ilana kovach (chemistry lab final review)
TRANSCRIPT
Ilana Kovach
Chemistry 1405 Lab Final
Review
Ilana Kovach
Ilana Kovach
Lab 1: Graphing & Data Analysis
Manual Excel
H2O (π. ππ β π. π)
π. ππβ. ππ
.9861g/ml
|. ππππ β π. π
π. π| β πππ
1.39% of Error
Mass 4.5ml
.9861 *4.5 = 4.437g
.9994g/ml
|. ππππ β π. π
π. π| β πππ
.06%Error
Mass 4.5ml
.9994 *4.5 = 4.497g
Ethanol (π. π β π. ππ)
π. ππ β π. ππ
.7791g/ml
|. ππππ β π. πππ
π. πππ| β πππ
1.25% Error
Mass 4.5ml
.7791 *4.5 = 3.506g
.7889g/ml
|. ππππ β π. πππ
π. πππ| β πππ
.013% Error
Mass 4.5ml
.7889 *4.5 = 3.550g
Ilana Kovach
Which Slope More Accurate? Prior to the experiment I knew excel was more accurate in calculations; however due to
actually determining the actual percentage of error in comparing the manual method
verses the Excel method proved my hypothesis. The percentage error is about 1% different
in comparison which might seem small but makes a difference. Small calculation
differences such as density can determine if a random substance floats or sinks. Both have
very small differences to the real number but from my experience of learning hand making
a graph is more prone to human errors. I am thankful for the person who incorporated the
option to transform my data in Excel into a graph.
Ilana Kovach
Lab 2: Calorie Content
Prelab:
1. Specific is the heat required to raise the temperature of the unit mass of a given
substance by a given amount
2. Energy needed in joules to raise a temp of 250ml of water from 22ΒΊC to 45ΒΊC
(Density of H2O =1.00g/ml & Specific Heat= 4.184J)
a. Q=mCβ²T
Q= (250*1) (4.184) (45-22)
Q=24,060J
b. Convert to Kcals
24060J * ππππ
π.ππππ± *
πππππ
πππππππ
5.75Kcal
3. A .50g sample of vegetable oil gives out 19.9kJ of Heat when burned. What is the
caloric value of the oil in kcal/g?
19.9kJ*ππππ
π.ππππ± *
πππππ
πππππππ = 4.76kcal
π.ππππππ
.ππ *2 = 9.52Kcal/g
4. Carbs(150), Protein (50) & fats(7)
(50*4) + (150*4) + (7*9) = 863kcal
5. Formula for calculating %Error
|π¬πππππππππβπ¨πππππ
π¨πππππ |*100 = Percentage of Error
Post lab: 1. Heating water isβ¦.
An endothermic process because the water is absorbing Heat. The water
Temperature rising shows it absorbed the Heat.
2. Based on Net heat calculations burning food isβ¦
An exothermic process because the food was releasing Heat.
3. NOT same caloric content as manufacture
a. Percent of error with the popcorn was 13.8% and marshmallow 91.3%
b. Each popcorn vary in size and marshmallow vary in size
c. Heat was released not in a trapped controlled area
d. Only one experiment needs more trials to be accurate
Ilana Kovach
Popcorn Marshmallow
a. Mass of food material
before heating 0.9578 1.2776 b. Mass of food material
(ashes) left after heating .7914 0.3089 c. Mass of the food material
burned
*(a-b)
0.1664 0.987
d. Volume of water sample 50ml 50ml
e. Mass of water
(Density 1.00)
*D=π
π
50g 50g
f. Tinitial
25.8ΒΊC 25.6ΒΊC g. Tfinal
36.0ΒΊC 30.0ΒΊC h. β²T
10.2ΒΊC 4.6ΒΊC i. Heat absorbed by water
in Joules 2133.84J (50)(4.184)(10.2)
1046J (50)(4.184)(10.2)
j. Convert
*JCal Kcal .51kcal 2133.84*
ππͺππ
π.ππππ±*
πππππ
πππππππ
.25kcal 2133.84*
ππͺππ
π.ππππ±*
πππππ
πππππππ
k. Caloric content of food
cal/g (J/C) 3.1kcal/g .51/.1664
.26kcal/g
.25/.987
l. Caloric content of food
cals/serving(manufacture) 3.6kcal/g 3.0kcal/g m. %of error |
π.πβπ.π
π.π |*100 =
3.8%
|.ππβπ.π
π.π |*100 =
91.3%
Ilana Kovach
Lab 3: Electronic Structure
Prelab:
1. Equipment for flame Testβ¦
Bunsen burner, Tube, Strikers, Metal ions, solutions and sample swabs
2. What is happening at atomic level to give rise to observed energy??
Excited electron drops and emits radiation (photon) depending on the amount of
energy released color results vary
3. Why does each salt have its distinctive flame test color?
Each element has different electron configuration which determines how much
energy is released when electrons fall from an excited state which result in a
different color.
Post lab: 1. Line spectra of different elements related to observed unaided eye
Closer light was to white the more color was saw.
2. Will everyone see all light in all emission spectra
No canβt see all colors without a spectroscope!
3. Representation of all transitions???
NO!!! ONLY see what is visible to the Human Eye
ROY G BIV (Increasing Energy)
Ilana Kovach
Name of Gas Color to Naked
Eye
Observed line spectra
Krypton White R,O,Y,G,V
Neon Red R,O,Y,G
Argon Pink R,O,Y,G,V
Xenon White R,O,Y,G,V
Helium Salmon Y,G,V,B,R
Hydrogen Pink R,O,Y,G,B,V
Gas Tube Emission Spectra
Flame Test
Metal Color of Flame Potassium Chloride Light Pink
Calcium Chloride Orange (Bright)
Copper Chloride Blue rimmed in Green
Strontium Chloride Red (bright)
Sodium Chloride Yellowish/Orange
Ilana Kovach
Lab 4: Nomenclature
Ionic Bonds
Metal & Nonmetal
Potassium Chloride
KCl
Naming is simply the name with the second word ending in -ide
Ionic Bonds
Transition Metals
Iron III oxide
Fe2O3
Naming transition metals use (#) to indicate what ion is used in the Formula
Covalent Bonds
No Metals
Diphosphorus Pentoxide
P2O5
Use prefixes such as:
Mono-, di-, tri-, tetra-, penta-, Hexa, Hepta-, Octo-
Ionic Bonds with Polyatomic
Ions
Ammonium Sulfate
(NH4)2SO4
Polyatomic Ions to Know:
NH4+, CH3COO-,
CN-, CO32-, ClO2
-
, ClO3-, ClO4
-, NO3
-, OH - ,PO43-
, SO42-
Ilana Kovach
Lab 5: VSEPR
Bent
Trigonal
Pyramidal
Bent
Linear
Trigonal Planar
Tetrahedral
Ilana Kovach
Lab 6: Making Hand Cream
MISSING pH Homogeneity Appearance Cooling
Effect
# 1
Nothing 8.5 Homogenous Snow White Cool YES
# 2
Mineral Oil 8 Homogenous Really White Yes &
Sticky
#3
Methyl
Stearate 8
Homogenous Thick & White Yes &
Sticky
#4
Glycerin 8 Homogenous White, Thick &
Creamy
No
#5
Triethanolam 8 Homogenous Yellowish No
Ilana Kovach
Missing Ingredients Effect on
Hand Cream Sample # 1: Everything Present This sample seemed the most similar to traditional lotion one would find in a local store
subtracting the fragrance.
Sample #2: Mineral Oil missing Mineral oil provides spreadibility. Mineral oil helps trap water in the skin by creating an
oily later on top of the skin. The reason this happens is due to the nonpolar properties. The
mineral oil missing seem to come out as a substance having a sticky feel rather than the
desired smooth feeling.
Sample #3: Methyl stearate missing The purpose of methyl stearate is to assist the desired texture which people associate with
lotion. I notice that the lotion missing this ingredient outcome still had the sticky texture
which would not be a desirable product. However it was still contain the thick/white
properties. This substance was also nonpolar.
Sample#4: Glycerin Missing Glycerin surprisingly is a polar ingredient. Reason I previously thought it wasnβt was due
to my mom owning pure glycerin suggested by her dermatologist. I associated glycerin to
the feeling of petroleum jelly. βIt never seems to come off by water alone if applied. The
lotion absent of glycerin seemed greasy, white/; thick and creamy. Researching glycerin
this makes more sense in that glycerin is polar and the way it reacts with skin is bring
water to the surface layer causing your deeper dermis to dry out. However the effects alone
are not as strong in a combination with an oil.
Sample #4: Triethanolamine missing Also called TEA. This chemical is an organic compound derived from ammonia when used
in low concentration to be an alkalizing agent and increase pH in cosmetics. Interesting
note when combined to stearic acid which is present in all the lotions forms a mild soap
thus this chemical acts as an emulsifier in lotion to breakdown bonds and allow water
(polar) Ingredients mix with nonpolar ingredients making ionic bonds. Also interesting the
lotion appearance outcome had a yellowish appearance which questions if color is due this
ingredient. Extra: emulsion by reducing surface tension so water soluble & oil soluble can
blend.
Ilana Kovach
Correlation between Polarity & Effects
on Hand Cream?
Sample 2(Mineral Oil) & 3(Methyl Stearate):
Non polar Properties missing The outcome correlation resulted in the substance being sticky with just the mineral oil
missing and methyl stearate missing could this be related to its properties of being
nonpolar missing? Understanding mineral oil the properties and methyl stearate they
contribute a large part to the texture in the desired smooth feeling so I believe the nonpolar
properties missing could possibly be culprit.
Sample 4(Glycerin) & 5(Triethanolam):
Polar properties missing Glycerin which has polar properties seem to contribute in making the substance than my
original thought. Seems glycerin may subtract excess water. However Triethanolamine
missing has polar properties MAINLY however has some nonpolar properties as well
which means it has an important factor in emulsification so the end result without it didnβt
seem to different⦠This was tricky I came to realization stearic acid also acts as an emulsifier so in reality we should have tested this substance subtracted from the mix as well simultaneously.
Most Important Ingredients of Lotion? I believe the most important foundational ingredients of the hand cream at minimum needs
to include water H2O and Lanoline Yellowish wax. The reason water is because
the purpose of water is to provide moisture; however moisture is not good enough if you
canβt absorb it and that is exact the purpose of lanolin which helps your skin actually
absorb the moisture. The basic of lotions purpose is to moisturize the skin and cure dry
skin.
Ilana Kovach
Calculate $ to make 10fl oz. of
hand cream 7.5g Triethenolumine 5β΅
3.0g Glycerin 11β΅
30.0g Stearic acid 43β΅
3.0g Methyl Stearate 12β΅
30.0g lanoline 11.11β΅
30ml Mineral oil 25β΅
10oz $2.77
A product of traditionally bought store lotion could sell for
$20.00
When determining Probability also consider:
Advertising & Packaging
Profit margin Per Bottle:ππ.ππβπ.ππ
ππ.ππ * 100
86%
Ilana Kovach
Lab 7: Physical &Chemical Changes
#
Observations
Physical
Change
Chemical
Change
Gas
Produced
(If Any)
1 3CuSO4 + 2Fe(s) 3Cu(s) + Fe2(SO4)3
a.) 30sec the steel wool changed color βpinkishβ
b.) 15min the color completely no longer black charcoal and
the copper (II) sulfate dissolved the wool.
N/A
2 Zn(s) + 2HCl(aq) ZnCl2 + H2(g)
Bubbles immediately meaning hydrogen gas must be present.
Hydrogen is flammable. In the experiment the Flame went out
However the reaction was expected to light the splint.
Hydrogen
Gas
3 ZnCl2 (aq) (β²Heat) ZnCl2(s)
The residue compound on the watch glass went from an Aqueous
solution to a solid state. This is essentially a βphase changeβ. N/A
4 AgNO3 + HCl HNO3 + AgCl
Liquid Became cloudy and salt began to form
N/A
5 CaCO3(s) + 2HCl CaCl2 + CO2 + H2O
The aqueous solution turned yellowish. Calcium Carbonate
reacted to form calcium chloride, Water and a nonflammable gas
proven by splint test of Carbon dioxide.
Carbon
Dioxide
Gas
6 4Cu(s) + O2(g) 2Cu2O
When the Copper Wire was exposed to the fire it produced a black
solid Copper(I) Oxide
N/A
7 2Mg(s) + O2(g) 2MgO
When Magnesium was exposed to the fire; the Mg ignited turning
from silver to white and shriveled up.
N/A
8 2KClO3(S) (MnO2(s) /β²Heat) 2KCl +3O2(g)
When splint placed in tube it light back on fire meaning the
product produced oxygen gas.
Oxygen
Gas
9 CaCO3 + β²Heat CaO +CO2 (g)
No change; but was supposed to⦠to breakdown calcium
carbonate into calcium Oxide and CO2 Gas. The splint test was
expected to blow out the flame to show CO2 was Produced.
Carbon
Dioxide
Gas
10 I2(s) (β²Heat) I2(g) I2(s)
Iodine Solid vapor rises and crystallized on the bottom of the dish.
(similar H2O phase changes) N/A
Ilana Kovach
Conclusion These experiments enhanced my understanding of physical and chemical Changes. In Experiment
#3 and #10 were examples of physical changes. The reason these two experiments were
demonstrating physical change is because they were showing a phase change; similar to that of
water changing states from solid, Liquid and gas. In experiment 3 the zinc chloride an aqueous
solution when heated up product resulted in Zinc chloride in a solid state. The residue compound
solidified on the watch glass. Experiment 10 the iodine solid when heated became a gas and then
crystallized on the bottom of the dish. Since the Iodine in the equation is merely just iodine
changing states of matter it would be considered a physical change. Chemical changes were further
comprehended in depth by clarifying physical changes distinctions and showing observations that
translate to being a chemical change.
In Experiment 1 Steel wool (Iron III) was placed in a blue solution Copper Sulfate. After 30sec the
steel wool changed a pinkish color and after 15min the color completely no longer black charcoal
and the copper II sulfate dissolved the wool. This reaction is an example of a single replacement
reaction. In Experiment 4 AgNO3 and Hydrochloric Acid liquid became cloudy and salt began
forming into HNO3 and AgCl.
Another way to see if a chemical reaction took place is by placing the metal in a Bunsen burner. For
instance in Experiment #6 when the copper wired was exposed to fire it produced a black solid
Copper (I) oxide. And Experiment #7 Magnesium was exposed to the fire it ignited and shriveled up
and turned into magnesium oxide. These types of experiments are clearly observable
The third method used when experiments used the splint test to see if any gases were produced such
as hydrogen gas, oxygen gas or Carbon dioxide gas. In Experiment #2 the zinc and Hydrochloric
acid when doing the splint test was supposed to relight the flame however our experiment did not
turn out with expectations. The product should have left a highly flammable gas hydrogen.
Experiment #5 was calcium carbonate and Hydrochloric acid. The splint test proves the equation
when the flame went out suggesting that carbon dioxide was in the product of the reaction. In
Experiment #8 KClO3 when heated the splint test suggested Oxygen gas must be in the product of
the reaction because it light back up. The final Experiment performed with the splint test method
was CaCO3 heated which during the actual experiment did not appear to have a reaction. However
since with knowledge of the equation balancing carbon dioxide was supposed to be produced and be
proven by blowing out the flame in the splint test.
As a whole, the various experiments give a more thorough comprehension of Chemical reactions
and how they work and what really happens when a chemical reaction is taking place. Overall each
chemical reaction has fundamental grounds in keeping an equilibrium of a balanced equation but
changing variables around is happening beyond the human eye.
Ilana Kovach
Lab 8: Balancing Equations & Types
Combination
A+ B AB
Decomposition
AB A + B
Single Replacement
A + BC AC + B
Double Replacement
AB + CD AD + CB
Combustion
CxHy + ZO2 (g) XCO2 (g) + π
π H2O (g)
ALL elements on the Reactant and Product side need to Equal (=)
Ilana Kovach
Lab 9: Stoichiometry Prelab:
1. Chemical Reactions for All 3 Theoretical Decomposition Reactions
NaHCO3 NaOH(s) + CO2 (g)
2NaHCO3 Na2O(s) + 2CO2 (g) + H2O (g)
2NaHCO3 Na2CO3(s) + CO2 (g) + H2O (g)
2. 4.2g of sodium bicarbonate is Equivalent to how many moles of sodium bicarbonate
4.2gNa2CO3 * π ππππ π΅ππ―πͺπΆπ
ππππ΅ππ―πͺπΆπ = .05moles NaHCO3
3. Using the number of moles calculated in question 2 How many moles & Grams
(theoretical Yield) of solid product can be produced in each of the possible
decomposition reactions
.05mol NaHCO3 * π ππππ π΅ππΆπ―
π ππππ π΅ππ―πͺπΆπ *
ππ.ππ π΅ππΆπ―
πππππ π΅ππΆπ― = 2.00g NaOH
.05mol 2NaHCO3 *π ππππ π΅πππΆ
π ππππ π΅ππ―πͺπΆπ *
ππ.ππ π΅ππΆ
πππππ π΅πππΆ = 1.58g Na2O
.05mol 2NaHCO3 *π ππππ π΅πππͺπΆπ
π ππππ π΅ππ―πͺπΆπ *
πππ.ππ π΅πππͺπΆπ
πππππ π΅πππͺπΆπ = 2.65g Na2CO3
4. Using decomposition Reaction #1 and the theoretical Yield calculated above, what
would be the percent yield be if the experimental yield was 1.85g π¬πππππππππππ
π»ππππππππππ * 100 = % Yield
π.ππ
π.ππ * 100 = 92.5%
Ilana Kovach
Experiment
a. Mass of Empty & Clean Crucible 39.3116g
b. Mass of the crucible & sodium Bicarbonate 43.9932g
c. Mass of Sodium Bicarbonate used in the
Experiment (B-A)
(43.9932- 39.3116)
4.6816g
d. Moles of Sodium Bicarbonate used in the
Experiment
4.6816g Na2CO3 * πππππ π΅πππͺπΆπ
ππππ΅ππ―πͺπΆπ
0.0057mol
e. Mass of the crucible and decomposition product
after 1st heating
42.263g
f. Mass of the crucible and decomposition product
after 2nd heating
42.2560g
g. Mass of Decompostion (F-A)
(42.2560g-39.3116g)
2.9444g
Ilana Kovach
Post lab:
1. Using the number of moles calculated in question 2 How many moles & Grams
(theoretical Yield) of solid product can be produced in each of the possible
decomposition reactions
.0557mol NaHCO3 * π ππππ π΅ππΆπ―
π ππππ π΅ππ―πͺπΆπ *
ππ.ππ π΅ππΆπ―
πππππ π΅ππΆπ― = 2.228g NaOH
.0557mol 2NaHCO3 *π ππππ π΅πππΆ
π ππππ π΅ππ―πͺπΆπ *
ππ.ππ π΅ππΆ
πππππ π΅πππΆ = 1.7267g Na2O
.0557mol 2NaHCO3 *π ππππ π΅πππͺπΆπ
π ππππ π΅ππ―πͺπΆπ *
πππ.ππ π΅πππͺπΆπ
πππππ π΅πππͺπΆπ = 2.9521g Na2CO3
2. Which Theoretical Yield Agrees with the Experiment Yield?
The Experimental yield was 2.9444g; the closest of the Theoretical yields is the
third one 2.952lg Na2CO3. Itβs a little odd that it was above the number of
theoretical compared to experimental but it was clearly the closest.
3. Which balanced chemical reaction actually occurred during the Experiment?
2NaHCO3 Na2CO3(s) + CO2 (g) + H2O (g)
4. Using decomposition Reaction #1 and the theoretical Yield calculated above,
what would be the percent yield be if the experimental yield was 1.85g
π.ππππ
π.ππππ * 100 = 99.7% Yield
Comment: Even though our Experiment was very close to the actual
number there could always be reasons why itβs not exactly 100%
#1 may we didnβt burn it enough but highly unlikely
#2 weighting is never perfect
Ilana Kovach
Lab 10: Boyleβs Law 1. Boyles Law?
The law stating that the pressure and volume of a gas have an inverse relationship
2. Mathematical Equation
P1V1= P2V2
3. New Pressure?
Given: (15mL, 21Ml & 740torr) (ππ)(πππ)
(ππ) = 925mmHg or Torr
4. Convert R (atm kPa) & (L mL)
8.206 * π³βπππ
πππβπ² *
ππππππ
ππ³ *
πππ.πππ·π
ππππ = 8313
ππ³βππ·π
πππβπ
5. Room Temperature Convert to Kelvin
23ΒΊC +273 = 296K
6. n= πΊππππ
πΉπ»
ππππ
(ππππ)(πππ) = 5.25 * 10-4 moles
7. 5.25* 10 -4 mol * π.ππβπππππππππππππ
π ππππ = 3.16 * 1020 molecules
8. (3.16 * 1020 molecules) * (π.πβππβππππ)
π ππππππππ = 1.20 * 10-3
volume of Gas in a 20mL syringe
9. % Empty Space ππ.ππππ³βπ.ππππππ³
ππππ³ * 100 = 99.994% empty space in 20mL
10.% Empty Space ππ.ππππ³βπ.ππππππ³
ππππ³ * 100 = 99.988% empty space in 10Ml
Ilana Kovach
Conclusion: Based on my previous knowledge I know gas molecules
are farther apart in relative comparison to solids and liquids. Gases have properties
that include having no definite shape or volume. Therefore the container itβs in such
as a room for instance fills the Entire space. The Kinetic Energy Laws states that gas
moves rapidly and are elastic. This Experiment brought statistics to Life. In that the
reality is gas is mainly made of empty space. My data stated in a 20ml space that
only 0.0012 represented the volume of the gas. The syringe in reality was mostly
empty space with 99.994% and when decreased to 10mL the amount of empty
space only decreased by .0l% stating 99.988% which is very little in
comparison to the amount we compressed the volume to change. If one takes a step
back in the thought process; empty space can be answered by the simplest
component of matter an atom. An atom structure contains the nucleus which
contains most of the mass with protons/neutrons and electrons surrounding the
nucleus mostly consist of space. This structure of an atom was proven by the
Rutherford experiment. The experiment performed shots against a gold foil and the
results were that a few times the shot was deflected; which would be the nucleus.
However most of the laser shots went right through. In reality mostly space applies
to all matter. However gas molecules are farther apart which vary depending on the
size of the container and if converted to be made into a solid it would be VERY
small. Why? Cause gas takes up very little space when condensed. Overall in
comparison a Gas contains mostly of space even more than solids and liquids
because of its unique properties.
Ilana Kovach
Lab 11: Acids, Bases & pH
I. pH of Various Solutions
Solution pH Paper pH meter
1.0M HCl 1 1.36 Vinegar 3 2.97 Ammonia 10 11.30 1.0M NaOH 11 13.07 Tap Water 6 6.5
Do pH values agree with 2 Different Methods?
YES, however the pH meter is more precise, there is slight difference but in general fairly close in pH values. Predict if Basic or Acidic?
Drain Cleaner = Basic Soft Drink = Acidic Bleach = Basic Lemon Juice = Acidic
Ilana Kovach
II. Antacids Antacid Mass (g) Drops of 1.0M HCl
CaCO3 0.127 64 NaHCO3 0.1002 55 Mg(OH)2 0.1000 41
1) CaCO3 + 2H+ H2O + CO2 + Ca2+ 2) NaHCO3 + H+
H2O+ CO2 + Na+ 3) Mg (OH) 2 + 2H+ 2H2O + Mg2+
Should it take the same # of drops to neutralize the same Mass?
0.127g CaCO3 * πππππ πͺππͺπΆπ
πππ.πππ πͺππͺπΆπ * πππππ π―+
π ππππ πͺππͺπΆπ * ππ π―+
πππππ π―+ = .0025g
.1002g NaHCO3 *
πππππ π΅ππ―πͺπΆπ
ππ.πππ π΅ππ―πͺππ * πππππ π―+
π ππππ π΅ππ―πͺπΆπ * ππ π―+
πππππ π―+ = .0012g
.1000g Mg(OH)2* πππππ π΄π(πΆπ―)π
ππ.πππ π΄π(πΆπ―)π * πππππ π―+
π ππππ π΄π(πΆπ―)π * ππ π―+
πππππ π―+ = .0034g
CaCo3 and Mg(OH)2 Chemical Equations based on the ratio of moles of hydrogen ions to the antacid needs more drops to neutralize. The experiment resulted in 64 drops which was CaCO3 was significantly higher than the NaHCO3 . However the Mg(0H)2 seemed to not perform based on expectations. There could be a number of reasons the number was lower. Maybe since this solution differed in waiting to dissolve verses fizz. The experiment wasnβt perfect but could easily been human error since drops can vary in size and are not always consistent. All together the concept taken is the chemical equation emphasizes the ratio of H+ ions compared to solution that was neutralized.
Extra Strength Acid contained 750mg of CaCO3 in a single
tablet: assume 20 drops per mL how many mL? Of l.0M HCl
would one of these tablets neutralize? .75g CaCO3 * π
π.πππππͺππͺπΆπ * πππ ππππ
π * πππ³
πππ ππππ = 19mL 1.0M HCl
Comment: That is a large #, so maybe antacid works despite professor Pribich thought of βthe placebo effectβ many drugs have the placebo effect that comes into play when determining if a drug/medication works. However from an unbiased viewpoint the experiment is suggesting it actually does have a REAL effect.
Ilana Kovach
III. Buffers
What was the Initial pH difference
6.96-4.19 = 2.77
PH difference after 1 drop? & 2?
3.82- 2.90 = .92 3.28- 2.44 = .84
Did the buffer do what the Buffer should do?
YES the buffer helped stabilize the pH. The ionized water without a pH buffer significantly dropped becoming very acidic when hydrochloric acid was added. When the pH 4 buffer was introduced and Hydrochloric acid was added the numbers was more stabilized near the pH of 4 just as expected should have been done.
Initial
pH
pH after 1
drop HCl
pH after 2
drops HCl
Deionized Water 6.96 2.90 2.44 pH 4 Buffer 4.19 3.82 3.28