Review of Basic Concepts,
Molarity, Solutions, Dilutions
and Beer’s Law
Chapter 4
4.5
Aqueous Solutions
In Chemistry, many reactions take place in
water. This is also true for Biological
processes.
 Reactions that take place in water are said
to occur in an aqueous solution.
 Three types of reactions take place in
aqueous solutions: Precipitation, AcidBase and Redox.

Properties of Aqueous Solutions




Solution- a homogeneous mixture of two or
more substances.
Solute- a substance in a solution that is present
in the smallest amount.
Solvent- a substance in a solution that is
present in the largest amount.
In an aqueous solution, the solute is a liquid or
solid and the solvent is always water.
Properties of Aqueous Solutions




All solutes that dissolve in water fit into one of
two categories: electrolyte or non-electrolyte.
Electrolyte- a substance that when dissolved in
water conducts electricity
Non-electrolyte- a substance that when
dissolved in water does not conduct electricity.
To have an electrolyte, ions must be present in
water.
Electrolytic Properties of
Aqueous Solutions

NaCl in water.
What happens?
 NaCl(s) → Na+(aq) + Cl–(aq)
 Completely dissociates

Strong vs. Weak Electrolytes

How do you know when an electrolyte is
strong or weak?

Take a look at how HCl dissociates in
water.

HCl(s) →
H+(aq) + Cl–(aq)
Electrolytic Properties of
Aqueous Solutions
Electrolytic Properties of Aqueous
Solutions
Hydrated Ions
Electrolytic Properties of
Aqueous Solutions
What about weak electrolytes?
 What makes them weak?


Ionization of acetic acid
 CH3COOH(aq)
↔ CH3COO–(aq) + H+(aq)
Electrolytic Solutions
Precipitation Reactions
Precipitation Reaction- a reaction that
results in the formation of an insoluble
product.
 These reactions usually involve ionic
compounds.
 Formation of PbI2:


Pb(NO3)2(aq) + 2KI(aq) →
PbI2(s) + 2KNO3(aq)
Preciptate
Precipitate
Precipitation Reactions




How do you know whether or not a precipitate
will form when a compound is added to a
solution?
By knowing the solubility of the solute!
Solubility- The maximum amount of solute that
will dissolve in a given quantity of solvent at a
specific temperature.
Three levels of solubility: Soluble, slightly soluble
or insoluble.
Precipitation Reactions
Determining Solubility

Determine the level of solubility for the
following:
(1) Ag2SO4
(2) CaCO3
(3) Na3PO4
Acid-Base Reactions




Acids- generally have a sour taste, change
litmus from blue to red, can react with certain
metals to produce gas, conduct electricity.
Bases- generally have a bitter taste, change
litmus from red to blue, feel slippery, conduct
electricity.
BrØnstead Acid- proton donor
BrØnstead Base- proton acceptor
Acid-Base Reactions

Acid or Base?

HCl(aq) + H2O(l) → H3O+(aq) + Cl–(aq)

NH3(aq) + H2O(l) → NH4+(aq) + OH–(aq)
Acid-Base Reactions

Look at the following compounds and
decide whether they are a BrØnstead Acid
or a BrØnstead Base.
HBr
 NO2–
 HCO3–

Acid-Base Reactions
Oxidation-Reduction Reactions
Can also be called Redox reactions.
 Considered electron-transfer reactions.
 Occur in steps called half-reactions.


Half-Reactions- Two parts to a redox
reaction that explicitly show the electrons
involved and where they are transferred.
Oxidation Reduction Reactions


Oxidation Reaction- refers to the half-reaction
that involves the loss of electrons.
Reduction Reaction- refers to the half-reaction
that involves the gain of electrons.
OILRIG


Oxidizing agent- the compound or ion in a
redox reaction that donates electrons.
Reducing agent- the compound or ion in a
redox reaction that accepts electrons.
Oxidation-Reduction Reactions
Concentration of Solutions



Concentration of a Solution- amount of solute
present in a given quantity of solvent or solution.
We will use Molarity, M for measurement.
Molarity can also be called Molar Concentration.
Molarity– the number of moles of solute per liter
of solution.

Molarity- moles of solutes/ liters of solution


Or n/v
Moles- grams of sample/ molecular weight of sample

Or g/ mw
Concentration of Solutions
How many moles are there in 24.0g of C?
moles = g/mw
moles = 24.0g C/ 12.0g C
moles = 2.0
There are 2.0 moles of C in 24.0g of C.

Concentration of Solutions

How many grams are in 2.0 moles of Boron?
moles= g/MW
2.0 moles = grams/ 10.81g Boron
2.0 moles x 10.81g Boron = grams
Grams = 21.62

There are 21.62 g of Boron in 2.0 moles of
Boron.
Concentration of Solutions

What is the Molarity of a 1L solution
containing 9.0g HCl?
9.00g HCl x 1 mol HCl/ 18.00g HCl
= 0.5 mol HCl
M = n/v
M = 0.5 mol HCl/ 1L
M = 0.5
The concentration of the solution is 0.5M.
Preparation of Solutions
Now that you know how to calculate M, n
and v, what does that mean?
 You can make your own solutions!
 What are the steps in making a proper
solution?

Preparation of Solutions
Concentration of Solutions

How many grams of Potassium Dichromate, K2Cr2O7, are required to
prepare a 250mL solution with a concentration of 2.16M?
250mL x 1L/ 1000mL = .250L
M= n/v
n= M x v
n= 2.16M x .250L
n= 0.54 mol
moles = g/MW
Grams = moles x MW
Grams = 0.54 mol K2Cr2O7 x 294.2 g K2Cr2O7
Grams = 159
159 grams of K2Cr2O7 are needed to prepare the requested solution.
Concentration of Solutions

In a biochemical assay, a chemist needs to add 0.381g of glucose to a
reaction mixture. Calculate the volume in millimeters of a 2.53M glucose
solution that she should use for this addition.
moles = g/MW
moles = 0.381g C6H12O6/ 180.2g C6H12O6
moles = 2.114 x 10 –2 mol C6H12O6
M = n/v
v = n/M
v = 2.114 x 10 –2 mol C6H12O6 / 2.53M C6H12O6
v = 8.36mL
She should use 8.36mL of the 2.53M glucose solution.
Preparation of Solutions

Explain the process of creating 1L of 3.0M KCl.
M = n/v
n=Mxv
n = 3.0M x 1L
n = 4.0 mol of KCl needed
moles= g/MW
Grams = moles x MW
Grams = 4.0 mol KCl x 36.0g KCl
Grams = 144g KCl
Weigh out 144g of KCl. Put in a 1L flask. Add enough dH20 to
dissolve KCl. Fill flask to 1L meniscus.
Dilution of Solutions



Dilution- the procedure for preparing a less
concentrated solution from a more concentrated
one.
Dilutions can be made in increments of 10, 20,
50 or any other value.
Serial Dilution- the process of diluting a solution
by removing part of it, placing this in a new flask
and adding water to a known volume in the new
flask.
Dilution of Solutions

When you want to dilute a solution, what
happens to the number of moles present
in the solution?
Do they increase?
 Decrease?
 Stay the same?

Dilution of Solutions
Dilution of Solutions
Dilution of solutions

Since moles are constant before and after
dilution, we can use the following formula
for calculations.
 MiVi
= MfVf
Dilution of Solutions

Describe how you would prepare 800mL of a 2.0M H2SO4 solution,
starting with a 6.0M stock solution of .
800mL x 1L/ 1000mL
= 0.800L
MiVi = MfVf
6.0M x Vi = 2.0M x 0.800L
6.0M x Vi = 1.6M x L
Vi = 1.6M x L/ 6.0M
Vi = 0.26L
0.26L of the 6.0M H2SO4 solution should be diluted to give a
final volume of 800mL.