ENG 201 - CHEMISTRY APPLICATIONS IN ENGINEERING
GENERAL CHEMISTRY
Table of Contents
Module 2:
2.1 Water as Universal Solvent
2.2 Ways of Expressing Solution
Concentrations
2.3 Colligative Properties
2.4 Acids and Bases: pH and pOH
Topic 2.1: Water as Universal Solvent
Role of Water as a Solvent
• It is one of the most important
substances on Earth
• it can dissolve or dissociate many
particles regardless of the type of
bond due to its polar nature
- Polar nature:
o Uneven charge
distribution
o Bent shape of the
molecule
Heterogenous Mixtures
• A mixture that has a non-uniform
composition
• It may differ in size or shape
Examples:
Homogenous Mixtures
• A mixture wherein, throughout the
mixture, the composition is uniform
• It is also called solution
Examples:
Solutions
Hydrogen Bond
• Unusually strong type of a dipole
force
• Strong H bonds occurs among polar
covalent molecules containing H and
either F, N, or O
• H from one molecule can associate
itself with the negative end of the
dipole of another molecule
Hydrogen bonding in water accounts for:
• High specific heat
• High boiling point
• Higher density of the liquid phase
relative to the solid
NOTE: A single water molecule can
participate in 4 hydrogen bonds with
other molecules
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Topic 2.2: Ways of Expressing
Solution Concentrations
Unsaturated Solution
• a solution in which more solute can
be dissolved.
Example:
Saturated Solution
• A solution in which the maximum
amount of solvent has been
dissolved.
• Any more solute added will sit as
crystals on the bottom of the
container.
Example:
Supersaturated
• A solution that contains more than
the average solvent that can be
dissolved at a given temperature.
Example:
Concentration Units
• Molarity (M)
• Mole Fraction (X)
• Mass Percent
• Molality (m)
Molarity
• is the amount of a substance in a
certain volume of solution.
𝑚𝑜𝑙𝑒𝑠 𝑠𝑜𝑙𝑢𝑡𝑒
𝑚𝑜𝑙𝑎𝑟𝑖𝑡𝑦 (𝑀) =
𝑙𝑖𝑡𝑒𝑟𝑠 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
•
Preparation of a solution to a
specified molarity can be attained
using the dilution equation.
𝑀1 𝑉1 = 𝑀2 𝑉2
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Mole Fraction
• Is the ratio of the number of moles of
one component of a solution or other
mixture to the total number of moles
representing all the components.
𝑚𝑜𝑙𝑒 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 (𝑥) =
•
𝑚𝑜𝑙𝑒𝑠 𝑖
𝑡𝑜𝑡𝑎𝑙 𝑚𝑜𝑙𝑒𝑠
By adding up the mole fractions of
all components of a solution, the
sum should always be 1
𝑋𝑎 + 𝑋𝐵 + ⋯ = 1
Molality
• is the amount of a substance
dissolved in a certain mass of
solvent.
𝑚𝑜𝑙𝑎𝑙𝑖𝑡𝑦 (𝑚) =
𝑚𝑜𝑙𝑒𝑠 𝑠𝑜𝑙𝑢𝑡𝑒
𝑘𝑔 𝑠𝑜𝑙𝑣𝑒𝑛𝑡
Conversions Between Concentration
Units
Mass Percent
• Is the ratio of the mass of solute that
is present in a solution, relative to
the mass of the solution, as a whole
𝑚𝑎𝑠𝑠 𝑝𝑒𝑟𝑐𝑒𝑛𝑡 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 =
•
𝑚𝑎𝑠𝑠 𝑠𝑜𝑙𝑢𝑡𝑒
𝑡𝑜𝑡𝑎𝑙 𝑚𝑎𝑠𝑠 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
When the amount of solute is very
small, the concentration is often
expressed as parts per million (ppm)
or parts per billion (ppb)
𝑔 𝑠𝑜𝑙𝑢𝑡𝑒
𝑥 106
𝑔 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
𝑔 𝑠𝑜𝑙𝑢𝑡𝑒
1 𝑝𝑝𝑏 =
𝑥 109
𝑔 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
1 𝑝𝑝𝑚 =
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Sample Problems
1. Copper sulfate is widely used as a dietary supplement for animal feed. A lab technician
prepares a “stock” solution of CuSO4 by adding 79.80 g of CuSO4 to enough water to
make 500.0 mL of solution. An experiment requires a 0.1000 M solution of CuSO4. MW
of CuSO4 is 159.6 g/mol
(a) What is the molarity of the CuSO4 “stock” solution prepared by the technician?
(b) How would you prepare 1.500 L of 0.1000 M solution from the stock solution?
2. Hydrogen peroxide is used by some water treatment systems to remove the
disagreeable odor of sulfides in drinking water. It is available commercially in a 20.0% by
mass aqueous solution. What is the mole fraction of H2O2?
MW of H2O2 = 34.02 g/mol
MW of H2O = 18.02 g/mol
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3. Glucose, C6H12O6, in water is often used for intravenous feeding. Sometimes sodium
ions are added to the solution. A pharmacist prepares a solution by adding 2.0 mg of
sodium ions (in the form of NaCl), 6.00 g of glucose, and 112 g of water.
(a) What is the molality of the glucose in solution?
(b) How many ppm of Na+ does the solution contain?
4. Convert 10 M of H2SO4 into molality. The density of H2SO4 is 1.84 g/mL and the molar
mass of H2SO4 is 98.09 g/mol.
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Topic 2.3: Colligative Properties
Colligative Properties
• Properties that depend primarily on
the concentration or the relative
number of solute particles present
rather than their nature. Such
properties include:
o Vapor pressure lowering
o Boiling point elevation
o Freezing point depression
o Osmotic pressure
Vapor pressure lowering
• Concentrated aqueous solutions
evaporate more slowly than pure
water.
•
Ex: The vapor pressure of water
above a 0.10 M solution of either
glucose or sucrose at 0 °C is about
0.008 mmHg less than that of pure
water
Raoult’s Law
P1 = 𝑋1 𝑃1𝑜
Freezing Point Depression
•
the solution of a nonvolatile solute,
when cooled, only freezes when its
temperature reaches below the
freezing point of the solvent
∆Tf = 𝑇𝑓𝑜 − 𝑇𝑓
∆Tf = 𝐾𝑓 ∗ 𝑚
Wherein:
Kf - molal freezing point constant
m - molality
Boiling Point Elevation
• The solution of a nonvolatile solute,
when heated, only boils when its
temperature exceeds the boiling
point of the solvent. The difference
in temperature is called the boiling
point elevation, ∆Tb
∆Tb = 𝑇𝑏𝑜 − 𝑇𝑏
∆Tb = 𝐾𝑏 ∗ 𝑚
Wherein:
Kb - molal boiling point constant
m - molality
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•
•
•
Osmotic Pressure
The spontaneous passage of a pure
solvent into a solution that is
separated via semipermeable
membrane, wherein solute particles
are not allowed to pass through is
called osmosis.
•
•
osmotic pressure, then the water
flows in the opposite direction. This
is called reverse osmosis
The pressure that is applied to the
solution to stop the influx of solvent
is called osmotic pressure (𝛱). If the
external pressure (P) exceeds the
osmotic pressure, then the water
flows in the opposite direction. This
is called reverse osmosis
π=
𝑛𝑅𝑡
𝑉
π – osmotic pressure
n – moles
r- gas law constant
T- Temperature
V- Volume
Sample Problems
1. A solution contains 82.0 g of glucose, C6H12O6, in 322 g of water. Calculate the vapor
pressure of the solution at 25°C (vapor pressure of pure water at 25°C = 23.76 mm Hg)
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2. An antifreeze solution is prepared containing 50.0 cm3 of ethylene glycol, C2H6O2 (d =
1.12 g/cm3), in 50.0 g of water. Calculate the freezing point of this 50-50 mixture
3. Calculate the osmotic pressure at 15°C of a solution prepared by dissolving 50.0 g of
sugar, C12H22O11, in enough water to form one liter of solution.
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Topic 2.4: Acids and Bases: pH and
pOH
Definition of Acids and Bases
• According to Svante Arrhenius:
o an acid is a species that
produces H+ ions in water
solution
o a base is a species that
produces OH- ions in water
solution
o
Amphiprotic
• a species that can either accept or
donate a proton Ex: Water molecule
Ion Product of Water
•
According to Johannes Brønsted
and Thomas Lowry:
o an acid is a proton (H+ ion)
donor
o a base is a proton (H+ ion)
acceptor
o in an acid-base reaction, a
proton is transferred from an
acid to a base
•
The Brønsted-Lowry reaction above
shows the amphiprotic nature of
water. However, it can also be
viewed as the ionization of a single
water molecule as seen below.
•
In the equilibrium constant
expression for reactions in a
solution:
o solutes enter as their molarity
[]
o the solvent, H2O in this case,
does not appear. Its
concentration is essentially
the same in all dilute
solutions
•
In pure water, both ions are formed
in equal numbers. Therefore,
Brønsted-Lowry Acid-Base Model
•
•
•
Conjugate base - the species formed
when a proton (H+) is removed from
an acid.
Conjugate acid - the species formed
when a proton (H+) is added to a
base
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•
In most water solutions, H+ and OHconcentrations are not equal. Both
ions have an inversely proportional
relationship. If [H+] is greater than
[OH-], the solution is termed acidic.
If [OH-] is greater than [H+], the
solution is termed as basic / alkaline.
pOH
•
power of the hydroxide ion\
•
•
•
if pH < 7.0 - acidic solution
if pH = 7.0 - neutral solution
if pH > 7.0 - basic solution
pH and pOH
pH
•
power of the hydrogen ion
Note: at 25 °C, pH + pOH = 14
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Sample Problem
Calculate, at 25°C:
1. the [H ] and pH of a tapwater sample in which [OH ] = 2.0 X 10E-07.
2. the [H ] and [OH ] of human blood at pH 7.40.
3. the pOH of a solution in which [H ] = (5.0)[OH ]
Ans:
1. [H+] = 5.0 X 10^-8 M, pH = 7.30
2. [H ] = 4.0 X 10^-8 M, [OH ] = 2.5 X 10^-7 M
3. pOH = 7.35
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References
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A%20a%20solution%20in%20which%20more%20solute%20can%20be%20dissolved.
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