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Aqueous Reactions and Solution Stoichiometry
Topics
1.
2.
3.
4.
5.
Electrolytes
Concentration and Dilution
Solubility Rules and Precipitation Reactions
Acid-Base Reactions
Oxidation-Reduction (Redox) Reactions
a. Oxidation numbers
b. Agents and half–reactions
c. Activity series
6. Solution Stoichiometry
a. Gravimetric analysis
b. Acid-base titration
Topic 1 – Electrolytes
Electrolytes are substance that will form electrically conductive solutions. The conductivity is proportional to the number of ions produced when the substance is dissolved. Strong electrolytes dissolve with
virtually 100% of the compound dissociated into ions. Non-electrolytes dissolve with virtually none of
the compound dissociated. Weak electrolytes form an equilibrium relationship between dissociated ions
and the parent compound when dissolved.
Example 1:
Compound
NaCl
Ba(OH)2
Sugar
SiO2 (glass)
Description
strong electrolyte
strong electrolyte
non-electrolyte
non-electrolyte
Notes
Very soluble, 100% dissociation
Sparingly soluble, but no Ba(OH)2(aq)
Dissolves as molecule, no dissociation
Does not dissolve at all
Or,
NaCl(s)  Na+(aq) + Cl-(aq)
Ba(OH)2(s)  Ba2+(aq) + OH-(aq)
C6H12O6(s)  C6H12O6(aq)
SiO2(s)  SiO2(s) (not soluble)
Topic 2 – Concentration and Dilution
Concentration is the ratio of amount of solute to amount of solution (or solvent). The amounts can be
measured in a variety of ways but the most common is solute in moles and solution in liters. This is
n( solute)
concentration expressed as Molarity: M 
V( solution)
Example 2:
What is the concentration of a solution made by diluting 3.57 g NaOH to 2.00 L?
M
n( solute)
V( solution)

0.0892 mol
 4.46 10  2 mol / L
2.00 L
= 4.46 x 10-2 M
M=?
N = 3.57 g x 1 mol/40.00 g = 0.0892 mol
V = 2.00 L
Example 3:
What volume of 0.050 M NaCl can be made from 75.0 g of NaCl?
M 
n( solute)
V( solution)
so, V( solution) 
n( solute)
M = 0.050 M = 0.050 mol·L-1
N = 75.0 g x 1 mol/58.44 g = 1.28 mol
V=?
M

1.28 mol L
0.050 mol
= 25.6 L ≈ 26 L
If I dilute a solution, it still has the same number of moles of solute as before the dilution. Based on
M=n/V,
nbefore  M beforeVbefore also nafter  M afterVafter and nbefore  nafter , so
M beforeVbefore  M afterVafter
Example 4:
or genericall y
M 1V1  M 2V2
How much 3.00 M HCl do I need to make 2.00 L of 0.10 M HCl?
M1 = 3.00 M
V1 = ?
M2 = 0.10 M
V2 = 2.00 L
V1 
M1V1 = M2V2
V1 = M2V2/ M1
M 2V2 0.10 M  2.00 L

 0.067 L  67 mL
M1
3.00 M
Topic 3 – Precipitation Reactions
When two solutions are combined, the dissolved electrolytes of one solution have the opportunity to react with electrolytes of the other solution. If the resulting product is also soluble, it will simply stay dissolved. We say it did not react. Alternatively, the resulting product may be insoluble and precipitate.
You must be able to recognize precipitation reactions by memorizing the following solubility rules:
Ion Species
Solubility Rules
Examples
Exceptions
Always soluble
nitrates, acetates,
ammoniums, alkalai salts
PbNO3 AgCH3COO
NH3CO3 NaSO4
None
Generally soluble
halides
BaCl2 CaF2
Ag
sulfates
MgSO4
sulfides
FeS CuS
Ag+ Hg22+ Pb2+ Sr2+
see always soluble above,
2+
2+
2+
Ca
Sr
Ba
carbonates
CaCO3
ZnCO3
see always soluble above
phosphates
CaPO4
ZnPO4
hydroxides
Mg(OH)2
see always soluble above
see always soluble above,
+
2+
2+
2+
NH4 Ca
Sr
Ba
Not soluble
Example 5:
CuSO4
Cu(OH)2
+
Hg2
2+
Pb
2+
What happens if a solution of magnesium nitrate is mixed with a solution of sodium
hydroxide?
Magnesium nitrate and sodium hydroxide are both soluble (given in the question) because according to the rules, all nitrates and all sodium salts are soluble. Thus the reactants can be written as follows:
Na+(aq) + OH-(aq) + Mg2+(aq) + NO3-(aq) →
The possibilities for a double replacement reaction are: NaNO3 and Mg(OH)2
However, according to the solubility rules, all nitrates are soluble and all hydroxides
are insoluble (Mg is not one of the exceptions). Thus, the complete ionic reaction is:
Na+(aq) + OH-(aq) + NO3-(aq) → Na+(aq) + NO3-(aq) + Mg(OH)2(s)
and by removing the spectator ions, the (AP Exam format) net ionic equation is:
Mg2+ + 2 OH- → Mg(OH)2
Beginning with the May, 2007 AP exam (Question 4), all reactions must be written in balanced, net ionic form.
Topic 4 – Acid-Base Reactions
Acids are substances that dissociate in water to produce H+ ions. Bases are substances that dissociate in
water to produce OH- ions. Strong acids and bases dissociate completely. Weak acids and bases are
soluble in their molecular form and only dissociate to a small extent. You must memorize the common
strong acids and bases:
Acids
HCl
HBr
HI
HClO3
HClO4
HNO3
H2SO4
Bases
hydrochloric
hydrobromic
hydroiodic
chloric
perchloric
nitric
sulfuric
NaOH
sodium hydroxide
and all other group I hydroxides
Ca(OH)2
calcium hydroxide
Sr(OH)2
strontium hydroxide
Ba(OH)2
barium hydroxide
(the “heavy” group II metal hydroxides)
You must also memorize several of the more common weak acids and bases:
HCH3COO
H3PO4
H2SO3
HCHOO
NH3
acetic acid
or
ethanoic acid
phosphoric acid
sulfurous acid
formic acid or
methanoic acid
ammonia (a weak base)
Acids and bases react with each other in neutralization reactions to form water and a salt. For example:
molecular equation
net ionic equation
Acids and bases react with each other in neutralization reactions to form water and a salt:
H2SO4 + 2NaOH → 2H2O + Na2SO4
2H+ + OH- → H2O
Mg(OH)2 + 2HCl → MgCl2 + 2H2O
Mg(OH)2 + 2H+ → Mg2+ + 2H2O (HCl is strong!)
Acids and bases react with each other in neutralization reactions to form a gas:
2HCl + Na2S → H2S + 2NaCl
2H+ S2- → H2S (formation of a gas, H2S)
HCl + NaHCO3 → NaCl + H2O + CO2
H+ + HCO3- → H2O + CO2
Recognize these reactions when you see them and be able to write net ionic equations.
Topic 5 – Oxidation-Reduction (Redox) Reactions
Redox reactions occur through the transfer of electrons between reactants. A species is said to be oxidized when it loses electrons. A species is said to be reduced when it gains electrons. Use this mnemonic device:
OIL RIG
Reduction Is Gaining
Oxidation Is Losing
Oxidation Numbers:
The assignment of oxidation numbers to atoms within a compound or ion is a useful bookkeeping exercise. Think of an oxidation number as the charge on an atom or, if it is not an ion, the “pretend” charge
on an atom in a covalent bond. For example, in water, H2O, the bonds are covalent, so there is no actual
charge on hydrogen or oxygen. However, we know that oxygen is highly electronegative and the electrons are held more closely to the oxygen than the hydrogen atoms. Thus, we can “pretend” that the oxygen has a charge (oxidation number) of 2- and each of the hydrogens has a charge (oxidation number)
of 1+. Note that the oxidation numbers add up to the charge on the water molecule, which is zero [ (+1)
+ (+1) + (-2) = (0) ].
7. The oxidation numbers
of all of the atoms in a compound or poly-atomic ion
must sum to the charge on
that compound or ion.
You must be able to use the following rules to determine oxidation numbers:
{
Example 6:
The oxidation number of elements in their elemental form is 0.
The oxidation number on any monatomic ion is the charge.
The oxidation number of fluorine is always -1
The oxidation number of hydrogen is always +1, except it is -1 when bonded to a metal.
5. The oxidation number of oxygen is usually -2, except when it is part of the
peroxide ion, O22- when it is -1.
6. The oxidation of other atoms is usually same as the charge if it was a
common ion.
Determine the oxidation number on each atom of SBr2.
atom
S:
Br:
Example 7:
1.
2.
3.
4.
count
1
2
oxidation #
+2
-1
total
+2
-2
0
rule
rule #7 – add to zero
rule #6 – Br is -1 if ion
Determine the oxidation number on each atom of Cr2O72-.
Cr
O
2
7
+6
-2
+12
-14
-2
rule 7 – add to -2
rule 5 – oxygen is -2
the charge on Cr2O72- is 2-
Agents and half-reactions:
You also need to be able to identify the atoms that are oxidized, the atoms that are reduced, the agent
that oxidizes the oxidized atom (oxidizing agent), and the agent that reduces the reduced atom (reducing
agent).
Example 8:
Name the element oxidized, the element reduced, the oxidizing agent, and the reducing agent in the following reaction: 2Al + Fe2O3 → Al2O3 + 2Fe
According to rule 1, the oxidation numbers of elemental Al and Fe are 0. Using the
rules as in examples 6 and 7 above, we determine that the oxidation number of oxygen in both compounds is –2, the oxidation number of Fe in Fe2O3 is +3, and the oxidation number of Al in Al2O3 is also +3. Thus,
Aluminum is oxidized from 0 to +3 (oxidation is losing electrons)
Iron is reduced from +3 to 0 (reduction is gaining electrons)
Fe2O3 is the agent that caused Al to be oxidized (Fe2O3 is oxidizing agent)
Elemental Al is the agent that caused Fe in Fe2O3 to be reduced (Al is reducing agent)
Finally, you must be able to write the oxidation and reduction half-reactions.
Example 9:
Write the oxidation and reduction half-reactions for Example 8 above.
Oxidation: Al → Al3+ + 3eReduction: Fe3+ + 3e- → Fe
Note that adding the half reactions gives the net reaction:
Al + Fe3+ = Al3+ + Fe
Activity series:
The activity series is just a summary of many experimental replacement reactions, and can be used to
predict when hydrogen and metal displacement reactions will occur. Given the activity series, tha following two rules apply:
1. Hydrogen displacement: Any metal above hydrogen in the activity series will displace hydrogen from water or from an acid. Metals below hydrogen will not react with either water or an acid.
2. Metal displacement: Any metal will react with a compound containing any metal ion listed below it.
Example 10: Predict the outcome of the reactions shown below and state the rule that pertains:
Mg(s) + HCl(aq) → ?
Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)
(magnesium lies above hydrogen in the series, so the hydrogen in HCl is displaced.)
Al(s) + Cu(NO3)2 → ?
2Al(s) + 3Cu(NO3)2(aq) → 3Cu(s) + 2Al(NO3)3(aq)
(aluminum is above copper in the series or more reactive,
so the aluminum displaces the copper.)
You don’t need to memorize the activity series, because the order is the same as the Standard Reduction
Potentials table that is given in the free response section of the AP exam. However you should understand the trend of activity as it relates to the periodic table, since you may need to make some predictions in the multiple-choice section where you only have access to the periodic table. For example,
know that aluminum is much more active than copper and magnesium is much more active than lead.
Remember, reactivity of the alkali metals is highest and general goes to a minimum at gold, and then
rises again to the right of gold.
Topic 6 – Solution Stoichiometry
Problems in solution stoichiometry require you to read carefully, work methodically, and be flexible in
your problem-solving strategies.
Gravimetric analysis:
Gravimetric analysis is an analytical technique based on separating solids from a solution or mixture.
Generally, in involves forming a precipitate, measuring its mass, and then using stoichiometry to make
calculations regarding the original solute or solution. It requires precipitation reactions to proceed to
completion (no slightly soluble precipitates).
Example 11: 137.5 mL of a solution containing lead ions is mixed with 1.0 M NaCl until no more
precipitate forms. The precipitate is filtered, washed, and dried, and then found to
have a mass of .7038 g.
a. What mass of lead was in the original sample?
b. What was the concentration of lead the original sample?
a. the equation for the precipitation reaction must be: Pb2+ + 2Cl- → PbCl2
0.7038 g PbCl 2 1mol PbCl 2
1mol Pb
207.2 g Pb



 0.5244 g Pb
1
278.1 g PbCl 2 1mol PbCl 2
1mol Pb
b. M = n/V = 0 .5244 g Pb  1mol Pb  1000mL  .01841M
137 .5 mL
207.2g Pb
1L
Acid-base titration:
Titration calculations are stoichiometry problems disguised with units of molarity instead of the usual
units of grams and moles. Always remember to convert between species using mole ratios from the balanced equation rather than molarity or mass.
Example 12: What is the concentration of a hydrochloric acid solution if 39.72 mL of 0.1015 M
NaOH is required to neutralize a 50.00 mL sample of the unknown acid solution?
The balanced equation is: H+ + OH- → H2O
39.72mL NaOH
1L NaOH
0.1015 mol NaOH 1mol OH 
1mol H  1mol HCl
1
1000 mL HCl








1
1000 mL NaOH
1L NaOH
1mol NaOH 1mol OH  1mol H  50.00 mL HCl
1L HCl
= .08063M
Note that this entire equation is actually a manifestation of M1V1 = M2V2 or (if a=acid
and b= base): MaVa = MbVb. Like this:
Ma = ?
Va = 50.00 mL
Mb = 0.1015 M
Vb = 39.72 mL
Ma 
Mb Vb 0.1015 M  39.72 mL

 0.08063 M
Va
50.00 mL
Caution!!! This only works if the acid-base mole ratio is 1:1