Unit 3
Solution Chemistry &
Reactions
Molarity
 Molarity
– the number of moles of solute in
one liter of solvent to make a solution




Usually the number of moles of a solid
dissolved in water
The solid can be any solid capable of any
amount of dissolving and/or dissociation
The solvent can be any pure liquid
Molarity is often referred to as a
concentration
Molarity
 Molarity
is abbreviated using M
 M = moles solute
Liters solvent
 Molarity can also be indicated by [1.0 M]

Brackets indicate a concentration
Calculate…
 You
have 11.24 grams of barium chloride
and you dissolve it into 250 mL of water.
Calculate the molarity of the solution.
 First, turn your grams into moles
 Second, turn your mL into L
 Third, plug into the molarity equation
 Solve for M
Calculate




You want to create 500 mL of a 0.4M solution
of Sodium hydroxide. Calculate the mass of
Sodium hydroxide required to create this
solution.
First, turn your mL into L
Second, plug into the equation and solve for
the number of moles necessary
Third, convert your moles into grams using a
conversion factor with the molar mass of
NaOH
Solution Stoichiometry
 Reactions
can occur between two
solutions – but it is still the moles of each
molecule (or ion) that are reacting.
 It is often necessary to find the number of
moles in a solution prior to using a
balanced equation to solve a
stoichiometry problem.
Solution Stoichiometry
 Calculate
the molarity of a hydrochloric
acid solution necessary to neutralize
(react completely) with 250 mL of a 0.75M
solution of Calcium hydroxide if 300 mL of
HCl are used.
Solution Stoichiometry
 2HCl
+ Ca(OH)2 → 2H2O + CaCl2
 0.75M = X moles Ca(OH) 2
0.250 L
 0.19 moles Ca(OH) 2
 0.19 mol Ca(OH) 2 x 2 mol HCl = 0.38 mol HCl
1
1 mol Ca(OH)2
 0.38 mol HCl = 1.27 M HCl
0.300 L
Serial Dilutions



Solutions often come in very high
concentrations that we do not use in actual
laboratory experiments. It is a cheap way to
obtain a lot of chemicals.
We can dilute the solution by adding water to
any concentration and volume we choose
The dilution formula ONLY works if there is NO
CHEMICAL REACTION

If there IS a reaction, you must use stoichiometry
Dilutions
 In

a dilution M1V1 = M2V2
Where M1 is the first molarity, V1 is the first
volume, M2 is the second molarity and V2 is
the second volume
Dilutions
 Hydrochloric
acid is sold in a
concentration of 12 M. You want 1.5 L of
a 1.0 M solution for a lab. How much 12M
acid must you measure out to dilute to
the proper concentration?
 M1 = 12 M, V1 = ?
 M2 = 1.0 M, V2 = 1.5 L
 (12 M)(V1) = (1.0 M)(1.5 L)
 V1 = 0.125 L or 125 mL
Oxidation and Reduction
(Redox)
 Electrochemistry
deals with
nonspontaneous Redox reactions
 Electrons are transferred
 Spontaneous redox rxns can transfer
energy



Electrons (electricity)
Heat
Non-spontaneous redox rxns can be
made to happen with electricity
Electrochemistry Terminology
 Oxidation
– A process in which an
element attains a more positive oxidation
state
Na(s)  Na+ + e Reduction – A process in which an
element attains a more negative
oxidation state
Cl2 + 2e-  2Cl LEO says GIR


Loss of electrons = oxidation
Gain of electrons = reduction
Electrochemistry Terminology
Oxidizing
agent
The substance that is reduced
is the oxidizing agent
 Reducing agent
The substance that is oxidized
is the reducing agent
Balancing a Redox reaction in
Acidic Solutions
1.
2.
Write separate equations for the oxidation
and reduction half-reactions.
For each half reaction
a.
b.
c.
d.
3.
4.
Balance all the elements except H and O
Balance O by using H2O
Balance H by using H+
Balance the charge using e-
If necessary, multiply one or both balanced
half reactions to equalize the e- transferred.
Add the half reactions canceling identical
species.
Examples – Acid Solution
MnO4-(aq)

5e- + 8H+(aq) + MnO4-(aq)  Mn2+(aq) + 4H2O(l)
5Fe2+(aq)  5Fe3+(aq) + 5e-

5Fe2+(aq) + 8H+(aq) + MnO4-(aq)  Mn2+(aq) + 5Fe3+(aq) + 4H2O(l)


+ Fe2+(aq)  Fe3+(aq) + Mn2+(aq)
H+(aq) + Cr2O72-(aq) + C2H5OH(aq)  Cr3+(aq) + CO2(g) + H2O(l)



4H2O(l) + C2H5OH(aq)  2CO2(g) + H2O(l) + 12H+(aq) + 12e12e- + 28H+(aq) + 2Cr2O72-(aq)  4Cr3+(aq) + 14H2O(l)
16H+(aq) + C2H5OH(aq) + 2Cr2O72-(aq)  2CO2(g) + 4Cr3+(aq) +
11H2O(l)
Reactions with Calculations
 Zinc
metal reacts with Copper (I) ions to
form Zinc ions and Copper metal.
Calculate how much copper metal is
formed if you start with 34.5 grams of Zinc.
Assume everything reacts completely.
Reactions with Calculations
 Permanganate
ions react with Lead (II)
ions to make both Manganese (II) and
Lead (IV) ions. How much permanganate
must you start with if you make 2.56 grams
of manganese (II) ions in the process of
the reaction?
Reactions with Calculations
 75
mL of a 2.5 M solution of Sodium
hydroxide reacts completely with an
unknown concentration of sulfuric acid. If
there are 25 mL of sulfuric acid, how
much sodium hydroxide is necessary to
neutralize all of the acid? (Neutralization
means react completely)