi Mf Vf=
46
e.g. How would you prepare 100.0 mL of 0.200 M HCl
from a stock solution of 4.00 M HCl?
MiVi = MfVf
Mi = 4.00 M Mf = 0.200 M Vf = 0.1000 L Vi = ? L
Vi =
MfVf
Mi
= 0.200 M x 0.1000 L
4.00 M
= 0.00500 L = 5.00 mL
Dilute 5.00 mL of acid with water to a total volume
of 100.0 mL.
47
Gravimetric Analysis
an analytical technique based on the measurement of mass.
(a) A solution containing a known amount of NaCl in a beaker.
(b) The precipitation of AgCl upon the addition of AgNO3 solution from a
measuring cylinder.
(c) AgCl precipitate is filtered through a vacuum filtration, which allows the
liquid (but not the precipitate) to pass through.
Basic steps for gravimetric analysis
e.g.
Solution Stoichiometry
Examples
A Gravimetric Calculation
A 10.00 mL solution containing Cl-
was treated with excess AgNO3 to
precipitate 0.4368 g of AgCl. What was the molarity of Cl-
in the
unknown?
Solution
The formula mass of AgCl is 143.321. Precipitate weighing 0.4368 g contains
Because 1 mol of AgCl contains 1 mol of Cl-
, there must have been
3.0483 x 10-3
mol of Cl-
in the unknown.
49
Introduction to Titrations
In a titration, increments of reagent solution (the titrant and it is usually delivered
from a buret) are added to analyte until their reaction is complete.
Requirements
-Reaction must be quick
-Reaction must proceed according to a defined
equation
-Availability of indicator
-Reactions must be complete or have a large equilibrium
constant
From the quantity of titrant required, the quantity of
analyte that must have been present can be calculated
Common titrations rely on acid-base, precipitation,
oxidation-reduction, or complex formation reactions.
51
Acid-Base Titrations
In titration, a solution of accurately known concentration, called
a standard solution, is added gradually to another solution of
unknown concentration, until the chemical reaction between the
two solutions is complete.
If we know the volumes of the standard and unknown
solutions, we can calculate the concentration of the unknown
one.
The acid often chosen for standardazation of bases is
potassium hydrogen phthalate (KHP), for which the molecular
formula is KHC8H4O4 (molar mass = 204.2 g)
The reaction between KHP and sodium
hydroxide is
KHC8H4O4(aq) + NaOH(aq) → KNaC8H4O4(aq) + H2O
Indicator: substance that changes its color at/or near the
equivalence point.
Slowly add base
to unknown acid
UNTIL
the indicator
changes color
(End point)
Equivalence point: the point at which the acid has completely
reacted with or been neutralized by the base.
End point: is marked by a sudden change in a physical
property of the solution, e.g. color change of a reactant or an
external indicator added.
53
e.g. What volume of a 1.420 M NaOH solution is
required to titrate 25.00 mL of a 4.50 M H2SO4
solution?
The chemical equation
volume acid moles acid moles base volume base
H2SO4 + 2NaOH 2H2O + Na2SO4
4.50 mol H2SO4
1000 mL soln
x
2 mol NaOH
1 mol H2SO4
x
1000 ml soln
1.420 mol NaOH
x25.00 mL = 158 mL
M
acid
rxn
coef.
M
base
Other method
At the equivalence point
moles of NaOH = 2 x moles of H2SO4
(M x V)NaOH 2= (M x V)H2SO4
1.420 x VNaOH = 2 x 4.50 x 25.00
VNaOH 158= mL
55
Example: The acidic substance in vinegar is acetic acid, HC2H3O2. When
4.00 g of a certain vinegar sample was titrated with 0.200 M NaOH, 25.56
mL of the NaOH solution were required to reach the equivalence point. What
is the percent composition by mass of HC2H3O2 in the vinegar?
A) 7.7%
B) 30.7%
C) 76.8%
D) 3.8%
E) 15.6%

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Ch4 Reactions in Aqueous Solution (updated)

  • 1. Reactions in Aqueous Solution Chapter 4 Dr. Sa’ib Khouri AUM- JORDAN Chemistry By Raymond Chang
  • 2. 2 Solution is a homogenous mixture of two or more substances. Solute is the substance present in a smaller amount. (In a solution may be more than one solute) Solvent is the substance present in a larger amount. Solution Solvent Solute Soft drink (l) Air (g) Soft Solder (s) H2O N2 Pb Sugar, CO2 O2, Ar, CH4 Sn Aqueous solutions: the solute is a liquid or a solid and the solvent is water General Properties of Aqueous Solutions A solution may be gaseous (air), solid (alloy), or liquid (seawater).
  • 3. 3 Electrolyte is a substance that, when dissolved in water, results in a solution that can conduct electricity. Nonelectrolyte does not conduct electricity when dissolved in water. nonelectrolyte weak electrolyte strong electrolyte Electrolytic Properties
  • 4. 4 Examples of strong electrolytes, weak electrolytes, and nonelectrolytes. Cations (+) and Anions (-) are responsible to conduct electricity in solution. Strong Electrolyte – 100% dissociation NaCl(s) Na + (aq) + Cl - (aq) H2O
  • 5. 5 Hydration is the process in which an ion is surrounded by water molecules arranged in a specific manner. d+ d- H2O Hydration helps to stabilize ions in solution and prevents cations from combining with anions The Na+ and Cl- ions are separated from each other and undergo hydration
  • 6. 6 A reversible reaction. The reaction can occur in both directions (chemical equilibrium)
  • 7. 7 Precipitation Reactions Precipitate – insoluble solid that separates from solution Molecular equation Ionic equation Net ionic equation Pb2+ + 2NO3 - + 2K+ + 2I- PbI2 (s) + 2K+ + 2NO3 - Here K + and NO3 - are spectator ions Pb(NO3)2 (aq) + 2KI (aq) PbI2 (s) + 2KNO3 (aq) Pb 2+ + 2I - PbI2 (s) e.g. Mixing of Pb(NO3)2 and KI solutions forms an insoluble yellow product, PbI2 through a double-displacement reaction:
  • 8. 8 Precipitation of Lead Iodide Pb 2+ + 2I - PbI2 (s)
  • 9. 9 is the maximum amount of solute that will dissolve in a given quantity of solvent at a specific temperature. Solubility
  • 10. 10 Examples of Insoluble Compounds CdS PbS Ni(OH)2 Al(OH)3 Practice Exercise Classify the following ionic compounds as soluble or insoluble: (a) CuS, (b) Ca(OH)2, (c) Zn(NO3)2, (d) Ag2SO4, (e) Na3PO4
  • 11. 11 Writing Net Ionic Equations 1. Write the balanced molecular equation. 2. Write the ionic equation showing the strong electrolytes completely dissociated into cations and anions. 3. Cancel the spectator ions on both sides of the ionic equation 4. Check that charges and number of atoms are balanced in the net ionic equation AgNO3 (aq) + NaCl (aq) AgCl (s) + NaNO3 (aq) Ag+ + NO3 - + Na+ + Cl- AgCl (s) + Na+ + NO3 - Ag+ + Cl- AgCl (s) e.g. Write the net ionic equation for the reaction between silver nitrate and sodium chloride.
  • 12.
  • 13. Write balance complete molecular equation, ionic equation, and net ionic equations for the reaction that occurs when Pb(NO3)2 and Na2S solutions are mixed. Identify the spectator ions in the reaction. e.g. Molecular equation Ionic equation Net ionic equation
  • 14. General Properties of Acids and Bases Acids • Acids have a sour taste; e.g. Vinegar owes its sourness to acetic acid, and lemons and other citrus fruits contain citric acid • Acids cause color changes in plant dyes, e.g. acids change the color of litmus from blue to red. • Acids react with certain metals to produce hydrogen gas 2HCl (aq) + Mg (s) → MgCl2 (aq) + H2 (g) Acid-Base Reactions
  • 15. Acids react with carbonates and bicarbonates to produce carbon dioxide gas 2HCl (aq) + CaCO3 (s) → CaCl2 (aq) + CO2 (g) + H2O (l) HCl (aq) + NaHCO3 (s) → NaCl (aq) + CO2 (g) + H2O (l) • Aqueous acid solutions conduct electricity Bases • Have a bitter taste • Feel slippery, e.g. soaps, which contain bases, exhibit this property. • Cause color changes in plant dyes • Aqueous base solutions conduct electricity
  • 16. 16 Arrhenius acid is a substance that produces H+ (H3O+) in water Arrhenius base is a substance that produces OH- in water Bronsted Acids and Bases Arrhenius’s definitions of acids and bases apply only to aqueous solutions
  • 17. 17 A Brønsted acid is a proton donor A Brønsted base is a proton acceptor Johannes Brønsted (Danish chemist) proposed broader definitions: Note: Brønsted’s definition do not require acids and bases to be in aqueous solution NH3 (g) + HCl (g) → NH4Cl (s) Here HCl (g) is a Brønsted acid NH3 (g) is a Brønsted base
  • 18. 18 Hydrochloric acid is a Brønsted acid because it donates a proton in water: HCl (aq) → H + (aq) + Cl - (aq) H+ ion is a hydrogen atom that has lost its electron; that is, it is just a bare proton and exists in the hydrated form: The hydrated proton, H3O + , is called the hydronium ion
  • 19. 19 Monoprotic acids: yields one hydrogen ion upon ionization HCl H+ + Cl- HNO3 H+ + NO3 - CH3COOH H+ + CH3COO- Strong acid Strong acid Weak acid Diprotic acids: gives up two H + ions, in two separate steps H2SO4 H+ + HSO4 - HSO4 - H+ + SO4 2- Strong acid Weak acid Triprotic acids: yields three H+ ions, in three separate steps H3PO4 H+ + H2PO4 - H2PO4 - H+ + HPO4 2- HPO4 2- H+ + PO4 3- Weak acid Weak acid Weak acid
  • 20. 20
  • 21. 21
  • 22. 22 Acid-Base Neutralization acid + base salt + water HCl (aq) + NaOH (aq) NaCl (aq) + H2O H+ + Cl - + Na+ + OH- Na + + Cl - + H2O H + + OH - H2O A neutralization reaction is a reaction between an acid and a base. Generally, aqueous acid-base reactions produce water and a salt (ionic compound) Acid-base reactions generally go to completion. Here only if we had started the reaction with equal molar amounts of the acid and the base.
  • 23. 23 Neutralization reaction involving a weak electrolyte weak acid + base salt + water HCN (aq) + NaOH (aq) NaCN (aq) + H2O HCN + Na + + OH - Na + + CN - + H2O HCN (aq) + OH - (aq) CN - (aq) + H2O The following are also examples of acid-base neutralization reactions, represented by molecular equations: HF(aq) + KOH(aq) ⟶ KF(aq) + H2O H2SO4(aq) + 2NaOH(aq) ⟶ Na2SO4(aq) + 2H2O HNO3(aq) + NH3(aq) ⟶ NH4NO3(aq) HNO3(aq) + NH+ 4 (aq) + OH− (aq) ⟶ NH4NO3(aq) + H2O
  • 24. 24 Acid-Base Reactions Leading to Gas Formation acid + base salt + water + CO2 2HCl (aq) + Na2CO3 (aq) 2NaCl (aq) + H2O +CO2 2H + + 2Cl - + 2Na + + CO3 2- 2Na + + 2Cl - + H2O + CO2 2H+ (aq) + CO3 2- (aq) H2O + CO2 Certain salts like carbonates (containing the CO3 2- ion), bicarbonates (containing the HCO3 1- ion), sulfites (containing the SO3 2- ion), and sulfides (containing the S2- ion) react with acids to form gaseous products. Similar reactions involving other mentioned salts are
  • 25. 25
  • 26. 26 Oxidation-Reduction, or Redox ,Reactions (electron transfer reactions) 2Mg 2Mg2+ + 4e- O2 + 4e- 2O2- Oxidation half-reaction (lose e-) Reduction half-reaction (gain e-) 2Mg + O2 + 4e- 2Mg2+ + 2O2- + 4e- 2Mg + O2 2MgO Example (Two elements combine to form a compound) The formation of magnesium oxide (MgO) from magnesium and oxygen. 2Mg(s) + O2(g) ⟶2MgO(s)
  • 27. 27 Zn(s) + CuSO4 (aq) ZnSO4 (aq) + Cu (s) (Zn is oxidized)Zn Zn2+ + 2e- (Cu2+ is reduced)Cu2+ + 2e- Cu Zn is the reducing agent Cu2+ is the oxidizing agent e.g. Copper wire reacts with silver nitrate to form silver metal. What is the oxidizing agent in the reaction? Cu (s) + 2AgNO3 (aq) Cu(NO3)2 (aq) + 2Ag (s) Cu Cu2+ + 2e- Ag+ + 1e- Ag (Ag+ is reduced) Ag+ is the oxidizing agent e.g. When metallic zinc is added to a solution containing copper(II) sulfate (CuSO4),
  • 28. 28
  • 29. 2 9 Oxidation number (or oxidation state) 1. Free elements (uncombined state) have an oxidation number of zero. Na, Be, K, H2, O2, P4 = 0 2. In monatomic ions (composed of only one atom), the oxidation number is equal to the charge on the ion. Li+ = +1; Fe3+ = +3; O2- = -2 3. The oxidation number of oxygen is usually –2. In H2O2 and O2 2- it is –1. 4.4 Signifies the number of charges the atom would have in a molecule (or an ionic compound) if electrons were transferred completely. Rules to assign oxidation numbers
  • 30. 30 4. The oxidation number of hydrogen is +1 except when it is bonded to metals in binary compounds. In these cases (e.g. NaH), its oxidation number is –1. 6. The sum of the oxidation numbers of all the atoms in a molecule or ion is equal to the charge on the molecule or ion. 5. Group IA metals are +1, IIA metals are +2 HCO3 - O = - 2 H = +1 For C: 3(- 2) + 1 + x = -1 x=+4 C = +4 e.g. What are the oxidation numbers of all the elements in HCO3 - ? 7. Oxidation numbers do not have to be integers. e.g. Oxidation number of oxygen in the superoxide ion, O2 -, is –½.
  • 31. 31 The Oxidation Numbers of Elements in their Compounds The more common oxidation numbers are in color
  • 32. 32 NaIO3 Na = +1 O = -2 3(-2) + 1 + x = 0 x=+5, I = +5 IF7 F = -1 7(-1) + x = 0 x=+7, I = +7 K2Cr2O7 O = -2 K = +1 7(-2) + 2(+1) + 2x = 0 x=+6, Cr = +6 e.g. What are the oxidation numbers of all the elements in each of these compounds? NaIO3 IF7 K2Cr2O7
  • 33. 33 Types of Redox Reactions Combination Reactions Decomposition Reactions A combination reaction is a reaction in which two or more substances combine to form a single product. Decomposition reactions are the opposite of combination reactions, or the breakdown of a compound into two or more components.
  • 34. 3 4 Combustion Reactions A combustion reaction is a reaction in which a substance reacts with oxygen, usually with the release of heat and light to produce a flame Example: Burning of propane (C3H8), a component of natural gas that is used for domestic heating and cooking: Here, we focus only on the oxidation number of O atoms, which changes from 0 to -2. Displacement Reactions In a displacement reaction, an ion (or atom) in a compound is replaced by an ion (or atom) of another element. 1. Hydrogen Displacement All alkali metals and some alkaline earth metals (Ca, Sr, and Ba), which are the most reactive of the metallic elements, will displace hydrogen from cold water C3H8(g) + 5O2(g) ⟶ 3CO2(g) + 4H2O(l) Example next slide
  • 35. 35 Example for combustion rxn in the last slide
  • 36. Zinc (Zn) and magnesium (Mg) do not react with cold water but do react with hydrochloric acid, as follows: 2. Metal Displacement A metal in a compound can be displaced by another metal in the elemental state Example: zinc replacing copper ions and copper replacing silver ions Metal displacement reactions find many applications, the goal of which is to separate pure metals from their ores e.g. vanadium is obtained by treating vanadium(V) oxide with metallic calcium V2O5(s) + 5Ca(l) ⟶2V(l) + 5CaO(s)
  • 37. 37 The metals are arranged according to their ability to displace hydrogen from an acid or water The activity series for metals According to this series, any metal above hydrogen will displace it from water or from an acid, but metals below hydrogen will not react with either water or an acid Any metal listed in the series will react with any metal (in a compound) below it. For example, Zn is above Cu, so zinc metal will displace copper ions from copper sulfate.
  • 38. 38 3. Halogen Displacement. Another activity series summarizes the halogens’ behavior in halogen displacement reactions: F2 > Cl2 > Br2 > I2 The power of these elements as oxidizing agents decreases as we move down Group 7A from fluorine to iodine, so molecular fluorine can replace chloride, bromide, and iodide ions in solution Molecular bromine, in turn, can displace iodide ion in solution:
  • 39. 39 Disproportionation Reaction an element in one oxidation state is simultaneously oxidized and reduced. Here the oxidation number of oxygen in the reactant (-1) both increases to zero in O2 and decreases to -2 in H2O This reaction describes the formation of household bleaching agents One reactant in a disproportionation reaction always contains an element that can have at least three oxidation states. The element itself is in an intermediate.
  • 40. 40
  • 41. 41
  • 42. 42 Concentration of solutions The concentration of a solution is the amount of solute present in a given quantity of solvent or solution. Molarity (M) = moles of solute (n) liters of solution(V) e.g. What mass of KI is required to make 500 mL of a 2.80 M (mol/L) KI solution? volume of KI solution (L) moles KI grams KI 500. mL = 232 g KI 166 g KI 1 mol KI x 2.80 mol KI 1 L soln x 1 L 1000 mL x Or directly: mass = V (L) X M (mol/L) X MM (g/mol)
  • 43. 43 Preparing a solution of known molarity (a) known amount of a solid solute is transferred into the volumetric flask; then water is added through a funnel. (b) The solid is slowly dissolved by gently swirling the flask. (c) After the solid has completely dissolved, more water is added to bring the level of solution to the mark. Knowing the volume of the solution and the amount of solute dissolved in it, we can calculate the molarity of the prepared solution.
  • 44. 44
  • 45. 45 Dilution of solutions: is the procedure for preparing a less concentrated solution from a more concentrated solution. Dilution Add Solvent Moles of solute before dilution (i) Moles of solute after dilution (f)= Mi Vi Mf Vf=
  • 46. 46 e.g. How would you prepare 100.0 mL of 0.200 M HCl from a stock solution of 4.00 M HCl? MiVi = MfVf Mi = 4.00 M Mf = 0.200 M Vf = 0.1000 L Vi = ? L Vi = MfVf Mi = 0.200 M x 0.1000 L 4.00 M = 0.00500 L = 5.00 mL Dilute 5.00 mL of acid with water to a total volume of 100.0 mL.
  • 47. 47 Gravimetric Analysis an analytical technique based on the measurement of mass. (a) A solution containing a known amount of NaCl in a beaker. (b) The precipitation of AgCl upon the addition of AgNO3 solution from a measuring cylinder. (c) AgCl precipitate is filtered through a vacuum filtration, which allows the liquid (but not the precipitate) to pass through. Basic steps for gravimetric analysis e.g. Solution Stoichiometry
  • 48. Examples A Gravimetric Calculation A 10.00 mL solution containing Cl- was treated with excess AgNO3 to precipitate 0.4368 g of AgCl. What was the molarity of Cl- in the unknown? Solution The formula mass of AgCl is 143.321. Precipitate weighing 0.4368 g contains Because 1 mol of AgCl contains 1 mol of Cl- , there must have been 3.0483 x 10-3 mol of Cl- in the unknown.
  • 49. 49
  • 50. Introduction to Titrations In a titration, increments of reagent solution (the titrant and it is usually delivered from a buret) are added to analyte until their reaction is complete. Requirements -Reaction must be quick -Reaction must proceed according to a defined equation -Availability of indicator -Reactions must be complete or have a large equilibrium constant From the quantity of titrant required, the quantity of analyte that must have been present can be calculated Common titrations rely on acid-base, precipitation, oxidation-reduction, or complex formation reactions.
  • 51. 51 Acid-Base Titrations In titration, a solution of accurately known concentration, called a standard solution, is added gradually to another solution of unknown concentration, until the chemical reaction between the two solutions is complete. If we know the volumes of the standard and unknown solutions, we can calculate the concentration of the unknown one. The acid often chosen for standardazation of bases is potassium hydrogen phthalate (KHP), for which the molecular formula is KHC8H4O4 (molar mass = 204.2 g) The reaction between KHP and sodium hydroxide is KHC8H4O4(aq) + NaOH(aq) → KNaC8H4O4(aq) + H2O
  • 52. Indicator: substance that changes its color at/or near the equivalence point. Slowly add base to unknown acid UNTIL the indicator changes color (End point) Equivalence point: the point at which the acid has completely reacted with or been neutralized by the base. End point: is marked by a sudden change in a physical property of the solution, e.g. color change of a reactant or an external indicator added.
  • 53. 53 e.g. What volume of a 1.420 M NaOH solution is required to titrate 25.00 mL of a 4.50 M H2SO4 solution? The chemical equation volume acid moles acid moles base volume base H2SO4 + 2NaOH 2H2O + Na2SO4 4.50 mol H2SO4 1000 mL soln x 2 mol NaOH 1 mol H2SO4 x 1000 ml soln 1.420 mol NaOH x25.00 mL = 158 mL M acid rxn coef. M base
  • 54. Other method At the equivalence point moles of NaOH = 2 x moles of H2SO4 (M x V)NaOH 2= (M x V)H2SO4 1.420 x VNaOH = 2 x 4.50 x 25.00 VNaOH 158= mL
  • 55. 55 Example: The acidic substance in vinegar is acetic acid, HC2H3O2. When 4.00 g of a certain vinegar sample was titrated with 0.200 M NaOH, 25.56 mL of the NaOH solution were required to reach the equivalence point. What is the percent composition by mass of HC2H3O2 in the vinegar? A) 7.7% B) 30.7% C) 76.8% D) 3.8% E) 15.6%