CH 40S MODULE 2 aqueous reactions LESSON 1 by nuhman10

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									                                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 net-
         ionic 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).




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 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



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 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 water-
      soluble. (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+),



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      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.   sodium sulphate (Na2SO4)
   b.   silver nitrate (AgNO3)
   c.   aluminum hydroxide (Al(OH)3)
   d.   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.



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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               6. nickel (II) iodide
2. lithium carbonate                7. barium phosphate
3. silver iodide                    8. lead (II) chloride
4. ammonium bromide                 9. copper (I) bromide
5. copper (II) chloride             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)


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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.


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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.




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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.   ammonium sulfate and rubidium carbonate
   b.   sodium hydroxide and nickel (II) chloride
   c.   strontium hydroxide and calcium iodide
   d.   ammonium phosphate and barium chloride
   e.   aluminum nitrate and magnesium sulfate
   f.   copper (II) chloride and sodium sulfide
   g.   magnesium bromide and potassium carbonate
   h.   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.



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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)


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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.



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    Acid         Formula                       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.


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                monoprotic            HCl(aq) → H+(aq) + Cl¯(aq)
                diprotic              H2SO4(aq) → 2 H+(aq) + SO42–(aq)
                triprotic             H3PO4(aq) → 3 H+(aq) + PO43–td>




       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                 Formula                               Common Uses
sodium hydroxide            NaOH      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                   Mg(OH)2   milk of magnesia, antacids, laxatives
hydroxide
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)


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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.




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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.


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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




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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.   Acetic acid and sodium hydroxide.
    2.   Sulphuric acid and potassium hydroxide.
    3.   Nitric acid and calcium hydroxide.
    4.   Phosphoric acid and lithium hydroxide.
    5.   Sulphuric acid and aluminum hydroxide.
    6.   Sulfurous acid and magnesium hydroxide.
    7.   Nitrous acid and barium hydroxide.
    8.   Hydrochloric acid and magnesium hydroxide.
    9.   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


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             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.


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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.




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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.




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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.




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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.




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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.




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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


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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.




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       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)   + 2e
                  -                                                                           -
Pb2+(aq)   + 2e        ‘       Pb(s)                      Cu(s)             Cu2+(aq) + 2e
                  -                                                                     -
  2+
Zn (aq) + 2e                   Zn(s)                      Ag(s)             Ag+(aq) +e

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.



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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+.




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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


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           (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-) = 1-
    x = 5+




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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


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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)              NaCl(aq) + BaSO4(aq)



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2.   H2(g) +    N2(g)        NH3(g)



3.     C(c) +      H2O(g)               CO(g) +    H2(g)




4.   AgNO3(aq) + FeCl3(aq)             AgCl(c) + Fe(NO3) 3(aq)




5.   H2CO3(aq)              H2O(1) + CO2(g)




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) +     NO3-(aq) + Fe2+(aq)              H2O(1) + NO(g) + Fe3+(aq)




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11. FeBr2(aq) + Br2(1)               FeBr3(aq)



                                         -       -
12. S2O32-(aq) + I2(c)            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)….
             2+                     2+
   Zn(s) + Cu (aq)               Zn (aq) + Cu(s)




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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.

                                    2+            -
      Zn(s)                   Zn         (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 reduction-
oxidation 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)
                                    -


Reduction:              A chemical change involving a gain of electrons.
                        (G.E.R. gain of electrons is reduction)
                                -


Oxidizing Agent:        removes electrons from another substance; i.e.,
                        oxidizes it by gaining electrons.
Reducing Agent:         gives up electrons to another substance; i.e.,
                         reduces it.



See next page for one last assignment…. And start preparing for the test! 




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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”




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