Types of Chemical Reactions
and Solution Stoichiometry
Chapter 4 of AP chemistry
4.1 Water (the universal solvent)


Water can be used to dissolve a wide variety
of substances. This is in large part due to the
polarity of the water molecule.
Let’s look at the structure of water
Polarity
• This polarity results in water being able to
“attack” the cations and the anions of ionic
compounds.
4.2 Strong, weak, and non-electrolytes



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Ions present in solution facilitate the
conduction of electricity.
Since pure water is a covalent molecule, it is
a very poor conductor of electricity.
Tap water conducts electricity because it
contains dissolved ions.
If you don’t believe me, ask yourself why
there’s a multi-million dollar water softening
industry, or wikipedia “hard water”.
Strong electrolytes, focus on acids


A strong electrolyte is an acid, base, or ionic
compound that dissolves fully.
Examples of strong acids (strong electrolytes)






Perchloric acid HClO4
Hydroiodic acid HI
Hydrobromic acid HBr
Hydrochloric acid HCl
Sulfuric acid H2SO4 (first dissociation only)
Nitric acid HNO3
I cut and pasted this list from wikipedia.
What’s an acid



You may have notices that the strong acids
were simply ionic compounds who’s cation
are hydrogen. That’s an acid in a nutshell.
If you want a technical definition feel free to
click here.
Acids release H+ ions (protons). These
protons tend to steal electrons from other
compounds, or metals and are attracted to
regions of negative charge (duh).
Strong electrolytes, focus on bases

Examples of strong bases (strong electrolytes)
 Potassium hydroxide (KOH)
 Barium hydroxide (Ba(OH)2)
 Cesium hydroxide (CsOH)
 Sodium hydroxide (NaOH)
 Strontium hydroxide (Sr(OH)2)
 Calcium hydroxide (Ca(OH)2)
 Lithium hydroxide (LiOH)
 Rubidium hydroxide (RbOH)
I cut and pasted this list from wikipedia.
What’s a base



A base is the counter part to the acid. A base
is an ionic compound who’s anion is OH-.
If you want a technical definition feel free to
click here.
Bases produce OH- ions. The OH- tend
accept/steal protons (H+) from compounds
and are attracted to regions of positive
charge (duh).
Weak electrolytes




These compounds do not dissolve
completely.
When tested use this rule of thumb: If it’s an
acid or a base and it’s not strong, it’s a weak
electrolyte.
Click here for a short and easy list of strong
and weak electrolytes. It’s a good page!!!
For now if an ionic compound is soluble think
of it as a strong electrolyte.
Non-electrolytes


Non-electrolytes do not form ions in water
and hence do not permit the water to conduct
electricity.
Common non-electrolytes include




Sucrose
Ethanol
Methanol
Pretty much all substances that remain
covalent molecules when dissolved in water.
4.3 Composition of Solutions



The concentration of a solution is very important.
The more sugar you dissolve in sweat tea the better
it will taste.
Concentration is expressed in “molarity” M
moles of solute
Molarity (M) 
Liters of solution

NOTE: Liters, not ml, of solution
Example

How many grams of NaOH would you need to add to
100.0 ml volumetric flask to make a 0.100 M NaOH
solution?
Given:
V = 100.0 ml
M = 0.100 M NaOH
moles of solute
Molarity (M) 
Liters of solution
Example
Given:
V = 100.0 ml
M = 0.100 M NaOH
Want:
Grams of solute
 First solve for moles of solute
x
0.100 M 
 x  0.0100 mol NaOH
0.10000 L

Next, convert moles to grams
40.00g
0.0100 mol NaOH *
 0.400g NaOH
1 mol
Examples problems
1.
2.
3.
Find the molarity of a solution composed of
85.84 g of CuSO4 in 250.0 ml of solution
If 450.0 ml of 2.000 M CaCO3 were
evaporated, how many grams of CaCO3
would be left behind?
A 10.00 ml volumetric flask weighs 25.18 g.
A NaCl solution of unknown molarity is filled
to the line and then completely evaporated.
If 28.62 g remain what was the molarity?
Answers: 1) 2.151M
2) 90.08g CaCO3
3) 5.886M
Dilutions


This isn’t hard at all. Simply follow the equation below
M1V1 = M2V2
Example:
If you take 300. ml of a 3.00 M AgNO3 and add 700. ml to it what is
the new molarity?
M1V1  M 2 V2
300.ml 3.00M   1000ml x

300.ml 3.00M 
x
 0.900M
1000ml
4.5 Precipitation reactions


A precipitation reaction occurs when two soluble
ionic solutions are mixed and a precipitate is formed.
Example:
Both NaCl and AgNO3 are soluble in water.
However, if you mix the two a precipitate of AgCl will
form. This is express as:
AgNO 3( aq)  NaCl( aq)  NaNO3( aq)  AgCl ( s )


The (aq) means aqueous (dissolved, still in solution)
The (s) stands for solid (precipitate)
Double replacement


Virtually all precipitation reactions are double
replacement reactions.
This is when cations and anions simply swap
partners.
4.6 Net ionic equations

Take the example below:
NH4 2 SO4(aq)  BaCl2(aq)  2NH4Cl(aq)  BaSO4( s)

The complete ionic equation is below (it’s long):
2NH4

( aq )
 2NH4
 SO 4

( aq )
2
( aq )
 Ba
2
( aq )
 2Cl ( aq) 
 2Cl ( aq)  BaSO 4 ( s )
-
-
4.6 Net ionic equations

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Writing the complete ionic equation is awful!
Note that NH4+ and Cl- appear on both sides of the
reaction and do nothing. Because they do nothing
they’re called spectator ions.
In the net ionic equation you simply leave out the
spectator ions:
Ba

2
( aq )
 SO 4
2
( aq )
 BaSO 4 ( s )
This is much easier to read and write.
Solubility rules
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It’s awful, but you’re just going to need to
memorize the solubility rules.
Check out this matching game.
To help this concept sink in, we’ll make nearly
100 mixtures.
YOU NEED TO HAVE THESE RULE
MEMORIZED!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Play the game, do whatever, but memorize
them! (pg 152 has the rules)
You also need to do stoichiometry


Example if you mix 325 ml 3.00M Ba(OH)2 and 600.
ml of 1.00 M Al(NO3)3 what precipitate would form?
What mass?
First, write the net ionic equation
Al

3
( aq )
 3OH

( aq )
 Al(OH) 3( s )
Next, determine the moles of each ion
# of moles
Multiply the OH- by three because each mole of
Al(OH)3 has three moles of OH- bonded to it.

mol OH  0.325L 3.002  1.95mol OH
-
mol Al  0.600L 1.001  0.600mol Al

3
Now find the limiting reagent. Recall the net ionic
equation:
Al
3
( aq )
 3OH

( aq )
 Al(OH) 3( s )
Limiting regent

YOU STILL NEED TO DIVIDE BY THE
COEFECENT
1.95mol OH 0.65
3
0.600mol Al
1

3
 0.600
Aluminum is the limiting reagent. Only 0.600
moles of Al(OH)3 will be produced.
Finished

If you know that 0.600 mole of Al(OH)3 will be
produced simply go from moles to grams.
78.00g
0.600mol AlOH3 *
 46.80 g AlOH3
1 mol

And you’re done
Sample problems
1.
If you mix 10.00 ml 18.00M H2SO4 and 600.
ml of 1.00 M Ba(NO3)2 what precipitate
would form? What mass?
2.
If you mix 1.25 L 0.50M FeCl3 and 100. ml
of 6.00 M Ba(OH)2 what precipitate would
form? What mass?
4.8 Acid-Base reactions

Acids produce H+ ions


Bases produce OH- ions


H
They donate protons
They accept protons
When an acid and base are mixed they produce
water. See below:

( aq)
 OH

( aq)

 H2O(l )
neutralization
Strong bases and acids

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A strong base, like NaOH, will strip a strong
or weak acid of all it’s protons.
Example: If H3PO4 (weak acid) is mixed with
an excess of NaOH what will occur?
Under normal circumstances H3PO4 would
not disassociate completely into H+ and
H2PO4-. Only a small % would disassociate.
However, as soon as a proton leaves H3PO4
it immediately is attacked by the OH- from
NaOH.
Neutralization


The OH- ions will continue to grab up H+ ions until
the OH- ions run out or until the H+ ions run out.
The second an H+ ion breaks off of the H3PO4 an
OH- ion gets it and becomes water.
H 3 PO 4 ( aq)   
 H
back and forth
H

( aq)
 OH

( aq)

 H 2 PO 4 ( aq)
-
( aq )

 H2O(l )
neutralization
Neutralization


Example: How much 1.00 standard NaOH would be
required to neutralize 50.0 ml of 18.00 M sulfuric
acid?
First write the equation:
2NaOH( aq)  H2SO4( aq)  2H2O(l )  Na 2SO4( aq)

Next, solve for moles of H2SO4:
mol
M
 mol  (M)(V)
V
mol  (18.00)(0. 0500)  0.900mol H 2SO 4
Neutralization

Now solve the stoichiometry for mole NaOH:
2NaOH( aq)  H2SO4( aq)  2H2O(l )  Na 2SO4( aq)
X
0.900mol H2SO4
2
1

 X  1.80 mol NaOH
X 0.900

Finally, solve for the volume of NaOH:
mol
1.80 mol NaOH
M
 1.00 
 V  1.80L
V
V
Measuring acid content, pH

The standard way that acidity is measured is
on the pH scale.





A pH of 7 in neutral.
1 is very acidic
14 is very basic
pH is measured with litmus paper or with a
pH meter.
We’re going to use a pH meter in lab today
Measuring acid content, indicators

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pH can also be measure using indicators.
An indicator is a chemical that will change
color when a specific pH is reached.
The indicator that we are using in lab today is
called phenolphthalein.
When the pH exceeds 8.2 (more or less
neutral) the color will change from clear to
fuchsia.
Measuring acid content, indicators


When your phenolphthalein turns this color
you know that you’ve arrived.
Watch out, it comes quick
We’re going to lab!!!


It has been said that soft drinks contain a lot
of acid. But exactly how much acid is in
there?
Ever wondered? Well, regardless, you’re
about to find out. Because…