General Chemistry
Principles and Modern Applications
Petrucci • Harwood • Herring
9th Edition
Chapter 5: Introduction to Reactions in
Aqueous Solutions
Yrd. Doç. Dr. Burak Esat
Fatih University
Contents
5-1
5-2
5-3
5-4
5-5
5-6
5-7
The Nature of Aqueous Solutions
Precipitation Reactions
Acid-Base Reactions
Oxidation-Reduction: Some General Principles
Balancing Oxidation-Reduction Equations
Oxidizing and Reducing Agents
Stoichiometry of Reactions in Aqueous
Solutions: Titrations
Focus on Water Treatment
5.1 The Nature of Aqueous Solutions
Electrolytes
• Some solutes can
dissociate into ions. These
solutes are called
electrolytes
• Electric charge can be
carried.
O2 molecules
as solute
The greatest number of molecules are the solvent molecules.
*In metals electrons carry the electric charge
*In aqueous solutions ions carry the
electric charge
Solutes molecules are present in much smaller numbers
1
The Electrical Conductivity of Ionic Solutions
The Role of Water as a Solvent:
The Solubility of Ionic Compounds
Electrical conductivity - The flow of electrical current in a solution is a
measure of the solubility of ionic compounds or a
measurement of the presence of ions in solution.
Electrolyte - A substance that conducts a current when dissolved in
water. Soluble ionic compound dissociate completely (?)
and may conduct a large current, and are called strong
Electrolytes.
NaCl(s) + H2O(l)
Na+(aq) + Cl -(aq)
Fig. 4.1
When sodium chloride dissolves into water the ions become solvated,
and are surrounded by water molecules. These ions are called “aqueous”
and are free to move through out the solution, and are conducting
electricity, or helping electrons to move through out the solution
Fig. 4.2
Fig. 4.3
2
Types of Electrolytes
• Strong electrolyte dissociates completely.
– Good electrical conduction.
– All soluble ionic compound (e.g. NaCl) and
few molecular compounds (e.g. HCl)
• Weak electrolyte partially dissociates.
Representation of Electrolytes using
Chemical Equations
A strong electrolyte:
MgCl2(s) → Mg2+(aq) + 2 Cl-(aq)
A weak electrolyte:
→ CH3CO2-(aq) + H+(aq)
CH3CO2H(aq) ←
– Fair conductor of electricity.
• Non-electrolyte does not dissociate.
A non-electrolyte:
– Poor conductor of electricity.
– Most molecular compounds
CH3OH(aq)
Three Types of Electrolytes
Fig. 4.4
3
Notation for Concentration
Determining Moles of Ions in Aqueous
Solutions of Ionic Compounds - I
Problem: How many moles of each ion are in each of the following:
MgCl2(s) →
Mg2+(aq)
+2
Cl-(aq)
In 0.0050 M MgCl2:
Stoichiometry is important.
[Mg2+] = 0.0050 M [Cl-] = 0.0100 M
[MgCl2] = 0 M
a)
b)
c)
d)
e)
4.0 moles of sodium carbonate dissolved in water
46.5 g of rubidium fluoride dissolved in water
5.14 x 1021 formula units of iron (III) chloride dissolved in water
75.0 ml of 0.56M scandium bromide dissolved in water
7.8 moles of ammonium sulfate dissolved in water
a) Na2CO3 (s)
H2 O
2 Na+(aq) + CO3-2(aq)
moles of Na+ = 4.0 moles Na2CO3 x
2 mol Na+
1 mol Na2CO3
= 8.0 moles Na+ and 4.0 moles of CO3-2 are present
Determining Moles of Ions in Aqueous
Solutions of Ionic Compounds - II
b)
RbF(s)
H2 O
Rb+(aq) + F -(aq)
1 mol RbF
moles of RbF = 46.5 g RbF x
104.47 g RbF
d) ScBr3 (s)
= 0.445 moles RbF
thus, 0.445 mol Rb+ and 0.445 mol F - are present
H2 O
Fe+3(aq) + 3 Cl -(aq)
c) FeCl3 (s)
moles of FeCl3 = 5,14 x 1021 formula units
x
= 0.0155 mol FeCl3
1 mol FeCl3
6.022 x 1023 formula units FeCl3
moles of Cl - = 0.0155 mol FeCl3 x 3 mol Cl
= 0.0465 mol Cl 1 mol FeCl3
and 0.0155 mol Fe+3 are also present.
Determining Moles of Ions in Aqueous
Solutions of Ionic Compounds - III
H2 O
Sc+3(aq) + 3 Br -(aq)
Converting from volume to moles:
1 L 0.56 mol ScBr3
x
= 0.042 mol ScBr3
103 ml
1L
Moles of Br - = 0.042 mol ScBr3 x 3 mol Br
= 0.126 mol Br 1 mol ScBr3
Moles of ScBr3 = 75.0 ml x
0.042 mol Sc+3 are also present
H2 O
e) (NH4)2SO4 (s)
2 NH4+(aq) + SO4- 2(aq)
2 mol NH4+
Moles of NH4+ = 7.8 moles (NH4)2SO4 x
= 15.6 mol NH4+
1 mol(NH4)2SO4
and 7.8 mol SO4- 2 are also present.
4
The Solubility of Ionic Compounds in Water
The solubility of ionic compounds in water depends upon the relative
strengths of the electrostatic forces between ions in the ionic compound
and the attractive forces between the ions and water molecules in the
solvent. There is a tremendous range in the solubility of ionic
compounds in water! The solubility of so called “insoluble” compounds
may be several orders of magnitude less than ones that are called
“soluble” in water, for example:
Solubility of NaCl in water at 20oC = 365 g/L
Solubility of MgCl2 in water at 20oC = 542.5 g/L
Solubility of AlCl3 in water at 20oC = 699 g/L
Solubility of PbCl2 in water at 20oC = 9.9 g/L
Solubility of AgCl in water at 20oC = 0.009 g/L
Solubility of CuCl in water at 20oC = 0.0062 g/L
The Solubility of Covalent Compounds in Water
The covalent compounds that are very soluble in water are the ones
with -OH group in them and are called “polar” and can have strong
polar (electrostatic)interactions with water. Examples are compound
such as table sugar, sucrose (C12H22O11); beverage alcohol, ethanol
(C2H5-OH); and ethylene glycol (C2H6O2) in antifreeze.
H
Methanol = Methyl Alcohol
H C O H
H
Other covalent compounds that do not contain a polar center, or the
-OH group are considered “nonpolar” , and have little or no
interactions with water molecules. Examples are the hydrocarbons in
gasoline and oil. This leads to the obvious problems in oil spills, where
the oil will not mix with the water and forms a layer on the surface!
Octane = C8H18
5-2 Precipitation Reactions
• Soluble ions can combine to form an
insoluble compound.
• Precipitation occurs.
Ag+(aq) + I-(aq) → AgI(s)
and / or
Benzene = C6H6
Net Ionic Equation
Overall Precipitation Reaction:
AgNO3(aq) +NaI (aq) → AgI(s) + NaNO3(aq)
Complete ionic equation:
Spectator ions
Ag+(aq) + NO3-(aq) + Na+(aq) + I-(aq) →
AgI(s) + Na+(aq) + NO3-(aq)
Net ionic equation:
Ag+(aq) + I-(aq) → AgI(s)
5
Silver Nitrate and Sodium Iodide
AgNO3(aq)
NaI(aq)
AgI(s)
Na+(aq) NO3-(aq)
Fig. 4.6
Solubility Rules
• Compounds that are mostly soluble:
• Compounds that are soluble:
– Alkali metal ion and ammonium ion salts
Li+, Na+, K+, Rb+, Cs+
NH4+
– Nitrates, perchlorates and acetates
NO3-
ClO4-
Solubility Rules
– Chlorides, bromides and iodides
Cl-, Br-, I-
• Except those of Pb2+, Cu+,Ag+, and Hg22+.
– Sulfates
SO42-
• Except those of Sr2+, Ba2+, Pb2+ and Hg22+.
• Ca(SO4) is slightly soluble.
CH3CO2-
6
Precipitation Reactions: A Solid Product is Formed
Solubility Rules
When ever two aqueous solutions are mixed, there is the possibility of
forming an insoluble compound. Let us look at some examples to see
how we can predict the result of adding two different solutions together.
• Compounds that are insoluble:
– Hydroxides and sulfides
HO-, S2-
• Except alkali metal (Group 1A) and ammonium salts
• Sulfides of alkaline earths are soluble (Group 2A)
• Hydroxides of Cu+, Ba2+, Sr2+ and Ca2+ are slightly
soluble.
– Carbonates ,phosphates, sulfites, oxalates and
chromates
– CO32-, PO43-,SO32-, C2O42-, CrO42• Except alkali metal and ammonium salts
Precipitation Reactions: Will a Precipitate Form?
Pb(NO3) (aq) + NaI(aq)
Pb+2(aq) + 2 NO3-(aq) + Na+(aq) + I-(aq)
When we add these two solutions together, the ions can combine in the
way they came into the solution, or they can exchange partners. In this
case we could have lead nitrate and sodium iodide, or lead iodide and
sodium nitrate formed, to determine which will happen we must look at
the solubility table to determine what could form. The table
indicates that lead iodide will be insoluble, so a precipitate will form.
Pb(NO3)2 (aq) + 2 NaI(aq)
PbI2 (s) + 2 NaNO3 (aq)
5-3 Acid-Base Reactions
If we add a solution containing potassium chloride to a solution
containing ammonium nitrate, will we get a precipitate?
KCl(aq) + NH4NO3 (aq)
=
K+(aq) + Cl-(aq) + NH4+(aq) + NO3-(aq)
By exchanging cations and anions we see that we could have potassium
chloride and ammonium nitrate, or potassium nitrate and ammonium
chloride. In looking at the solubility table it shows all possible
products as soluble, so there is no net reaction!
• Latin acidus (sour)
– Sour taste
• Arabic al-qali (ashes of certain plants)
– Bitter taste
KCl(aq) + NH4NO3 (aq) = No Reaction!
If we mix a solution of sodium sulfate with a solution of barium nitrate,
will we get a precipitate? From the solubility table it shows that barium
sulfate is insoluble, therefore we will get a precipitate!
Na2SO4 (aq) + Ba(NO3)2 (aq)
• Svante Arrhenius 1884 Acid-Base theory.
BaSO4 (s) + 2 NaNO3 (aq)
7
Acids
Solvation: The Hydrated Proton
• Acids provide H+ in aqueous solution.
• Strong acids:
HCl(aq)
→
H+(aq) + Cl-(aq)
←
→
H+(aq) + CH3CO2-(aq)
• Weak acids:
CH3CO2H(aq)
Bases
Recognizing Acids and Bases.
• Bases provide OH- in aqueous solution.
– CH3CO2H or HC2H3O2
• Strong bases:
NaOH(aq)
• Acids have ionizable hydrogen ions.
HO
Na+(aq) + OH-(aq)
→
2
• Weak bases:
NH3(aq) + H2O(l)
• Bases have OH- combined with a metal ion.
KOH
or are identified by chemical equations
←
→
OH-(aq) + NH4+(aq)
Na2CO3(s) + H2O(l)→ HCO3-(aq) + 2 Na+(aq) + OH-(aq)
8
Selected Acids and Bases
Acid - Base Reactions: Neutralization Rxns.
An Acid is a substance that produces H+ (H3O+) ions when dissolved
in water.
A Base is a substance that produces OH - ions when dissolved in water.
Acids and bases are electrolytes, and their strength is categorized in
terms of their degree of dissociation in water to make hydronium or
hydroxide ions. Strong acids and bases dissociate completely, and are
strong electrolytes. Weak acids and bases dissociate weakly and are
weak electrolytes.
The generalized reaction between an Acid and a Base is:
HX(aq) + MOH(aq)
Acid
+
Base
MX(aq) + H2O(L)
=
Salt
+
Water
Acids
Bases
Strong
Strong
Hydrochloric acid, HCl
Hydrobromic acid, HBr
Hydriodic acid, HI
Nitric acid, HNO3
Sulfuric acid, H2SO4
Perchloric acid, HClO4
Weak
Weak
Hydrofluoric acid, HF
Phosphoric acid, H3PO4
Acetic acid, CH3COOH
(or HC2H3O2)
Carbonic acid H2CO3
Ammonia, NH3
Acetate ion, CH3COOCarbonate ion CO3-2
Bicarbonate ion HCO3-
Writing Balanced Equations for
Neutralization Reactions - I
Problem: Write balanced chemical reactions (molecular, total ionic, and
net ionic) for the following chemical reactions:
a) calcium hydroxide(aq) and hydriodic acid(aq)
b) lithium hydroxide(aq) and nitric acid(aq)
c) barium hydroxide(aq) and sulfuric acid(aq)
Plan: These are all strong acids and bases, therefore they will make
water and the corresponding salts.
Solution:
a)
Ca(OH)2 (aq) + 2HI(aq)
CaI2 (aq) + 2H2O(l)
Ca2+(aq) + 2 OH -(aq) + 2 H+(aq) + 2 I -(aq)
Ca2+(aq) + 2 I -(aq) + 2 H2O(l)
2 OH -(aq) + 2 H+(aq)
Sodium hydroxide, NaOH
Potassium hydroxide, KOH
Calcium hydroxide, Ca(OH)2
Strontium hydroxide, Sr(OH)2
Barium hydroxide, Ba(OH)2
Writing Balanced Equations for
Neutralization Reactions - II
b) LiOH(aq) + HNO3
(aq)
LiNO3 (aq) + H2O(l)
Li+(aq) + OH -(aq) + H+(aq) + NO3-(aq)
Li+(aq) + NO3-(aq) + H2O(l)
OH -(aq) + H+(aq)
c)
Ba(OH)2 (aq) + H2SO4 (aq)
Ba2+(aq) + 2 OH -(aq) + 2 H+(aq) + SO42-(aq)
H2O(l)
BaSO4 (s) + 2 H2O(l)
BaSO4 (s) + 2 H2O(l)
2 H2O(l)
9
5-7 Stoichiometry of Reactions in
Aqueous Solutions: Titrations.
An Acid-Base Titration
• Titration
– Carefully controlled addition of one solution to
another.
• Equivalence Point
– Both reactants have reacted completely.
• Indicators
– Substances which change colour near an
equivalence point.
Finding the Concentration of Base from
an Acid - Base Titration - I
Problem: A titration is performed between sodium hydroxide and
potassium hydrogenphthalate (KHP) to standardize the base solution,
by placing 50.00 mg of solid potassium hydrogenphthalate in a flask
with a few drops of an indicator. A buret is filled with the base, and the
initial buret reading is 0.55 ml; at the end of the titration the buret
reading is 33.87 ml. What is the concentration of the base?
Plan: Use the molar mass of KHP (204.2 g/mol) to calculate the
number of moles of the acid, from the balanced chemical equation,
the reaction is equal molar (moles of acid= moles of base) , so we
know the moles of base, and we can determine the volume from the
difference in the buret readings, we can calculate the molarity of the
base.
Solution:
HKC8H4O4 (aq) + OH -(aq)
KC8H4O4-(aq) + H2O(aq)
Potassium Hydrogenphthalate HKC8H4O4
O
O
C
C
O K+
O
O
C
O
K+
O
H
C
O
H+
10
Finding the Concentration of Acid from
an Acid - Base Titration
Volume (L) of base (difference in
buret readings)
Finding the Concentration of Base from
an Acid - Base Titration - II
moles KHP =
M (mol/L) of base
= 0.00024486 mol KHP
Volume of base = Final buret reading - Initial buret reading
= 33.87 ml - 0.55 ml = 33.32 ml of base
Moles of base
molar ratio
mole of acid = mole of base; therefore 0.00024486 moles of
acid will react with 0.00024486 moles of base in a volume of 33.32 ml.
Moles of acid
volume (L) of acid
molarity of base =
M (mol/L) of acid
Acid + Base
0.00024486 moles
= 0.07348679 moles per liter
0.03332 L
molarity of base = 0.07349 M
An Aqueous Strong Acid-Strong Base
Reaction on the Atomic Scale
Fig. 4.9
1.00 g
50.00 mg KHP
x
1000 mg
204.2 g KHP
1 mol KHP
Salt + H2O
An Acid-Base Reaction That Forms
a Gaseous Product
The reaction of acid with carbonates or bicarbonates will
produce carbon dioxide gas that is released from solution
as a gas in the form of bubbles that leave the solution.
Fig. 4.10
11
More Acid-Base Reactions
Gas Forming Reactions
• Limestone and marble.
CaCO3(s) + 2 H+(aq) → Ca2+(aq) + H2CO3(aq)
But: H2CO3(aq) → H2O(l) + CO2(g)
CaCO3(s) + 2 H+(aq) → Ca2+(aq) + H2O(l) + CO2(g)
2NO2- + 2H+
2HNO2
CN- + H+
HCN (g)
NO(g) + NO2 (g) + H2O
12