MODULE 2
AQUEOUS REACTIONS
Lesson 1 – Solubility and Precipitation Reactions
Introduction
Precipitation reactions are one type of aqueous reaction that is
important in both nature and industry. The formation of structures
in limestone caves and some gemstones are the result of
precipitation reactions. The scales that form in hot water pipes or
kettles are also the result of precipitation reactions. Water
purification plants often use precipitate formation followed by
filtration as a method for removing impurities from drinking water.
In this lesson we will determine which substances tend to form
precipitates in water.
Learning Outcomes
When you have completed this lesson, you will be able to:

Operationally define precipitate.

Identify ionic substances that precipitate from aqueous
solution using experimentation.

Write balanced equations for double displacement and netionic precipitation reactions.
Aqueous Solutions of Ionic Compounds
If you recall from Chemistry 30S, water is a polar molecule because
the hydrogen atoms and the oxygen atom do not share electrons
evenly. Oxygen is more electronegative than hydrogen resulting in
the electrons in the hydrogen - oxygen bonds lying more towards
the oxygen than they do towards the hydrogen. The oxygen end of
the water molecule has a partial negative charge and the hydrogen
end has a partial positive charge. The bent shape of the water
molecule combined with the two polar bonds makes water a polar
molecule (figure below).
1
MODULE 2
2010
When we dissolve an ionic compound in water the substance
dissociates into aqueous positive and negative ions. The positive
ions are attracted to the negative oxygen side of the water
molecule and the negative ions are attracted to the positive
hydrogen side. The ions become surrounded by water molecules, or
hydrated. The substance no longer exists as a compound, but as
freely moving ions, moving throughout the solution. For sodium
chloride, we represent the dissolving in the equation
NaCl(s) → Na+(aq) + Cl−(aq)
The (aq) indicates the ions are dissolved in water, or aqueous.
Visit the following website to see an animation of dissolving:
http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/flashfiles/ther
mochem/solutionSalt.html
Precipitates
Some ionic solids are not very soluble in water because the
attraction between the ions is greater than the attraction between
water and the individual ions. When these substances form in a
solution, the solution goes cloudy and the solid settles to the
bottom. These are called precipitation reactions.
When a solution of aqueous silver nitrate and aqueous sodium
chloride are mixed, the two clear colourless solutions form a cloudy
mixture. The cloudy mixture clears as a white solid settles to the
bottom of the container.
Before we mix the two solutions above, they appear like the
diagram below.
Notice from the diagram that the sodium and nitrate ions remain in solution. Since
the ions separate in solution, they are allowed to react separately. The sodium ions
CHEM 40S
RETSD Adult Ed Page 2
MODULE 2
2010
and nitrate ions do not participate in the reaction; they just "stand around watch the
show". Ions that do not participate in a precipitation reaction, and still remain moving
freely in solution are called spectator ions. Sodium and nitrate ions are the
spectators in this reaction.
Solubility Rules
Several ionic substances have a low solubility in water because the force of attraction
between the ions is greater than the force water molecules use to dissociate the ions.
These compounds do not dissociate in water, rather they will usually form suspensions
and settle to the bottom.
Chemists have done many experiments to determine the solubility of substances. These
experiments have resulted in some general rules that can be used to determine if a
substance is soluble in water.
General Solubility Rules for Selected Ionic Compounds in Water at 25°C
The following "rules" are a set of generalizations about the solubility of ionic
compounds in water at 25°C. It is important to note that the rules apply to compounds.
Individual ions cannot be soluble or insoluble, and positive ions must always be
associated with negative ions in a compound.
1. 1) Compounds of Li+, Na+, K+, Cs+, Fr+ (i.e., all metals in Main Group 1 of the
periodic table), H+, and NH4+ ions are water-soluble. (Potassium perchlorate
(KClO4) is an exception and is somewhat insoluble.)
2. Most compounds of chlorides (Cl–), bromides (Br–), and iodides (I–) are watersoluble. (Important exceptions are compounds formed between these ions and
copper(I) (Cu+), silver (Ag+), lead (Pb2+), and mercury(I) (Hg22+) which are
insoluble at room temperature.)
3. Compounds containing the nitrate (NO3–), acetate (C2H3O2–), perchlorate (ClO4–
), and chlorate (ClO3–) ions are water-soluble. (Potassium perchlorate (KClO4)
is an exception and is somewhat insoluble; silver acetate (AgC2H3O2) is
considered sparingly soluble.)
4. Most compounds of sulfates (SO42–) are water-soluble. (Important exceptions
are compounds formed between sulfates and calcium (Ca2+), strontium (Sr2+),
CHEM 40S
RETSD Adult Ed Page 3
MODULE 2
2010
barium (Ba2+), and lead (Pb2+) ions, which are water-insoluble. Silver sulfate
(Ag2SO4) is considered sparingly soluble.)
5. Most compounds of sulfides (S2–) are water-insoluble. (Important exceptions are
the sulfides of Main Group 1 and Main Group 2 elements and of the ammonium
ion (NH4+).)
6. Most compounds containing the hydroxide ion (OH–) are water-insoluble.
(Important exceptions are the hydroxides of metals from Main Group 1 of the
periodic table, and of compounds formed between hydroxide ion and H+, Sr2+,
Ba2+, and NH4+.)
7. Most compounds containing the carbonate (CO32–), phosphate (PO43–), and
sulfite (SO32–) ions are water-insoluble. (Important exceptions are the
compounds formed with H+, NH4+, and with metals from Main Group 1 of the
periodic table.)
There is a table in your text book, page 797 which sums this all up nicely in a table
format. You may use this any time, for tests or exams.
Predicting Solubility
As mentioned, we can use the rules to predict if a compound is soluble or not. The chart
you downloaded uses the words "low solubility" rather than insoluble because a small
amount does actually dissolve.
Example 1. Use the solubility rules to predict if the following compounds are soluble.
a.
b.
c.
d.
sodium sulphate (Na2SO4)
silver nitrate (AgNO3)
aluminum hydroxide (Al(OH)3)
lead (II) chloride (PbCl2)
Solution.
Make your prediction by first finding the negative ion in the chart. Find the
corresponding positive ion and determine whether the compound is soluble.
a) sodium sulphate (Na2SO4)
We find the sulfate ion in the chart  then look for where it intersects with
sodium… soluble. (also, sodium is an alkali metal and is therefore soluble with
everything  rule #1)
b) silver nitrate (AgNO3)
Find the nitrate ion in the chart  look for where it intersects with silver.
Soluble.
CHEM 40S
RETSD Adult Ed Page 4
MODULE 2
2010
c) not soluble
d) not soluble
Assignment 1
Using the Ion Solubility Chart, identify each of the following compounds as soluble or
insoluble (low solubility).
1. tin (III) sulphate
2. lithium carbonate
3. silver iodide
4. ammonium bromide
5. copper (II) chloride
6. nickel (II) iodide
7. barium phosphate
8. lead (II) chloride
9. copper (I) bromide
10. calcium sulfide
Precipitation Reactions
A precipitation reaction is a chemical reaction between ions in solution that results in the
formation of a precipitate. The equation of the reaction between aqueous silver nitrate
and aqueous sodium chloride is
AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)
You may recognize this as a double replacement reaction. All precipitation reactions are
double replacement reactions. The equation above is called the precipitation equation or
molecular equation because it shows the ions together as if they formed neutral
molecules.
The molecular equation is not an accurate representation of what is actually occurring. It
does give us an indication of the contents of each solution, but we know that the ions are
not together in the solutions. A more accurate representation is the ionic equation. The
ionic equation shows all ions in their dissociated form.
The equation for the dissociation of silver nitrate in water is
AgNO3(aq) → Ag+(aq) + NO3−(aq)
The equation for the dissociation of sodium chloride in water is
NaCl(aq) → Na+(aq) + Cl−(aq)
CHEM 40S
RETSD Adult Ed Page 5
MODULE 2
2010
These two dissociation equations reflect the ions present before the two solutions are
mixed. The ions are the reactants in the reaction. The ionic equation for the reaction
between aqueous silver nitrate and aqueous sodium chloride is
Ag+(aq) + NO3−(aq) + Na+(aq) + Cl−(aq)→ AgCl(s) + Na+(aq) + NO3−(aq)
Notice we have not separated the silver chloride into ions since after the reaction is
complete because the silver and chloride ions are not able to move freely. The ions are
held into a solid ionic compound.
The ionic equation also includes the spectator ions, whereas the net ionic equation shows
the actual reaction that occurs, without the spectator ions. We do not include the
spectator ions in the net ionic equation because they do not participate in the reaction.
They remain unchanged throughout the reaction. By eliminating the spectators from the
ionic equation we get
The net ionic equation for the reaction between aqueous silver nitrate and aqueous
sodium chloride is
Ag+(aq) + Cl− (aq) → AgCl(s)
Example 2. Write the complete set of equations (molecular, ionic and net ionic) for the
reaction between aqueous lead (II) nitrate and aqueous potassium iodide.
Solution.
Write the formulas of the initial compounds.
lead (II) nitrate = Pb(NO3)2
potassium iodide = KI
Step 1: Identify the ions present:
The ions present are: Pb2+, NO3−, K+, I−
Step 2: Write the formulas of the 2 possible compounds formed from the ions.
The ions could combine to form PbI2 and KNO3
Pb2+ and K+ will not combine nor will I− and NO3− because the ions with like charges
will repel each other.
Step 3: Determine which, if any, of these compounds would form a precipitate.
CHEM 40S
RETSD Adult Ed Page 6
MODULE 2
2010
The compound that has low solubility will result in a precipitate.
From the chart, we know that potassium ions and nitrate ions will be soluble with
essentially any ion, so potassium nitrate is not a precipitate.
From the chart, we can see that lead (II) ions will have a low solubility with iodide ions.
Lead (II) iodide will likely form a precipitate.
Write the molecular equation and balance:
Pb(NO3)2(aq) + 2 KI(aq) → PbI2(s) + 2 KNO3(aq)
Be sure to include the states.
Write the balanced ionic equation:
Pb2+(aq) + 2 NO3−(aq) + 2 K+(aq) + 2 I− (aq) → PbI2(s) + 2 K+(aq) + 2 NO3− (aq)
Notice the coefficient for the nitrate ion is 2. This because the dissociation of each lead
(II) nitrate results in 2 nitrate ions.
Eliminate the spectators from the ionic equation
and write the balanced net ionic equation
Pb2+(aq) + 2 I−(aq) → PbI2 (s)
Common Precipitation Reactions
Hot water pipes and kettles can get clogged because precipitates form with magnesium
and calcium ions in “hard” water combine with carbonates and oxides in the water. As
these precipitates form, they deposit in the pipes and kettles.
Kidney stones form when calcium ions, usually from dairy products, and oxalates or
phosphates form precipitates in the kidney and urinary tract. These precipitates can form
large crystals and block the urinary tract. One way to avoid kidney stones is to drink a
lot of water. The added water increases the amount of these precipitates that can
dissolve.
Limestone formations in caves are due to the precipitation of calcium carbonate carried
by water draining into the cave.
CHEM 40S
RETSD Adult Ed Page 7
MODULE 2
2010
Assignment 1 - 2
1. Identify if the following compounds are soluble or not.
a. barium hydroxide
b. aluminum nitrate
c. magnesium phosphate
2. Write balanced equations, total ionic equations, and net ionic equations for the mixing
of the following solutions. If no reaction occurs, write “no reaction”. Show states (aq or
s).
a.
b.
c.
d.
e.
f.
g.
h.
ammonium sulfate and rubidium carbonate
sodium hydroxide and nickel (II) chloride
strontium hydroxide and calcium iodide
ammonium phosphate and barium chloride
aluminum nitrate and magnesium sulfate
copper (II) chloride and sodium sulfide
magnesium bromide and potassium carbonate
barium hydroxide and iron (III) nitrate
Lesson 2 – Acids and Bases and Neutralization Reactions
Knowledge of acids and bases are important to be a good consumer. The commercials
on TV tell you all about your "acid indigestion" and how to stop heartburn. Companies
tell you their soaps, shampoos and other personal care products are "pH balanced" so
they won't harm your skin or hair.
Knowledge of acid base systems is important in cooking and choosing foods
that will prevent upset stomachs.
In this module we will briefly study the properties of acids and bases and some
acid-base reactions.
When an accident happens where an acid is spilled, a base, is often added to
the acid to neutralize that acid. Neutralization involves the conversion to water,
which is neutral.
CHEM 40S
RETSD Adult Ed Page 8
MODULE 2
2010
Learning Outcomes
When you have completed this lesson, you will be able to:






Describe the uses of common acids and bases in daily experience.
Operationally define acids and bases in terms of observable and measurable
properties.
Define acids and bases in terms of Arrhenius' theory.
Write formulas from memory for common acids and bases.
Write balanced equations for neutralization reactions.
Compare the neutralization of polyprotic acids and polyhydroxic bases.
Defining Acids and Bases
Svante Arrhenius noticed that acidic and basic solutions were electrolytes. He concluded
because of this that acids and bases must ionize or dissociate in water.
According to Arrhenius, an acid is defined as a substance that releases
hydrogen ions in water.
Example 1. Hydrochloric acid:
HCl(aq) → H+(aq) + Cl–(aq)
Example 2. Acetic acid (vinegar):
HC2H3O2(aq) → H+(aq) + C2H3O2–(aq)
We can often easily recognize the formula of an acid because it will usually
begin with one or more hydrogen atoms.
According to Arrhenius, a base is defined as a substance that releases
hydroxide ions in water.
Example 3. Sodium hydroxide:
NaOH(s) → Na+(aq) + OH–(aq)
CHEM 40S
RETSD Adult Ed Page 9
MODULE 2
2010
Example 4. Aqueous ammonia:
Ammonia is a little troublesome. It must first be shown to react with water then
dissociate.
NH3(g) + H2O(l) → NH4OH(aq) → NH4+(aq) + OH–(aq)
Aqueous ammonia is often called ammonium hydroxide.
Many bases, with the exception of ammonia, will contain one or more hydroxide
ions.
Properties of Acids
You have learned in previous science courses that acidic solutions in water have the
following properties:




Taste sour. For example, lemons (citric acid) and vinegar (acetic acid).
Burn when touching skin.
Neutralize basic solutions.
React with carbonates to produce carbon dioxide gas. For example, when you
add vinegar to baking soda (sodium bicarbonate), fizzing occurs. This fizzing is
the production of carbon dioxide.
Acid + carbonate → salt + water + carbon dioxide gas

Corrosive to metals. Many acids react with active metals to produce hydrogen
gas. This is a single replacement reaction where the hydrogen ion from the acid
is replaced by the metal ion.
Acid + metal → salt + hydrogen gas
Here is a group of common acids. You must know each acid's name and formula.
To help you remember the names and formulas, here are a few general naming
rules for acids



compounds ending in –ide become hydro...ic acids. For example, hydrogen
chloride (HCl) becomes hydrochloric acid.
compounds ending in –ate become –ic acids. For example, hydrogen sulfate
(H2SO4) becomes sulfuric acid.
compounds ending in –ite become –ous acids. For example, hydrogen sulfite
(H2SO3) becomes sulfurous acid.
CHEM 40S
RETSD Adult Ed Page 10
MODULE 2
Acid
Formula
2010
Common Uses or Commonly Found
Hydrochloric HCl
Commonly known as muriatic acid or stomach acid. Produced in and secreted
by the stomach, and used commonly in laboratories. It is also used in food
processing, as a cleaning agent, in the activation of oil wells, in the
production of other chemicals, and in recovering magnesium from sea water.
Sulfuric
H2SO4
Battery acid. The sulfur compounds, released by onions when cut, form
sulphuric acid when dissolved in water causing teary eyes. Used in fertilizer
manufacturing, petroleum refining, the production of metals, paper, paint,
dyes, detergents, and many chemical raw materials. Pure sulfuric acid is a
dense, oily liquid that has a high boiling point. When you combine water with
concentrated sulfuric acid, a large amount of heat is released. Concentrated
sulfuric acid is a good dehydration agent.
Nitric
HNO3
Pure nitric acid is an unstable liquid. Concentrated nitric acid is more stable
and is 70% by mass of the acid dissolved in water. It is used in making
rubber, chemicals, dyes, plastics, drugs, and explosives.
Acetic
HC2H3O2 Also known as vinegar.
carbonic
H2CO3
CO2 dissolved in water, as in carbonated beverages, forms carbonic acid
nitrous
HNO2
One component of acid rain. Used in the manufacture of fertilizers and in
organic reactions
sulfurous
H2SO3
A component of acid rain. Used to manufacture sulfuric acid and in the pulp
and paper industry.
phosphoric
H3PO4
In carbonated beverages. Dilute phosphoric acid is not toxic and has a sour
taste. It is used as a flavouring agent in beverages and is used as a cleaning
agent for dairy equipment. It is also used in the manufacture of detergents,
ceramics, and phosphorous-containing chemicals.
hypochlorous HClO
When chlorine gas is dissolved in water hypochlorous acid is a product.
Hypochlorous acid is used as a disinfectant in water supplies and swimming
pools. Also used in the manufacture of bleaches.
Acids such as hydrochloric, nitric, and acetic are called monoprotic (mono =
one, protic = proton) acids. This means they release one hydrogen ion or proton
per molecule in water. Acids that release more than one hydrogen ion in water
are called polyprotic (poly = more than one, protic = proton) acids. For example,
carbonic, sufuric and sulfurous acids are diprotic (di = 2 ) because they release
2 hydrogen ions and phosphoric acid is triprotic (tri = 3) because it releases 3
hydrogen ions in water.
CHEM 40S
RETSD Adult Ed Page 11
MODULE 2
monoprotic
HCl(aq) → H+(aq) + Cl¯(aq)
diprotic
H2SO4(aq) → 2 H+(aq) + SO42–(aq)
triprotic
H3PO4(aq) → 3 H+(aq) + PO43–td>
2010
Properties of Bases
In previous science courses we have seen that basic solutions have the following
properties:






taste bitter (I don't recommend going around tasting bases, unless you want to try
some baking soda), similar to soap
slippery touch
pH > 7
caustic, meaning bases degrade animal tissues
neutralize acids
turn litmus blue, phenolphthalein pink and bromothymol blue blue
Here is a list of common bases. You must know the name and formula for each.
Base
sodium hydroxide
Formula
NaOH
Common Uses
Also called lye. Used in oven cleaners and drain cleaners. It is also
used in the manufacturing of soap.
potassium hydroxide KOH
Also known as pot ash. Used for soap making, drain cleaners and
making alkaline batteries.
magnesium
hydroxide
Mg(OH)2
milk of magnesia, antacids, laxatives
calcium hydroxide
Ca(OH)2
Also called slaked lime or hydrated lime. Used in antacids, making
plaster and concrete, removing hair from leather hides, and steel
making.
ammonia
NH3
Used in window and general household cleaners, manufacturing of
fertilizers and explosives.
aluminum hydroxide Al(OH)3
Used in antacids, antiperspirants, ceramics
Bases such as sodium hydroxide and potassium hydroxide are called
monohydroxic (mono = one, hydroxic = hydroxide ion) bases. This means they
release one hydroxide ion per molecule in water. Bases that release more than
one hydroxide ion in water are called polyhydroxic (poly = more than one)
CHEM 40S
RETSD Adult Ed Page 12
MODULE 2
2010
bases. For example, calcium hydroxide and magnesium hydroxide are
dihydroxic (di = 2) bases because they release 2 hydroxide ions and aluminum
hydroxide is trihydroxic (tri = 3) because it releases 3 hydroxide ions in water.
monohydroxic
NaOH(s) → Na+(aq) + OH–(aq)
dihydroxic
Mg(OH)2(s) → Mg2+(aq) + 2 OH–(aq)
trihydroxic
Al(OH)3(s) → Al3+(aq) + 3 OH–(aq)
Neutralization Reactions
A neutralization reaction is special double replacement reaction between an acid and a
base. The products of this reaction are water and a salt. A salt is defined as a compound
composed of the negative ion of an acid and the positive ion of the base.
In general, the reaction for a neutralization reaction is given by
acid + base → salt + water
For example, when hydrochloric acid and sodium hydroxide are mixed the
following reaction occurs:
HCl(aq) + NaOH(aq) → NaCl(aq) + HOH(l)
The hydrogen from the acid combines with the hydroxide from the base to form
HOH, or water. The remaining metal and non-metal ions combine to form the
aqueous salt.
Neutralization Equations
On the previous page, we saw the equation
HCl(aq) + NaOH(aq)→ NaCl(aq) + HOH(l)
This is equation is known as the neutralization equation or the molecular
equation. It shows the reactants and products as molecules.
CHEM 40S
RETSD Adult Ed Page 13
MODULE 2
2010
The reaction between the hydrogen and hydroxide ions is seen more easily in
the ionic equation. Recall, the ionic equation shows the reactants and products
as aqueous ions.
H+(aq) + Cl–(aq) + Na+(aq) + OH–(aq) → Na+(aq) + Cl–(aq) + H2O(l)
The aqueous sodium ions and chloride ions are spectators in this reaction, since
they start as separate aqueous ions and end as the same.
The net ionic equation shows the reaction that occurs, omitting the spectator
ions. The net ionic equation for this reaction omits the sodium and chloride ions
from the ionic equation.
H+(aq) + OH–(aq) → H2O(l)
The net result of a neutralization reaction is the reacting of the hydrogen ion
from an acid with the hydroxide of a base, to form water.
Example 1. Write the complete set of equations (molecular, ionic and net ionic)
showing the reaction between potassium hydroxide and sulphuric acid.
Solution.
Step 1: Write the formulas for each reactant.
potassium hydroxide = KOH
sulphuric acid = H2SO4
Step 2: Write the molecular equation.
Water is one product and the salt is the negative ion from the acid, SO 42–, and
the positive ion from the base, K+. The salt will be potassium sulphate, K2SO4.
We can now write the molecular equation.
CHEM 40S
RETSD Adult Ed Page 14
MODULE 2
2010
KOH(aq) + H2SO4(aq) → H2O(l) + K2SO4(aq)
You will notice the equation is not balanced. We must balance the equation
2 KOH(aq) + H2SO4(aq) → 2 H2O(l) + K2SO4(aq)
Step 3: Use the molecular equation to write the balanced ionic equation.
KOH will produce K+ ions and OH– ions. Remember there are 2 KOH in the
balanced molecular equation, so we must make 2 of each.
H2SO4 will produce 2 H+ ions and a single SO42–.
The balanced ionic equation is
2 K+(aq) + 2 OH–(aq) + 2 H+(aq) + SO42–(aq) → 2 H2O(l) + 2 K+(aq) + SO42–(aq)
Step 4: Eliminate the spectators and write the balanced net equation.
The spectators are K+(aq) and SO42–(aq)
The net equation is
2 OH–(aq) + 2 H+(aq) → 2 H2O(l)
we need to reduce this down by dividing each coefficient by 2.
OH–(aq) + H+(aq) → H2O(l)
Notice the net equation for this reaction is the same as the first reaction.
For every neutralization reaction the net equation will be
CHEM 40S
RETSD Adult Ed Page 15
MODULE 2
2010
H+(aq) + OH–(aq) → H2O(l)
Assignment 2-2
Write the complete set of reactions that occur when the following acid and bases
are reacted.
1.
2.
3.
4.
5.
6.
7.
8.
9.
Acetic acid and sodium hydroxide.
Sulphuric acid and potassium hydroxide.
Nitric acid and calcium hydroxide.
Phosphoric acid and lithium hydroxide.
Sulphuric acid and aluminum hydroxide.
Sulfurous acid and magnesium hydroxide.
Nitrous acid and barium hydroxide.
Hydrochloric acid and magnesium hydroxide.
Aluminum hydroxide and nitric acid.
Lesson 3 – Acids and Bases Quantity Calculations
We can use the stoichiometry of the neutralization reactions to determine how much acid
must be added to a given amount of base in order to neutralize the base. We can also use
the stoichiometry to determine the concentration of a given amount of acid or base
solution.
Learning Outcomes
When you have completed this lesson, you will be able to:


Calculate the concentration or volume of an unknown acid or base from the
concentration and volume of a known acid and base required for neutralization.
Compare the neutralization of polyprotic acids and polyhydroxic bases.
Defining Neutralization
When an acidic solution is added to a basic solution, we can predict the amount of that
acid needed to neutralize the base. We know that at the point of neutralization, the
amount of hydrogen ions and hydroxide ions will be equal.
That is, at neutralization
CHEM 40S
RETSD Adult Ed Page 16
MODULE 2
2010
moles of H+ ions from the acid = moles of OH– ions from the base
If we are mixing solutions of known concentrations,
moles of H+ ions = (concentration of H+ ions)×(volume of the acid) or,
molesH = CH×VA
and
moles of OH– ions = (concentration of OH– ions)×(volume of the base) or,
molesOH = COH×VB
By substituting
CH×VA = COH×VB
Solving neutralization problems:
Step 1: Write the balance neutralization equation.
Step 2: Substitute given values into the equation above and solve for the required
amount , or find moles of given amount, then use stoichiometry to find the moles,
then required amount of unknown.
Example 1. Calculate the concentration of hydrochloric acid, if 25.0 mL is just
neutralized by 40.0 mL of a 0.150 mol/L sodium hydroxide solution.
Solution
We are given:



25.0 mL hydrochloric acid
40.0 mL sodium hydroxide (base)
0.150 mol/L sodium hydroxide (base)
We must find the concentration of the acid
Step 1 : Write the balanced neutralization reaction.
The reactants are hydrochloric acid, HCl, and sodium hydroxide, NaOH.
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
Step 2 : Substitute given values into the equation.
CHEM 40S
RETSD Adult Ed Page 17
MODULE 2
2010
The concentration of the base and hydroxide ions are equal because sodium hydroxide is
monohydroxic (that is, one sodium hydroxide produces one hydroxide ion).
CH = ?
VA = 25.0 mL = 0.0250 L
COH = 0.150 mol/L
VB = 40.0 mL = 0.0400 L
Since the hydrochloric acid is monoprotic (that is, one hydrochloric acid molecule
produces one hydrogen ion), the concentration of hydrochloric acid equals the hydrogen
ion concentration.
or by stoichiometry,
The concentration of the hydrochloric acid is 0.240 mol/L.
Example 2. What volume of 1.00 mol/L potassium hydroxide is needed to neutralize
750. mL of a 0.500 mol/L nitric acid solution?
Solution.
We are given:



the volume of nitric acid is 750. mL, or 0.750 L
the concentration of the nitric acid is 0.500 mol/L
the concentration of the potassium hydroxide (base) is 1.00 mol/L
We must find the volume of the potassium hydroxide (base) solution.
Step 1: Write the balanced neutralization equation.
HNO3(aq) + KOH(aq) → KNO3(aq) + H2O(l)
Step 2: Calculate the volume of base needed to neutralize the acid.
CHEM 40S
RETSD Adult Ed Page 18
MODULE 2
2010
Since the nitric acid is monoprotic (that is, one nitric acid produces one hydrogen ion),
the concentration of hydrogen ions equals the nitric acid concentration.
Similarly, the concentration of the base and hydroxide ions are equal because potassium
hydroxide is monohydroxic.
CH = 0.500 mol/L
VA = 0.750 L
COH = 1.00 mol/L
VB = ?
or by stoichiometry,
Find the number of moles of acid then, using stoichiometry, find the number of moles of
base.
The volume of base needed is 0.375 mol/L
Neutralization With Polyhydroxic Bases and
Polyprotic Acids
Example 3. What is the concentration of a sample of magnesium hydroxide, if 225 mL
of the base is neutralized by 125 mL of a 0.200 mol/L hydrochloric acid solution?
Solution.
We are given:



the volume of hydrochloric acid is 125 mL, or 0.125 L
the concentration of the hydrochloric acid is 0.200 mol/L
the volume of the magnesium hydroxide (base) is 225 mL, or 0.225 L
We must find the concentration of the magnesium hydroxide (base) solution.
CHEM 40S
RETSD Adult Ed Page 19
MODULE 2
2010
Step 1: Write the balanced neutralization equation.
The reactants are hydrochloric acid, HCl, and magnesium hydroxide, Mg(OH)2.
2 HCl(aq) + Mg(OH)2(aq) → MgCl2(aq) + 2 H2O(l)
Step 2: Substitute into the equation.
Since the hydrochloric acid is monoprotic (that is, one hydrochloric acid produces one
hydrogen ion), the concentration of hydrogen ions equals the hydrochloric acid
concentration.
CH = 0.200 mol/L
VA = 0.125 L
COH = ?
VB = 0.225 L
Substituting these values into the equation, we will solve for hydroxide ion
concentration.
This is the concentration of the hydroxide ion, not the base in this case because the
magnesium hydroxide is dihydroxic.
Mg(OH)2 → Mg2+(aq) + 2 OH–(aq)
To find the base concentration
We can also solve this using stoichiometry.
Find the number of moles of acid then, using stoichiometry, find the number of moles of
base.
CHEM 40S
RETSD Adult Ed Page 20
MODULE 2
2010
Example 4. Calculate the volume of 0.100 mol/L carbonic acid needed to neutralize
25.0 mL of 0.200 mol/L sodium hydroxide.
Solution.
We are given:



the concentration of carbonic acid is 0.100 mol/L
the volume of the sodium hydroxide is 25.0 mL, or 0.0250 L
the concentration of the sodium hydroxide (base) is 0.200 mo/L
We must find the volume of the carbonic acid solution.
Step 1: Write the balanced neutralization equation.
The reactants are carbonic acid, H2CO3 and sodium hydroxide, NaOH.
H2CO3(aq) + 2 NaOH(aq) → Na2CO3(aq) + 2 H2O(l)
Step 2: Substitute into the equation.
The concentration of the base and hydroxide ions are equal because sodium hydroxide is
monohydroxic.
Since the carbonic acid is diprotic (that is, one carbonic acid produces 2 hydrogen ions),
the concentration of hydrogen ions equals twice the carbonic acid concentration.
CH = 2(H2CO3) = 2(0.100 mol/L) = 0.200 mol/L
VA = ?
COH = 0.200 mol/L
VB = 0.0250 L
Substituting these values into the equation, we will solve for volume of the acid.
or by stoichiometry,
Find the number of moles of base then, using stoichiometry, find the number of moles of
acid.
CHEM 40S
RETSD Adult Ed Page 21
MODULE 2
2010
0.0250 L, or 25 mL of acid are needed to neutralize the base.
Assignment 2-3
Answer the following questions, showing all your work.
1. What volume of 0.240 mol/L sulfuric acid can be neutralized by 50.0 mL of
0.360 mol/L sodium hydroxide?
2. Concentrated hydrochloric acid (11.7 mol/L) is added to a spill of 5.00 L of
sodium hydroxide solution with a concentration of 2.00 mol/L. What volume of
acid is required to neutralize the spill?
3. A clumsy chemistry teacher (not this one, of course) spills 75.0 mL of
concentrated sulfuric acid (18.0 mol/L). He has a stock solution of 1.00 mol/L
sodium hydroxide on hand to neutralize the spill. How much base does he need
to neutralize the spill?
4. What volume of a 2.00 mol/L solution of phosphoric acid is required to bring the
pH of 150 mL of a 2.50 mol/L solution of calcium hydroxide down to 7.0?
5. If 250 mL of a 6.0 mol/L solution of carbonic acid neutralizes 750 mL of a
solution of ammonia (use “ammonium hydroxide”), what is the concentration of
the ammonia?
6. Calculate the concentration of potassium hydroxide if 1.40 mL of the base is
needed to neutralize 22.5 mL of 0.175 mol/L nitric acid.
7. If 0.750 mL of an antacid containing magnesium hydroxide is completely
neutralizes 0.100 L of 0.100 mol/L hydrochloric acid solution, what is the
concentration of the antacid?
8. What is the concentration of a solution of phosphoric acid if 325 mL is required
to neutralize 250. mL of a 0.250 mol/L solution of potassium hydroxide?
9. Calculate the concentration of nitrous acid if 25.0 mL of the acid is needed to
neutralize 19.0 mL of 0.830 mol/L potassium hydroxide.
CHEM 40S
RETSD Adult Ed Page 22
MODULE 2
2010
LESSON 4 – OXIDATION / REDUCTION
LESSON 1 OUTCOMES
After working through this lesson, you should be able to:
1. define oxidation, define reduction
2. be able to assign correct oxidation numbers to each atom in a chemical equation,
using your memorized rules for assigning oxidation numbers
3. identify a reaction as redox or not redox
4. identify the substance oxidized, the substance reduced, and the oxidizing agent
and the reducing agent in a redox reaction
5. write half reactions for oxidation and reduction
OXIDATION
The term oxidation was first applied to the combining of
oxygen with other elements. Strong oxidizers are those that readily
accept electrons, that is, they will take or steal electrons from many
substances. There are many known instances of this behaviour.
Iron rusts and carbon burns. Burning is another name for rapid
oxidation.
Chemists recognized that other nonmetallic elements unite with substances in a
manner similar to that of oxygen. Hydrogen, antimony, and sodium all burn in
chlorine, and iron will burn in fluorine. Since these reactions were similar,
chemists formed a more general definition of oxidation. Remember, elements on
the right side of the periodic table want to gain electrons. Oxygen, chlorine and
fluorine form negative ions; they want to take electrons. In the reactions
mentioned above, electrons were removed from each free element by the
reactants O2 and Cl2. Thus, oxidation is defined as the process by which electrons
are removed from an atom or ion.
REDUCTION
A reduction reaction was originally limited to the type of reaction in which ores
were “reduced” from their oxides. An ore is a mineral compound from which a
useful metal can be extracted. Oxides were said to be “reduced” by the removal
of oxygen. Iron ore was “reduced” to iron by carbon monoxide. Oxygen was
removed and the free element was produced. The free element can be produced in
CHEM 40S
RETSD Adult Ed Page 23
MODULE 2
2010
other ways. An electric current passing through molten sodium chloride produces
sodium.
The similarities between oxidation and reduction reactions led chemists to
formulate a more general definition of reduction. By definition, reduction is the
process by which electrons are apparently added to atoms or ions.
OXIDIZING AND REDUCING AGENTS
In an oxidation-reduction reaction electrons are transferred. All the electrons
exchanged in an oxidation-reduction reaction must be accounted for. Therefore,
oxidation and reduction must occur at the same time in a reaction. Electrons are
lost and gained at the same time and the number lost must equal the number
gained.
The substance in the reaction which gives up electrons (and thus helps another
substance get reduced) is called the reducing agent. The reducing agent contains
the atoms which are oxidized (the atoms which lose electrons). Zinc is a good
example of a reducing agent. It is oxidized to the zinc ion, Zn2+.
The substance in the reaction which gains electrons (and thus aids another
substance in becoming oxidized) is called the oxidizing agent. It contains the
atoms which are reduced (the atoms which gain electrons). Dichromate ion,
Cr2O72-, is a good example of an oxidizing agent. It is reduced to the chromium
(III) ion, Cr3+.
…… If this sounds confusing, keep reading into the next section on oxidation
numbers, then come back to this point and analyze each atom in the dichromate
ion for oxidation number. (You will see that Cr goes from an oxidation # of +6
to an ox. # of +3. It gains 3 electrons, which reduces (GER) its oxidation # from
+6 to +3.)
Note:
If a substance gives up electrons readily it is said to be a strong reducing agent. Its
oxidized form, however, is normally a poor oxidizing agent. If a substance gains
electrons readily it is said to be a strong oxidizing agent. Its reduced form is a
weak reducing agent.
CHEM 40S
RETSD Adult Ed Page 24
MODULE 2
2010
The idea above relates back to strong acids and bases, which have
correspondingly weak conjugates..
Four metals and their ions, Zn, Pb, Cu, and Ag are listed below in the order of
their tendency to react. These lists are written as reduction and oxidation half
reactions.
Reaction of Ions
Reaction of Metals
Ag+(aq) +e-
Ag(s)
Zn(s)
Zn2+(aq) + 2e-
Cu2+(aq) + 2e-
Cu(s)
Pb(s)
Pb2+(aq)
Pb(s)
Cu(s)
Zn(s)
Ag(s)
Cu2+(aq) + 2eAg+(aq) +e-
Pb2+(aq)
+ 2e
-
‘
Zn2+(aq) + 2e-
+ 2e
-
Notice that the first list is ordered from silver to zinc and the second list is
ordered from zinc to silver. The metal ion with the greatest tendency to gain
electrons (Ag+(aq)) reacts to produce the metal which is least likely to lose
electrons (Ag(s)). The metal which is most likely to lose electrons (Zn(s)) reacts to
produce the metal ion with the least tendency to gain electrons (Zn2+(aq) ).
LESSON 4, ASSIGNMENT 1
1. Which chemical species in the previous list was the strongest oxidizing
agent?
2. Which chemical species in the previous list was the strongest reducing agent?
3. Which chemical species in the previous list was the weakest oxidizing agent?
4. Which chemical species in the previous list was the weakest reducing agent?
OXIDATION NUMBERS
How is it possible to determine whether an oxidation-reduction reaction has taken
place? Answer: by determining whether an electron shifts have taken place during
the reaction. To indicate electron changes, check the oxidation numbers of the
atoms in the reaction. The oxidation number is the charge an atom appears to
have when assigned a certain number of electrons. Any change in oxidation
numbers in the course of a reaction indicates an oxidation reduction reaction has
taken place.
-
CHEM 40S
RETSD Adult Ed Page 25
MODULE 2
2010
Example:
Suppose iron, as a reactant in a reaction, has an oxidation number of 2+. If iron
appears as a product with an oxidation number other than 2+, say 3+, or 0, then a
redox reaction has taken place.
ASSIGNING OXIDATION NUMBERS
Oxidation numbers are assigned according to the apparent charge of the element.
To determine the apparent charge consult the electron dot structure for the
substance. Shared electrons are assigned to the more electronegative element. The
algebraic sum of the oxidation numbers of a compound is zero.
Example 1:
What is the oxidation number of the sodium atom and the sodium ion?
Solution:
The atom (Na) is not charged and the number of electrons is equal to the number
of protons. The apparent charge of the sodium atom is 0 and the oxidation
number is 0. The sodium ion (Na+) shows that there is one more proton than
electron. Since its apparent charge is 1+, its oxidation number is 1+.
CHEM 40S
RETSD Adult Ed Page 26
MODULE 2
2010
Example 2:
What are the oxidation numbers of hydrogen, oxygen, and sulfur in H2SO4?
Solution:
A possible electronic structure for sulfuric acid, H2SO4 is
O
H O S O H
¨
O
Here is an explanation of how oxidation numbers work – don’t worry, you don’t
have to reason through the assigning of these numbers – there are rules to follow,
given on the next page. This just tells you where the rules come from:
It can be seen, from the electronic structure, that the oxygen atoms share electrons
with both sulfur and hydrogen atoms. Since oxygen is more electronegative than
sulfur, the shared electrons are arbitrarily assigned to each oxygen atom. Thus,
we can assume oxygen to have a full valence shell, or a total of 10 electrons,
giving each atom an apparent charge of –2 (8 protons, 10 electrons). Each
hydrogen atom, less electronegative than the oxygen atoms, will have its electron
assigned to oxygen. The hydrogen oxidation number in this compound is 1+. The
sulfur atom, with a resulting apparent charge of 6+, is assigned an oxidation
number of 6+. The total of the oxidation numbers of all the atoms in the compound
must be zero. In H2SO4, one sulfur atom has an oxidation number of 6+, four
oxygen atoms have the oxidation number 2-, and two hydrogen atoms have 1+.
The apparent charge of the compound is zero – it is neutral overall.
RULES FOR ASSIGNING OXIDATION NUMBERS
The following general rules have been made so it is easier to determine the
oxidation numbers.
Rule 1:
The oxidation number of any free element is zero. This statement is true for
all atomic and molecular structures: monatomic, diatomic, and
polyatomic.
Rule 2:
The oxidation number of monatomic ion is equal to the charge on the ion.
Some atoms have several possible oxidation numbers.
Rule 3:
The oxidation number of each hydrogen atom in most compounds is 1+.
There are some exceptions. In compounds such as lithium hydride
CHEM 40S
RETSD Adult Ed Page 27
MODULE 2
2010
(LiH), hydrogen being the most electronegative atom, has an oxidation
number of 1-.
Rule 4:
The oxidation number of each oxygen atom in most compounds is 2- (H2O). In
peroxides, each oxygen is assigned 1-.
Rule 5:
The sum of the oxidation numbers of all the atoms in a particle must equal the
apparent charge of that particle.
Rule 6:
In compounds, the elements of group IA and IIA and aluminum have positive
oxidation numbers numerically equal to their group number in the periodic
table.
Example 1:
What are the oxidation numbers of the elements in Na2SO4?
Solution:
According to rule 6, the oxidation number of sodium is 1+. According to rule
4, the oxidation number of oxygen is 2-. According to rule 5, the total of all
oxidation numbers in the formula unit is 0. Letting x = oxidation number of
sulfur,
2(1+)+x+4(2-)=0
x = 6+
Example 2:
What are the oxidation numbers of the elements in NO3-?
Solution:
This is an ion with a charge of negative one. The total charge must equal –1.
According to rule 4, the oxidation number of oxygen is 2-. According to rule 5, the
total of oxidation numbers in the ion is 1-. Letting x = the oxidation number of
nitrogen,
x + 3(2-) = 1x = 5+
CHEM 40S
RETSD Adult Ed Page 28
MODULE 2
2010
LESSON 4, ASSIGNMENT 2
In the following, give the oxidation number for the indicated atoms.
1. S in Na2SO3
2. Mn in KMnO4
3. N in NO2
4.
N in Ca(NO3)2
5. S in HSO4
-
6. C in Na2CO3
IDENTIFYING OXIDATION REDUCTION REACTIONS
Oxidation numbers can be used to determine whether oxidation and reduction
(electron transfer) occur in a specific reaction. Even the simplest reaction may be a
redox reaction.
The direct combination of sodium and chlorine to produce sodium chloride is a
simple example.
2Na(c) + Cl2(g)
2NaCl (c)
As a reactant, each sodium atom has an oxidation number of 0. In the product, the
oxidation number of each sodium atom is 1+. Similarly, each chlorine atom as a
reactant has an oxidation number of 0. As a product, each chlorine atom has an
oxidation number of 1-. Since a change in oxidation number has occurred, an
oxidation reduction reaction has taken place.
-
2Na(c) + Cl2(g)
2NaCl(c)
The change in oxidation number can result only from a shift of electrons between
atoms. This shift of electrons alters the apparent charge (oxidation number).
A gain of electrons means the substance is reduced. It also means that the apparent
charge is algebraically lowered, and the oxidation number is lowered. In contrast,
a loss of electrons is oxidation. When an atom is oxidized, its oxidation number
CHEM 40S
RETSD Adult Ed Page 29
MODULE 2
2010
increases.
The equation for the reaction can be used to determine whether the reaction is a
redox reaction. It can also be used to find the substance oxidized, the substance
reduced, and the oxidizing and reducing agents. Since the oxidation number of
sodium in the equation
2Na(s) + Cl2(g)
2NaCl(c)
changed from 0 to 1+, sodium is oxidized. Sodium is also the reducing agent. A
reducing agent always looses electrons and is, therefore, always oxidized as it
reduces the other substance.
Example:
For the following reaction, tell what is oxidized, what is reduced, and identify the
oxidizing and reducing agents.
16H+(aq) + 2MnO4-(aq) + 5C2O42-(aq)
2Mn2+(aq) + 8H2O(1) + 10CO2(g)
Solution
Manganese is reduced (7+
2+) and carbon is oxidized (3+
4+). The
permanganate ion (MnO4-) is the oxidizing agent because it contains manganese, and
manganese is reduced.
The oxalate ion (C2O42-) is the reducing agent; it contains carbon, which is
oxidized.
LESSON 4, ASSIGNMENT 3
Some of the following unbalanced reactions are oxidation-reduction reactions, and
some are not. In each case answer the following:
a. Is the reaction redox?
b. If yes, name the element reduced, the element oxidized, the oxidizing
agent, and the reducing agent. You do not have to balance these
reactions.
1. BaCl2(aq) + Na2SO4(aq)
CHEM 40S
NaCl(aq) + BaSO4(aq)
RETSD Adult Ed Page 30
MODULE 2
2.
3.
H2(g) +
N2(g)
C(c) +
NH3(g)
H2O(g)
CO(g) +
H2(g)
4.
AgNO3(aq) + FeCl3(aq)
5.
H2CO3(aq)
6.
H2O2(aq) + PbS(c)
PbSO4(c) + H2O(1)
7.
KCl(c) + H2SO4(aq)
KHSO4(aq) + HCl(g)
8.
HNO3(aq) + H3PO3(aq)
NO(g) + H3PO4(aq) + H2O(1)
9.
HNO3(aq) + I2(c)
HIO3(aq) + NO2(g) + H2O(1)
10. H+(aq) +
CHEM 40S
AgCl(c) + Fe(NO3) 3(aq)
H2O(1) + CO2(g)
NO3-(aq) + Fe2+(aq)
H2O(1) + NO(g) + Fe3+(aq)
RETSD Adult Ed Page 31
2010
MODULE 2
11. FeBr2(aq) + Br2(1)
12. S2O32-(aq) + I2(c)
2010
FeBr3(aq)
S4O62-(aq) + I- (aq)
13. H2O2(aq) + Mn+7(aq)
O2(g) + Mn2+(aq)
14. Na2S(aq) + AgNO3(aq)
Ag2S(c) + NaNO3(aq)
Read Page 604 – 609 in your textbook: “Identifiying Oxidation-Reduction
Reactions”, and do practice problems #14 and 15 on page 610.
LESSON 5 – BALANCING REDOX EQUATIONS
LESSON 1 OUTCOMES
After working through this lesson, you should be able to:
1. balance redox reactions using the half reaction method
HALF REACTIONS
Metals can change to ions and metallic ions can change to free metals. Consider the net
equation written below – note the phase notation and the charges (or absence of)….
Zn(s) + Cu2+(aq)
CHEM 40S
Zn2+(aq) + Cu(s)
RETSD Adult Ed Page 32
MODULE 2
2010
Copper ions, with a net charge of 2+ , have changed to copper atoms, with a net charge of
zero. In order for this to happen each copper ion must gain two electrons.
Cu2+(aq) + 2e
Cu(s)
Zinc atoms, with no net charge, have changed to zinc ions with a charge of 2+. Each zinc
atom must lose two electrons.
Zn(s)
-
Zn2+ (aq) + 2e
Overall, each copper ion has removed two electrons from each zinc atom.
Any such reaction which involves the transfer of electrons can be written as two separate
equations, one expressing the loss of electrons and one expressing the gain of electrons.
For every reaction of this type electrons are conserved; i.e., the number of
electrons gained equals the number of electrons lost.
A reaction involving either a gain or a loss of electrons with the electrons written in
the equation is called a half reaction. Two such half reactions make up a total reaction.
The half reaction which involves a gain of electrons is called a reduction half
reaction. The half reaction which involves a loss of electrons is called an
oxidation half reaction. The combined total reaction is called a reductionoxidation reaction or redox reaction.
You will now learn how to ensure that these equations are BALANCED in terms
of electrons lost and gained.
Definitions Review
Oxidation;
A chemical change involving a loss of electrons.
(L.E.O. loss of electrons is oxidation)
A chemical change involving a gain of electrons.
(G.E.R. gain of electrons is reduction)
removes electrons from another substance; i.e.,
oxidizes it by gaining electrons.
gives up electrons to another substance; i.e.,
reduces it.
-
Reduction:
-
Oxidizing Agent:
Reducing Agent:
See next page for one last assignment…. And start preparing for the test! 
CHEM 40S
RETSD Adult Ed Page 33
MODULE 2
2010
Read Page 613 – 616 in your textbook: “Balancing Redox Equations – the Half
Reaction Method”, and do practice problems 19 & 20 on page 616.
ALSO READ THE INTERNET PRINTOUT FOLLOWING THIS LESSON,
ENTITLED “BALANDING REDOX REACTIONS BY THE HALF-REACTION
METHOD”
CHEM 40S
RETSD Adult Ed Page 34