Chemistry 1A
Chapter 5
Water, H2O
Water
Attractions
Liquid
Water
Solutions
• A solution, also called a
homogeneous mixture, is a mixture
whose particles are so evenly
distributed that the relative
concentrations of the components
are the same throughout.
• Water solutions are called
aqueous solutions.
Solution (Homogeneous
Mixture)
Solute and Solvent
• In solutions of solids dissolved in
liquids, we call the solid the solute
and the liquid the solvent.
• In solutions of gases in liquids, we
call the gas the solute and the
liquid the solvent.
• In other solutions, we call the minor
component the solute and the
major component the solvent.
Solution of an Ionic Compound
Solution of an Ionic
Compound (cont.)
Liquid-Liquid Solution
Precipitation Reactions
• In a precipitation reaction, one
product is insoluble in water.
• As that product forms, it emerges,
or precipitates, from the solution as
a solid.
• The solid is called a precipitate.
• For example,
Ca(NO3)2(aq) + Na2CO3(aq)
→ CaCO3(s) + 2NaNO3(aq)
Precipitation Questions
•
•
•
•
Describe the solution formed at
the instant water solutions of two
ionic compounds are mixed
(before the reaction takes
place).
Describe the reaction that takes
place in this mixture.
Describe the final mixture.
Write the complete equation for
the reaction.
Solution of Ca(NO3)2
Solution of Ca(NO3)2 and Na2CO3 at the
time of mixing, before the reaction
Product Mixture for the reaction
of Ca(NO3)2 and Na2CO3
Complete Ionic Equation
Spectator Ions
• Ions that are important for
delivering other ions into solution
but that are not actively involved in
the reaction are called spectator
ions.
• Spectator ions can be recognized
because they are separate and
surrounded by water molecules
both before and after the reaction.
Net Ionic Equations
• An equation written without
spectator ions is called a net
ionic equation.
Ca2+(aq) + CO32−(aq) → CaCO3(s)
Writing Precipitation
Equations
• Step 1: Determine the formulas for 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. If
either possible product is insoluble, a
precipitation reaction takes place, and
you may continue with step 3. If neither is
insoluble, write “No reaction”.
Water Solubility
• Ionic compounds with the following ions are
soluble.
– NH4+, group 1 metal ions, NO3−, and C2H3O2−
• Ionic compounds with the following ions are
usually soluble.
– Cl−, Br−, I− except with Ag+ and Pb2+
– SO42− except with Ba2+ and Pb2+
• Ionic compounds with the following ions are
insoluble.
– CO32−, PO43−, and OH− except with NH4+ and
group 1 metal cations
– S2− except with NH4+ and group 1 and 2 metal
cations
Writing Precipitation
Equations (cont)
• Step 3: Follow these steps to
write the complete equation.
– Write the formulas for the reactants
separated by a “+”.
– Separate the formulas for the reactants and
products with a single arrow.
– Write the formulas for the products separated
by a “+”.
– Write the physical state for each formula.
• The insoluble product will be followed by (s).
• Water-soluble ionic compounds will be followed by
(aq).
– Balance the equation.
Writing Precipitation
Equations (cont)
• Write the complete ionic equation.
– Describe aqueous ionic compounds
as ions.
– Describe the solid with a complete
formula.
– Make sure that you have correctly
done the following
• Balanced the equation
• Included charges on ions
• Included states
Writing Precipitation
Equations (cont)
• Write the net ionic equation.
– Eliminate ions that are in the
same form on each side of the
complete ionic equation.
– Rewrite what’s left and balance.
Skills to Master (1)
• Convert between names and
symbols for the common elements.
• Identify whether an element is a
metal or a nonmetal.
• Determine the charges on many of
the monatomic ions.
• Convert between the name and
formula for polyatomic ions.
Skills to Master (2)
• Convert between the name and
formula for ionic compounds.
• Balance chemical equations.
• Predict the products of double
displacement reactions.
• Predict ionic solubility.
Arrhenius Acid
Definition
• An acid is a substance that
generates hydronium ions, H3O+
(often described as H+), when
added to water.
• An acidic solution is a solution
with a significant concentration of
H3O+ ions.
Characteristics of
Acids
• Acids have a sour taste.
• Acids turn litmus from blue to red.
• Acids react with bases.
Strong and Weak Acids
• Strong Acid = due to a completion
reaction with water, generates close
to one H3O+ for each acid molecule
added to water.
• Weak Acid = due to a reversible
reaction with water, generates
significantly less than one H3O+ for
each molecule of acid added to
water.
Strong Acid and Water
When HCl dissolves in water, hydronium
ions, H3O+, and chloride ions, Cl−, ions form.
Solution of a Strong Acid
Acetic Acid
Weak Acid and Water
Acetic acid reacts with water in a reversible
reaction, which forms hydronium and acetate ions.
Solution of Weak Acid
Strong
and
Weak
Acids
Recognizing Acids
• From name
– (something) acid
– (metal) hydrogen sulfate
– (metal) dihydrogen phosphate
• From formula
– HX(aq), HaXbOc, HC2H3O2, RCO2H
(RCOOH)
– (symbol for metal)HSO4
– (symbol for metal)H2PO4
Strong and Weak Acids
• Strong Acids
– Monoprotic – HCl(aq), HBr(aq),
HI(aq), HNO3, HClO4
– H2SO4
• Weak Acid
– The rest
Arrhenius Base
Definitions
• A base is a substance that
generates OH− when added to
water.
• A basic solution is a solution with
a significant concentration of OH−
ions.
Characteristics of
Bases
• Bases have a bitter taste.
• Bases feel slippery on your fingers.
• Bases turn litmus from red to blue.
• Bases react with acids.
Strong Bases
• Anions in ionic compounds
except
– Neutral anions - Cl−, Br−, I−, NO3−,
ClO4−
– Acidic anions – HSO4−, H2PO4−
• Uncharged, molecular bases –
NH3
Strong Bases
• Strong Base = due to a completion reaction with
water, generates close to one (or more) OH− for
each formula unit of base added to water.
– Metal hydroxides (e.g. NaOH)
NaOH(s) → Na+(aq) + OH −(aq)
– Metal hydrides (e.g. LiH)
LiH(s)
→
Li+(aq) + H−(aq)
→
H−(aq) + H2O(l)
H2(aq) + OH−(aq)
– Metal amides (e.g. NaNH2)
NaNH2(s)
→
Na+(aq) + NH2−(aq)
NH2−(aq) + H2O(l)
→
NH3(aq) + OH−(aq)
Weak Base
• Weak Base = due to a reversible
reaction with water, generates
significantly less than one OH− for
each formula unit of base added to
water.
– All other bases
Ammonia and Water
Ammonia reacts with water in a reversible
reaction, which forms ammonium and hydroxide
ions.
Ammonia Solution
pH
• Acidic solutions have pH values
less than 7, and the more acidic
the solution is, the lower its pH.
• Basic solutions have pH values
greater than 7, and the more
basic the solution is, the higher its
pH.
pH Range
Neutralization
Reactions
• Reactions between Arrhenius
acids and Arrhenius bases are
called neutralization reactions.
HNO3(aq) + NaOH(aq)
→ H2O(l) + NaNO3(aq)
Aqueous Nitric Acid
Mixture of HNO3 and NaOH Before Reaction.
Strong Acid and Strong
Base Reaction
The hydronium ion, H3O+, from the strong acid
reacts with the hydroxide ion, OH−, from the
strong base to form water, H2O.
Mixture of HNO3 and NaOH
After the Reaction
Electrolytes
• Strong electrolytes ionize (strong
acids) or dissociate (water-soluble
ionic compounds) completely when
added to water, causing the water to
conduct electric currents strongly.
• Weak electrolytes ionize (weak acids
and ammonia) incompletely when
added to water, causing the water to
conduct electric currents weakly.
• Nonelectrolytes (such as alcohols
and sugars) do not form ions in
solution, and therefore, do not cause
water to conduct electric currents.
Steps to Neutralization
Equations
• Do you have an acid and a base?
• If you have a strong acid or a strong
base or if both are strong, write a single
arrow. If both are weak, write a double
arrow.
• Write the formulas and states for the
products.
– If the base is ammonia,
NH3(aq) + HX(aq) → NH4X(aq)
– Otherwise,
AB + CD → AD + CB
• Balance the complete equation.
Steps to Neutralization
Equations (2)
• Write the complete ionic equation.
– Describe strong electrolytes as ions.
– Describe everything else with a
complete formula.
– Make sure that you have correctly
done the following
• Balanced the equation
• Included charges on ions
• Included states
Steps to Neutralization
Equations (3)
• Write the net ionic equation.
– Eliminate ions that are in the same
form on each side of the complete
ionic equation.
– Rewrite what’s left and balance.
Steps to Neutralization
Equations (4)
• Check to be sure the following is
true.
– Strong acids described as H+.
– Weak acids described with a complete
formula.
– H2SO4 described as H+ and HSO4−.
– Pure (s), (l), or (g) described with
complete formula.
– Ions in strong electrolytes on both
sides of the equation are eliminated.
Reaction between an Acid
and a Hydroxide Base.
• The reaction has the double
displacement form.
AB + CD → AD + CB
– The positive part of the acid is H+.
• The hydroxide base can be soluble
or insoluble.
• The products are water and a
water-soluble ionic compound.
Reaction between an Acid
and a Carbonate Base.
• The reaction has the double
displacement form.
AB + CD → AD + CB
– The positive part of the acid is H+.
• The products are water, carbon
dioxide, and a water-soluble ionic
compound. The H2O and the CO2
come from the decomposition of the
initial product H2CO3.
Arrhenius Acid-Base
Reactions?
NH3(aq) + HF(aq)
base
acid
H2O(l) + HF(aq)
neutral
acid
NH3(aq) + H2O(l)
base
neutral
H2PO4−(aq) + HF(aq)
acid
acid
NH4+(aq) + F−(aq)
H3O+(aq) + F−(aq)
NH4+(aq) + OH−(aq)
H3PO4(aq) + F−(aq)
Acid and Base Definitions
• Acid
– Arrhenius: a substance that generates
H3O+ in water
– Brønsted-Lowry: a proton, H+, donor
• Base
– Arrhenius: a substance that generates OHin water
– Brønsted-Lowry: a proton, H+, acceptor
• Acid-Base Reaction
– Arrhenius: between an Arrhenius acid and
base
– Brønsted-Lowry: a proton (H+) transfer
Brønsted-Lowry Acids and
Bases
NH3(aq) + HF(aq)
base
acid
H2O(l) + HF(aq)
base
acid
NH3(aq) + H2O(l)
base
acid
H2PO4−(aq) + HF(aq)
base
acid
NH4+(aq) + F−(aq)
H3O+(aq) + F−(aq)
NH4+(aq) + OH−(aq)
H3PO4(aq) + F−(aq)
Why Two Definitions for
Acids and Bases? (1)
• Positive Aspects of Arrhenius Definitions
– All isolated substances can be classified as
acids (generate H3O+ in water), bases
(generate OH- in water), or neither.
– Allows predictions, including (1) whether
substances will react with a base or acid, (2)
whether the pH of a solution of the substance
will be less than 7 or greater than 7, and (3)
whether a solution of the substance will be
sour.
• Negative Aspects of Arrhenius Definitions
– Does not include similar reactions (H+ transfer
reactions) as acid-base reactions.
Why Two Definitions for
Acids and Bases? (2)
• Positive Aspects of Brønsted-Lowry
Definitions
– Includes similar reactions (H+ transfer reactions) as
acid-base reactions.
• Negative Aspects of Brønsted-Lowry
Definitions
– Cannot classify isolated substances as acids (generate
H3O+ in water), bases (generate OH− in water), or
neither. The same substance can sometimes be an
acid and sometimes a base.
– Does not allow predictions of (1) whether substances
will react with a base or acid, (2) whether the pH of a
solution of the substance will be less than 7 or greater
than 7, and (3) whether a solution of the substance will
be sour.
Conjugate Acid-Base Pairs
Brønsted-Lowry Acids and
Bases
NH3(aq) + HF(aq)
base
acid
H2O(l) + HF(aq)
base
acid
NH3(aq) + H2O(l)
base
acid
H2PO4−(aq) + HF(aq)
base
acid
NH4+(aq) + F−(aq)
acid
base
H3O+(aq) + F−(aq)
acid
base
NH4+(aq) + OH−(aq)
acid
base
H3PO4(aq) + F−(aq)
acid
base
Amphoteric Substances
Can be a Brønsted-Lowry acid in one reaction and
a Brønsted-Lowry base in another?
HCO3−(aq) + HF(aq)
CO2(g) + H2O(l) + F−(aq)
base
acid
HCO3−(aq) + OH−(aq)
CO32−(aq) + H2O(l)
acid
base
H2PO4−(aq) + HF(aq)
H3PO4(aq) + F−(aq)
base
acid
H2PO4−(aq) + 2OH−(aq) → PO43−(aq) + 2H2O(l)
acid
base
Oxidation
• Historically, oxidation meant
reacting with oxygen.
2Zn(s) + O2(g)
→
2ZnO(s)
Zn → Zn2+ + 2e−
or 2Zn → 2Zn2+ + 4e−
O + 2e− → O2−
or O2 + 4e− → 2O2−
Oxidation Redefined (1)
• Many reactions that are similar to
the reaction between zinc and
oxygen were not considered
oxidation.
• For example, both the zinc-oxygen
reaction and the reaction between
sodium metal and chlorine gas
(described on the next slide) involve
the transfer of electrons.
Oxidation and Formation of
Binary Ionic Compounds
Similar to Oxidation
of Zinc
2Na(s) + Cl2(g)
Na
→
2NaCl(s)
→ Na+ + e−
or 2Na → 2Na+ + 2e−
Cl + e− → Cl−
or Cl2 + 2e− →
2Cl−
Oxidation = Loss of Electrons
Oxidation Redefined (2)
• To include the similar reactions in
the same category, oxidation
was redefined as any chemical
change in which at least one
element loses electrons.
Zinc Oxide Reduction
• The following equation describes one of the
steps in the production of metallic zinc.
ZnO(s) + C(g) → Zn(s) + CO(g)
• Because zinc is reducing the number of bonds
to oxygen atoms, historically, zinc was said to
be reduced.
• When we analyze the changes taking place, we
see that zinc ions are gaining two electrons to
form zinc atoms.
Zn2+ + 2e− → Zn
• The definition of reduction was broadened to
coincide with the definition of oxidation.
According to the modern definition, when
something gains electrons, it is reduced.
Reduction
• The loss of electrons (oxidation) by
one substance is accompanied by
the gain of electrons by another
(reduction). Reduction is any
chemical change in which at least
one element gains electrons.
Memory Aid
Oxidizing and
Reducing Agents
• A reducing agent is a substance that
loses electrons, making it possible for
another substance to gain electrons and
be reduced. The oxidized substance is
always the reducing agent.
• An oxidizing agent is a substance that
gains electrons, making it possible for
another substance to lose electrons and
be oxidized. The reduced substance is
always the oxidizing agent.
Identifying Oxidizing and
Reducing Agents
2Zn(s) + O2(g)
→
2ZnO(s)
Zn → Zn2+ + 2e−
O + 2e− → O2−
• Zinc atoms lose electrons, making it possible
for oxygen atoms to gain electrons and be
reduced, so zinc is the reducing agent.
• Oxygen atoms gain electrons, making it
possible for zinc atoms to lose electrons and
be oxidized, so O2 is the oxidizing agent.
Partial Loss and Gain
of Electrons
N2(g)
+ O2(g) → 2NO(g)
• The N-O bond is a polar covalent bond
in which the oxygen atom attracts
electrons more than the nitrogen atom.
• Thus the oxygen atoms gain electrons
partially and are reduced.
• The nitrogen atoms lose electrons
partially and are oxidized.
• N2 is the reducing agent.
• O2 is the oxidizing agent.
Redox Terms (1)
Redox Terms (2)
• Oxidation-Reduction Reaction
– an electron transfer reaction
• Oxidation
– complete or partial loss of electrons
• Reduction
– complete or partial gain of electrons
• Oxidizing Agent
– the substance reduced; gains electrons,
making it possible for something to lose them.
• Reducing Agent
– the substance oxidized; loses electrons,
making it possible for something to gain them.
Questions Answered by
Oxidation Numbers
Is the reaction
redox?
If any atoms change their
oxidation number, yes.
What’s oxidized?
The element that increases its
oxidation number
What’s reduced?
The element that decreases its
oxidation number
What’s the
reducing agent?
The substance with the element
oxidized
The substance with the element
reduced
What’s the
oxidizing agent?
Steps for Determination
of Oxidation Numbers
• Step 1: Assign oxidation numbers to as
many atoms as you can using the
guidelines described on the next slide.
• Step 2: To determine oxidation numbers
for atoms not described on the pervious
slide, use the following guideline.
– The sum of the oxidation numbers for each
atom in the formula is equal to the overall
charge on the formula. (This includes
uncharged formulas where the sum of the
oxidation numbers is zero.)
Oxidation Numbers
uncharged element
0
no exceptions
monatomic ions
no exceptions
combined fluorine
charge
on ion
-1
no exceptions
combined oxygen
-2
-1 in peroxides
covalently bonded
hydrogen
+1
no exceptions
Single Displacement
Single Displacement
Reaction
Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s)
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
oxidation:
Zn(s) → Zn2+(aq) + 2e−
reduction:
Cu2+(aq) + 2e− → Cu(s)
Single Displacement
Reaction Example
Voltaic Cell
• The system in which two halfreactions for a redox reaction are
separated allowing the electrons
transferred in the reaction to be
passed between them through a
wire is called voltaic cell.
Voltaic Cell
Electrodes
• The electrical conductors placed in
the half-cells are called electrodes.
• They can be active electrodes,
which participate in the reaction, or
passive electrodes, which transfer
the electrons into or out of a halfcell but do not participate in the
reaction.
Anode
• The anode is the site of oxidation.
• Because oxidation involves loss of
electrons, the anode is the source of
electrons. For this reason, it is described
as the negative electrode.
• Because electrons are lost forming more
positive (or less negative) species at the
anode, the surroundings tend to become
more positive. Thus anions are attracted
to the anode.
Cathode
• The cathode is the site of reduction.
• By convention, the cathode is the
positive electrode.
• Because electrons come to the cathode
and substances gain these electrons to
become more negative (or less positive),
the surroundings tend to become more
negative. Thus cations are attracted to
the cathode.
Other Cell
Components
• A device called a salt bridge can
be used to keep the charges
balanced.
• The portion of the
electrochemical cell that allows
ions to flow is called the
electrolyte.
Leclanché Cell
or Dry Cell
Anode oxidation:
Zn(s) → Zn2+(aq) + 2e−
Cathode reduction:
2MnO2(s) + 2NH4+(aq) + 2e−
→ Mn2O3(s) + 2NH3(aq) + H2O(l)
Overall reaction:
Zn(s) + 2MnO2(s) + 2NH4+(aq)
→ Zn2+(aq) + Mn2O3(s) + 2NH3(aq) + H2O(l)
Dry Cell Image
Alkaline Batteries
Anode oxidation:
Zn(s) + 2OH−(aq)
→ ZnO(s) + H2O(l) + 2e−
Cathode reduction:
2MnO2(s) + H2O(l) + 2e−
→ Mn2O3(s) + 2OH−(aq)
Overall reaction:
Zn(s) + 2MnO2(s) → ZnO(s) + Mn2O3(s)
Electrolysis
• Voltage, a measure of the strength of an
electric current, represents the force that moves
electrons from the anode to the cathode in a
voltaic cell.
• When a greater force (voltage) is applied in the
opposite direction, electrons can be pushed
from what would normally be the cathode
toward the voltaic cell’s anode. This process is
called electrolysis.
• In a broader sense, electrolysis is the process
by which a redox reaction is made to occur in
the nonspontaneous direction.
2NaCl(l) → 2Na(l) + Cl2(g)
Primary and
Secondary Batteries
• Batteries that are not
rechargeable are called primary
batteries.
• A rechargeable battery is often
called a secondary battery or a
storage battery.
Nickel-Cadmium
Battery
Anode reaction:
Cd(s) + 2OH−(aq)
Cd(OH)2(s) + 2e−
Cathode reaction:
NiO(OH)(s) + H2O(l) + e−
Ni(OH)2(s) + OH−(aq)
Net Reaction:
Cd(s) + 2NiO(OH)(s) + 2H2O(l)
Cd(OH)2(s) + 2Ni(OH)2(s)
Lead Acid Battery
Pb(s) + HSO4−(aq) + H2O(l)
PbSO4(s) + H3O+(aq) + 2e−
Cathode reaction:
PbO2(s) + HSO4−(aq) + 3H3O+(aq) + 2e−
PbSO4(s) + 5H2O(l)
Net Reaction:
Pb(s) + PbO2(s) + 2HSO4−(aq) + 2H3O+(aq)
2PbSO4(s) + 4H2O(l)
Conversions to Moles
Molarity
• Converts between moles of solute and
volume of solution
Equation Stoichiometry (1)
Equation
Stoichiometry (1)
• Tip-off - The calculation calls for you to
convert from amount of one substance to
amount of another, both of which are
involved in a chemical reaction.
• General Steps
1. If you are not given it, write and balance
the chemical equation for the reaction.
2. Start your dimensional analysis in the
usual way.
Equation
Stoichiometry (2)
3. Convert from the units that you are
given for substance 1 to moles of
substance 1.
– For pure solids and liquids, this means
converting mass to moles using the
molar mass of the substance.
– Molarity can be used to convert from
volume of solution to moles of solute.
Equation
Stoichiometry (3)
4. Convert from grams of substance 1 to moles
of substance 1.
5. Convert from moles of substance 2 to the
desired units for substance 2.
–
–
For pure solids and liquids, this means converting
moles to mass using the molar mass of substance
2.
Molarity can be used to convert from moles of
solute to volume of solution.
6. Calculate your answer and report it with the
correct significant figures (in scientific
notation, if necessary) and unit.
Titration
• Titration involves the addition of one
solution (solution 1) to another solution
(solution 2) until a chemical reaction
between the components in the solutions
is complete. Solution 1 is called the
titrant, and we say that it is used to
titrate solution 2. The completeness of
the reaction is usually shown by a
change of color caused by a substance
called an indicator.
Titration
Apparatus
Steps for Titration (1)
• A specific volume of the solution
to be titrated (solution 2) is added
to an Erlenmeyer flask. For
example, 25.00 mL of a
phosphoric acid solution of
unknown concentration might be
added to a 250-mL Erlenmeyer
flask.
Steps for Titration (2)
• A solution of a substance that reacts with
the solute in the solution in the
Erlenmeyer flask is added to a buret.
This solution in the buret, which has a
known concentration, is the titrant. The
buret is set up over the Erlenmeyer flask
so the titrant can be added to the
solution to be titrated. For example, a
1.02 M NaOH solution might be added to
a buret, which is set up over the
Erlenmeyer flask containing the
phosphoric acid solution.
Steps for Titration (3)
• An indicator is added to the
solution being titrated. The
indicator is a substance that
changes color when the reaction is
complete. In our example,
phenolphthalein, which is a
common an acid-base indicator, is
added to the phosphoric acid
solution in the Erlenmeyer flask.
Phenolphthalein
• Phenolphthalein is a substance
that has two forms. In acidic
conditions, it is in the acid form,
which is colorless. In basic
conditions, an H+ ion is removed
from each phenolphthalein
molecule, converting it to its base
form, which is red.
Steps for Titration (4)
• The titrant is slowly added to the
solution being titrated until the indicator
changes color, showing that the
reaction is complete. This stage in the
procedure is called the endpoint. In
our example, the NaOH solution is
slowly added from the buret until the
mixture in the Erlenmeyer flask
changes from colorless to red. At this
point, 34.2 mL of 1.02 M NaOH has
been added.
Making Solutions
• From pure solids
• From a pure or almost pure acid
(e.g. 17 M HC2H3O2 or 18 M
H2SO4)
• From any other more
concentrated solution
Making a Solution
from Pure Solid
Making a Solution From
Concentrated Acid
Add Concentrated Acid
to Water
Dilution Problems
mol solute concentrated = mol solute dilute
⎛ --- mol solute conc ⎞
--- L conc soln ⎜
⎟ = --- L dil soln
L conc soln
⎝
⎠
Vconc M conc = Vdilute M dilute
M C VC = M D VD
⎛ --- mol solute dil ⎞
⎜
⎟
L dil soln
⎝
⎠
Making a Solution from a More
Concentrated Solution (not
concentrated acid)