MTChem1 (Lab) – Module 3 | PostLaboratory Discussion – Types of Chemical
Reactions and Solutions
October 1, 2022
Sodium bicarbonate – baking soda
Acetic acid – vinegar
The gas produced – carbon dioxide
By Jan Marc E. Dimatulac
TOPIC OUTLINE
I.
II.
III.
Chemical Reaction
Solutions
Problem Solving
CHEMICAL REACTION
Set-up A (Vinegar and Baking Soda)
Type of Reaction
•
Two types of reaction occurred:
o Double displacement reaction
▪ Partner switching
o Decomposition reaction
Double displacement/replacement
Evidences of Chemical Change
•
•
Bubbles – formation of gas
Formation of gas indicates that gas is
produced and gas is included in the
product.
•
•
The sodium in sodium bicarbonate will
combine with the acetate in the acetic
acid producing sodium acetate.
For the next product, the remaining
hydrogen from the acetic acid, will
combine with the carbonate from baking
soda to produce carbonic acid.
Decomposition reaction
•
Chemical Equation
NaHCO3 + HC2H3O2 → NaC2H3O2 + H2O
+ CO2
Sodium bicarbonate + acetic acid → sodium
acetate + water + carbon dioxide.
•
Carbonic acid is an unstable compound,
which indicates that it can immediately
and can easily undergo a type of
reaction. It is not the final product.
Carbonic
acid
will
undergo
decomposition reaction to produce water
and carbon dioxide.
Type of Reaction
Fermentation
•
Set-up B (Sugar Solution Water and Yeast)
Evidences of Chemical Change
•
Bubbles – indicates the formation of gas.
Chemical Equation
C12H22O11 + H2O + Yeast → 4C2H5OH +
4CO2
Table sugar + Hot water + Yeast will produce
Ethanol and carbon dioxide (in form of gas)
Yeast – microorganism; it will be using table
sugar to produce and synthesize ethanol and
carbon dioxide.
Ex. Someone baking – will set aside dough first
because they use this to make the dough rise
because it is synthesizing the sugar in the dough
to produce carbon dioxide.
Happens when sugar is converted to
alcohol or acid, and a gas.
Set-up B (Sugar Solution Water and Yeast)
Evidences of Chemical Change
•
Bubbles – indicates the formation of gas.
Chemical Equation
H2O2 + Yeast → H2O + O2
Hydrogen peroxide + yeast (microorganism) will
produce water and oxygen.
Type of Reaction
Type of Reaction – Combustion
Distinct number of bubbles are observable in the
setup, as well as increase in temperature in the
container.
Two types of combustion:
Decomposition
Catalase Test
•
Used to detect the existence of an
enzyme,
catalase
(from
a
microorganism), and to differentiate
catalase positive from catalase negative
organisms.
o Catalase positive – has catalase
in them.
o Catalase negative – catalase is
not synthesized by the organism.
Steps for catalase test:
1. Prepare glass slide
2. Use hydrogen peroxide, a droplet only
3. Using a wooden stick, you will get a
colony or a sample of microorganism
from the medium.
4. Introduce it to the droplet of hydrogen
peroxide.
5. Observable white particle indicates that
there is a production of gas because
there is formation of bubbles.
The production of gas indicates that the
microorganism tested is positive to the enzyme
catalase.
•
•
Complete combustion
Incomplete combustion
Complete Combustion
C18H38 (Paraffin) + O2 → CO2 + H2O
•
•
•
•
Involvers complete burning of fuel
Takes place when enough oxygen is
present
Produces blue flame, also known as
Truplic (?)
Products: carbon dioxide and water
Complete Combustion
C18H38 (Paraffin) + O2 → C + CO + H2O
•
•
•
•
Involves partial burning of fuel
Takes place when there is not enough
oxygen present
Red-Orange flame (e.g., when lighting up
candles)
Products: carbon monoxide, carbon
(black soot), water
In the heating of the spoon, the reaction
observed is an incomplete combustion because
there is the production of the black soot.
Set-up B (Heating Spoon)
SOLUTIONS
Solubility of Solids
SET-UP A (SALT)
Evidences of Chemical Change
•
•
Formation of black soot.
Formation of black smoke.
•
Water
Salt with different solvents.
•
•
Soluble
o Salt is NaCl (+ and - charge) while
water has both the charges (+ and
- charge).
o Since sodium is positively
charged/has a positive charge, it
will interact with the negative
charge of water; the negative
charge of chlorine will interact with
the negative charge of water,
making both soluble.
Polar solvent + Polar ionic
o “Like dissolves like”
o Since water is considered as a
polar solvent, and it is combined
with a polar ionic solute, it will be
soluble.
Alcohol
•
•
Insoluble
Semi polar + Polar ionic
o Alcohol is a polar solvent, but it
means that there is less charge in
them, making it harder for the
charges to interact with the
charges with the salt.
Ethanol
•
•
Polarity Index: 5.2
Boiling Point: 78.0 °C
Water
•
•
Polarity Index: 10.2
Boiling Point: 100.0 °C
Polarity Index – is how much polar and how
much charges do the solvent contain.
Acetone and Ethanol have very big (small)
polarity index indicating that it cannot dissolve
salt, and it has very little charges.
Unlike water which has a high polarity index,
indicating that it can dissolve polar ionic
solutions.
The lower the polarity index, the lower the
boiling point is.
Acetone
•
•
Insoluble
Weak polar + Polar ionic
o Acetone is a weak polar; there are
lesser charges in them, making it
difficult to interact with the charges
of the salt.
SET-UP B (TAWAS)
•
Tawas with different solvents.
Water
•
•
Soluble
Polar solvent + Polar ionic
o Tawas is polar ionic, like salt.
Alcohol
•
•
Acetone
•
•
Polarity Index: 5.1
Boiling Point: 56.0 °C
Insoluble
Semi polar + Polar ionic
o Alcohol is semi polar (containing
weak charges), similar with
acetone.
Acetone
•
Insoluble
•
o Atoms in the molecules have
nearly similar electronegativity
values.
Weak polar + Polar ionic
Tawas
•
•
•
Chemical name: Aluminum Potassium
Sulfate
Another name: Potash alum
Chemical formula: KAl (SO4)2
POLAR
SOLVENT
SOLVENT
AND
NON-POLAR
Polar Solvents
•
•
•
•
Definition
o Polar solvents are liquids that
have a large dipole moment
▪ E.g., water
Ability to dissolve compounds
o Dissolve polar compounds
Charge separation
o Partial positive charges and partial
negative
charges
can
be
observed.
Electronegativity
o Molecules contain atoms with
higher
differences
in
electronegativity.
Non-Polar Solvents
•
•
•
•
Definition
o Nonpolar solvents are liquids that
have no dipole moments.
Ability to dissolve compounds
o Dissolve nonpolar compounds
Charge separation
o No charge separation
Electronegativity
Dipole moment
•
Represented and looks like an arrow,
wherein it starts with a positively charged
element and then it points towards the
negatively charged element.
Example 1:
H2O – LEWIS STRUCTURE
•
•
•
Positively charged hydrogen, negatively
charged oxygen.
We have an arrow wherein it starts with a
positively charged element which is the
hydrogen, and it points towards the
negative charge which is the oxygen.
For us to identify if it is dipole or no dipole,
both arrows need to point towards the
same direction, and in this example, they
are both pointing towards the negative
charge, making it POLAR because it has
a large dipole moment.
Example 2:
CO2 – LEWIS STRUCTURE
•
•
•
•
Again, in DIPOLE moments the positively
charged will be pointing towards the
negative charge.
Negative charge in CO2 is oxygen, and
the positive charge is carbon.
C, the positive charge, is both pointing
outwards; the opposite way. Unlike in
H2O, a polar solvent, they are both
pointing the same way which is the single
negative charge.
CO2 has NO DIPOLE because they are
both pointing the opposite direction,
cancelling these two arrows.
Electronegativity
Always remember that the EN of polar solvents
should be GREATER THAN OR EQUAL TO 0.5
(≥ 0.5)
Example 1: Water (H2O)
EN
H = 2.1
O = 3.5
Difference: 1.4
•
1.4 is greater than 0.5, and that indicates
that water is a polar solvent.
Solubility of Liquids
SET-UP C
Water + Ethanol
•
Miscible
o Ethanol has a functional group
(OH) or the hydroxyl group.
•
o Hydroxyl group – considered as
a hydrophilic group which
means that it is water loving.
o This results to the hydroxyl group
attaching to water with a hydrogen
bond, causing them to dissolve
each other; thus, their solubility
with each other is miscible.
Polar solvent + Polar solvent
o Likes dissolves like.
Water + Acetone
•
•
Miscible
Polar solvent + Polar solvent
Water + Oil
•
•
Immiscible
Polar solvent + Nonpolar solvent
o An oil contains the so-called
hydrophobic molecules.
o Hydrophobic – water fearing.
Instead of mixing with water, they
repel it, causing them to not mix
with each other.
NOTE: Polar solvents have large dipole
moments (aka “partial charges”); they contain
bonds between atoms with very different
electronegativities, such as oxygen and
hydrogen. Non polar solvents contain bonds
between atoms with similar electronegativities,
such as carbon and hydrogen (think
hydrocarbons, such as gasoline).
SET-UP D
Ethanol + Acetone
•
•
Miscible
Polar solvent + Polar solvent
o Likes dissolve like.
Ethanol + Oil
•
•
Immiscible
Polar solvent + Nonpolar solvent
o Ethanol is polar, oil is nonpolar.
Acetone + Oil
•
•
Immiscible
Polar solvent + Nonpolar solvent
PROBLEM SOLVING
Mark and Johnny are two medical technologists
working in a district hospital. They were asked
to prepare the following solutions in the
laboratory: Mark prepares 25.5 grams of
sodium hydroxide pellets, dissolved in a 750
mL of solution. While Johnny was asked to
prepare 2.00 L of 0.01M solution of HCl.
2 significant figures, Complete Solution
Problem #1
Calculate the molar concentration of the NaOH
solution that Mark has prepared.
•
•
Given
o 25.5 g NaOH
o 750 mL or 0.75 L
None of the given follows the unit of
measurement that should be used in our
formula so we have to convert.
o mL / 1000 = L
o g / Molar Mass = mol
750 𝑚𝐿
1𝐿
×
= 𝟎. 𝟕𝟓𝑳
1
1000 𝑚𝐿
25.5𝑔 𝑜𝑓 𝑁𝑎𝑂𝐻 ×
1 𝑚𝑜𝑙 𝑜𝑓 𝐶𝑎 (𝑂𝐻)2
39.998 𝑔 𝑁𝑎𝑂𝐻
= 𝟎. 𝟔𝟑𝟕𝟓𝟑𝟏𝟖𝟕𝟔 𝒎𝒐𝒍 𝑵𝒂𝒐𝑯
•
Looking for molar concentration:
𝑀𝑜𝑙𝑎𝑟𝑖𝑡𝑦 (𝑀) =
𝑀𝑜𝑙𝑎𝑟𝑖𝑡𝑦 (𝑀) =
*Mostly the immiscible ones they contain oil
which is a nonpolar solvent.
𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
𝐿 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡
0.637531876 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻
0.75 𝐿
𝑴𝒐𝒍𝒂𝒓𝒊𝒕𝒚 (𝑴) = 𝟎. 𝟖𝟓
𝒎𝒐𝒍
𝒐𝒇 𝑵𝒂𝑶𝑯
𝑳
Problem #2
Calculate the moles of HCl needed by Johnny.
•
•
Given
o 2.00 L
o 0.01M solution of HCl (M = Molar
Concentration)
Since we are talking about molar
concentration, we will use the formula for
molar concentration.
𝑀𝑜𝑙𝑎𝑟𝑖𝑡𝑦 (𝑀) =
0.01 𝑀 =
𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
𝐿 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡
𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
2.00 𝐿
0.01 𝑀 (2.00 𝐿) = 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
𝒎𝒐𝒍𝒆𝒔 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒆 = 𝟎. 𝟎𝟐 𝒎𝒐𝒍𝒆𝒔 𝒐𝒇 𝑯𝑪𝒍
*M (Molar Concentration) = mol/L
Problem #3
If Johnny has 2.00 L of 0.01M HCl solution, what
would be the final concentration if he dilutes it
further to 3.5L?
•
•
Given
o C1 (First Concentration) = 0.01M
o V1 (Volume) = 2.00 L
o C2 (Final Concentration) = x
o V2 (Final Volume) = 3.5 L
In looking for dilution, we use the formula:
𝐶1 𝑉1 = 𝐶2 𝑉2
(0.01 𝑀)(2.00 𝐿) = 𝐶2 (3.5 𝐿)
0.02 𝑚𝑜𝑙 = 𝐶2 (3.5 𝐿)
𝑪𝟐 = 𝟎. 𝟎𝟎𝟓𝟕
𝒎𝒐𝒍
𝒐𝒇 𝑯𝑪𝒍
𝑳
Scientific notation: 5.7x10-3 M or mol/L of
HCl