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Experiment 10
Noah McNally
Chemistry 112 – Section 201
Dates of Experiment: March 8 and March 22
Experiment 10 – Acid-base Titrations: Part A – Analysis of vinegar and Part B – Analysis of a
Carbonate/Bicarbonate mixture
Unknown Number: __ 796 __
Abstract
The purpose of this experiment was to standardize two titrant solutions in order to
identify the cheapest brand of vinegar and the molarities of NaHCO 3
and Na 2
CO
3
within an
unknown solution. To come to these conclusions, the principles and methods of acid-base
titrations were used. The experiment explored concepts such as the creation of stock solutions,
standardizing a secondary standard, reactions between acids and bases, and the titration of these
acids and bases, along with observing their change in pH. Initially, the titrant (NaOH, a base)
was prepared in a stock solution and standardized with the use of 100% pure KHP and the
indicator phenolphthalein. This was done with several titrations to determine the exact molarity
of the stock solution prepared. With this standardization, the NaOH was used to volumetrically
analyze the concentration of two brands of vinegar. By determining these concentration values
and comparing them to the price and total volume of each vinegar, the better priced vinegar was
determined.
With the use of a standardized solution of HCl, an unknown solution of sodium
bicarbonate and sodium carbonate was able to be analyzed. Through the use of two separate sets
of titrations, both the total alkalinity and the individual molar amount of NaHCO 3
were
determined. Success in this experiment was contingent upon accurately standardizing the NaOH
and HCl solutions, determining accurate concentrations and the best buy for each of the vinegar
solutions, and determining the accurate molarities of sodium bicarbonate and sodium carbonate.
Procedure
To prepare a one-liter sample of 0.1 M NaOH, a more concentrated sample was taken and
diluted. The smaller amount of concentrated NaOH was poured into a 1 - liter plastic bottle, and
distilled water was poured to the level of the bottle’s shoulder and then was mixed thoroughly.
After mixing the stock solution, the analytical balances were used to weigh about 201.5 mg
of KHP into each of three 200 mL tall-form beakers, recording the exact weight of KHP in each.
This was followed by adding about 25 mL of distilled water, followed by a stir bar and 2 drops
of phenolphthalein solution. The flask was then put onto the stirrer.
A 50 mL buret was rinsed with some of the stock NaOH solution and then filled with the
NaOH solution. At this point, the pH sensor was set up. The Lab-quest was connected to the
computer via a USB port, and the pH was connected to channel 1 of the Lab-quest. The pH probe
was removed from the storage solution and immediately, the bulb was rinsed with water from the
squirt bottle. Quickly, the pH probe was secured to a ring stand using a clamp, and then the pH
probe was lowered into the solution that would soon be titrated. The bulb was not allowed to
touch anything in the beaker, as it was very fragile. The pH was recorded before adding titrant, at
Experiment 10
the endpoint, and after the titration was completed, and the initial volume reading of the buret
was recorded before any titration began.
Once the initial measurements were made, the first solution was titrated until a consistent
faint pink endpoint was reached. This titration was conducted by holding the stopcock and
slowly letting out the titrant drop by drop, going very slowly as the endpoint was neared. The
final volume reading on the buret and the pH at the endpoint were recorded. The titration was
repeated with the other two KHP samples until consistent results were obtained.
Next, the vinegar samples were analyzed. Two vinegar sample were collected, noting the
brand name, the original volume of the bottle, and its price. 2 mL of the first vinegar brand was
pipetted into each beaker, and then water and 2 drops of phenolphthalein were added. The first
solution was titrated, and the initial volume, final volume, and pH at the endpoint were recorded
Two more titrations were repeated with the same brand of vinegar. Finally, three more titrations
were conducted with the second brand of vinegar, noting all of the same measurements for the
brand and during the titration.
After the analysis of the two vinegar samples was complete, a stock solution of HCl was
produced and standardized. This was done by first preparing 250 mL of 0.1 M HCl by diluting
25 mL 1 M HCl to 250 mL in a small plastic bottle with distilled water. The solution was shaken
to mix.
Three of the 200 mL tall-form beakers were obtained, and about 104.8 mg of Na 2
CO
3
were
added to each, recording the exact amount added. A stir bar was then added to each. About 100
mL of boiled water was obtained in a clean 100 mL graduated cylinder, and about 25 mL of this
boiled water was added to each of the beakers containing Na 2
CO
3
. The hot plate was set to
around 250° C, and two 50 mL burets were cleaned and prepared for titrations by clamping them
to the ring stand as before. A funnel was used to pour the two different titrants into each buret.
One of the burets was rinsed and filled with the stock solution of HCl, and 2 drops of
phenolphthalein was added to each of the beakers containing Na 2
CO
3
that were to be titrated.
The initial volume on the HCl buret was recorded, and the titration was conducted by slowly
allowing the acid solution to drain drop by drop. Once the initial pink color had just disappeared,
the endpoint was a little over halfway reached. At this point, 2 drops of bromcresol green was
added to the solution, and the titration was continued.
Because of the presence of CO 2
being produced, it was difficult to reach the accurate
endpoint. Therefore, the solution was titrated until there was a slight change from original blue
color to a greenish tinge. The solution was then boiled, and once this was done, it should have
returned to blue. A few more drops of titrant were added to reach the endpoint. The final volume
and final pH were recorded. The Vernier temperature probe was used throughout the titration as
was previously done for the first nine titrations. Two more good titrations were repeated in the
same fashion. This completed the standardization of the HCl solution.
To determine the total alkalinity of an unknown solution, 2 mL of the unknown were pipetted
into each of the 200 mL tall form beakers with a volumetric pipet, followed by adding 25 mL of
boiled water to each, along with 2 drops of phenolphthalein and a stir bar to each. The three
solutions were titrated with the standardized HCl solution. The bromcresol green was added at its
appropriate time, once the phenolphthalein endpoint was reached. To determine the amount of
NaHCO 3
, 2 mL of the unknown were again pipetted into each of the three beakers, 15 mL of
water was added to each, along with 2 drops of phenolphthalein. The second buret was rinsed
and filled with the stock NaOH solution produced previously. About 15 mL of the NaOH
solution and about 5 mL of 10% CaCl 2
was added to the beakers containing the unknowns. Each
Experiment 10
Acid Mass in
aliquot taken
0.1021 g 0.1032 g
Grams of Acid
per Liter
51.07 g/L 51.61 g/L
Mass of Vinegar 2.010 g 2.010 g
% Acetic Acid 5.080 % 5.134 %
Cost of Vinegar
in Moles per
Dollar
0.207 moles per dollar 0.649 moles per dollar
Better Buy? Brand 2: Great Value Apple Cider Vinegar
Standardization of HCl Stock Solution
Trial 1 Trial 2 Trial 3
Mass of Na 2
CO
3
0.1048 g 0.1048 g 0.1052 g
Moles of Na 2 CO 3 9.888 ´ 10
mol 9.888 ´ 10
mol 9.925 ´ 10
mol
Initial HCl Volume 0.80 mL 21.70 mL 0.95 mL
Final HCl Volume 21.70 mL;
21.29 to 21.70 mL
42.1 mL;
21.70 to 22.55 mL
21.29 mL;
22.55 to 23.65 mL
Volume HCl Used 21.31 mL 21.25 mL 21.44 mL
Concentration of HCl 0.09280 M 0.09306 M 0.09258 M
Mean Molarity 0.09281 M
Mean Normality 0.09281 N
pH – Endpoint 4.
Titrations for Total Alkalinity
Trial 4 Trial 5 Trial 6
Volume of Unknown 2 mL 2 mL 2 mL
Initial buret Volume 1.60 mL 21.10 mL 0.61 mL
Final buret Volume
22.70 mL;
21.25 to 22.20 mL
43.5 mL;
22.20 to 23.30 mL
21.25 mL;
23.30 to 24.51 mL
Volume HCl used 22.05 mL 23.50 mL 21.76 mL
Total mmol HCl used 2.046 mmol 2.181 mmol 2.020 mmol
Average HCl used 2.082 mmol
pH – Endpoint 3.
Titrations for NaHCO 3
Trial 7 Trial 8 Trial 9
Volume of Unknown 2 mL 2 mL 2 mL
Total Volume of NaOH
(at beginning and after
addition of HCl)
15.45 mL 15.50 mL 15.50 mL
Total Volume of HCl 5.60 mL 6.50 mL 6.10 mL
Total mmol NaOH used 1.661 mmol 1.666 mmol 1.666 mmol
mmol HCl need to titrate
excess NaOH
0.520 mmol 0.603 mmol 0.566 mmol
Experiment 10
mmol NaOH used in
converting NaHCO 3
to
Na 2
CO
3
1.141 mmol 0.603 mmol 1.100 mmol
mmol NaHCO 3
converted to Na 2
CO
3
1.141 mmol 0.603 mmol 1.100 mmol
mmol HCl consumed by
NaHCO 3
1.141 mmol 0.603 mmol 1.100 mmol
mmol HCl consumed by
Na 2
CO
3
0.941 mmol 1.019 mmol 0.982 mmol
mmol Na 2 CO 3 0.471 mmol 0.510 mmol 0.491 mmol
Molarity of NaHCO 3
0.5705 M 0.5315 M 0.5500 M
Average Molarity
NaHCO 3
0.5507 M
Molarity of Na 2
CO
3
0.236 M 0.255 M 0.246 M
Average Molarity
Na 2
CO
3
0.246 M
Sample Calculations:
Total Volume of Titrant Used:
Final Volume – Initial Volume = Volume used to titrate
18.61 mL – 8.30 mL = 10.31 mL
Concentration of NaOH (standardization):
Moles KHP = Moles NaOH used to titrate
!"# %&'
( )*+&
= Concentration (M)
,../0 × 23
45
6"#
- 32372 (
= 0.09570 M
Average Normality:
For NaOH, since 1 mol NaOH lends 1 mol OH
, Molarity = Normality
0.1075 M NaOH = 0.1075 N NaOH
Standard Deviation:
2
)
(∑ (mean − each individual)
J
K
LM 2
, where N is the total amount of individuals in the
set.
Experiment 10
3 ..tJ3 6"#
2333 6(
´ 246 mL = 0.207 moles per dollar
Standardization of HCl: Similar calculations as that done for standardization of NaOH
Total mmol HCl added to titrate excess NaOH:
HCl molarity (mmol/mL) ´ volume HCl added (mL) = mmol HCl
0.09281 (mmol/mL) ´ 5.50 mL = 0.520 mmol HCl
Excess NaOH: Equal to total mmol HCl added to titrate it (above calculation), since it is a 1:
stoichiometric ratio
mmol of NaOH used in converting NaHCO 3
to Na 2
CO
3
Total NaOH added to solution (mL) – Excess NaOH (mL) = mmol used to convert
1.661 mmol – 0.520 mmol = 1.141 mmol NaOH
Since there are again 1:1 ratios from the chemical equations (discussed in Discussion section),
the millimolar amount of NaHCO 3
converted to Na 2
CO
3
and mmol HCl consumed by NaHCO 3
are also both equal to 1.141 mmol.
mmol HCl consumed by Na 2
CO
3
mmol HCl added in total alkalinity titration (average value) – mmol HCl consumed by NaHCO 3
= mmol
2.082 mmol – 1.141 mmol = 0.941 mmol HCl
mmol Na 2
CO
3
66"# &v# f"6dl6Wh mw )*Jv+ 7
J
= mmol Na 2
CO
3
3 .,t2 66"# &v#
J
= 0.471 mmol Na 2
CO
3
Molarity of NaHCO 3
or Na 2
CO
3
66"# "e mgfcm"XkW "c fcm"XkW
J. 333 6(
= Molarity (M)
2 .2t2 66"# )*&v+ 7
J. 333 6(
= 0.5705 M NaHCO 3
Experiment 10
Discussion
One of the many applications of chemistry is the ability to manipulate chemical
properties or reactions to analyze certain unknown quantities. One of these analytical processes
is called volumetric analysis, in which the concentration of an unknown substance is determined
through a process called titration. Titrations utilize one solution of known concentration and
volume that is added to an analyte (or the solution being analyzed) of unknown concentration.
The solution that is added to the analyte is normally called the titrant. While there are many
forms of titrations, this experiment specifically used acids and bases as the titrants and analytes.
While the end goal is to discover a concentration, the procedural goal of a titration is to
reach the titration’s endpoint, the point at which the desired color change occurs. This is used to
indicate when enough volume of titrant has been added to reach the equivalence point of the
reaction, the point at which the moles of titrant and analyte have neutralized each other. This is
different from the endpoint, but since strong acids and bases are used in the titration process, the
pH changes so drastically with incredibly small amounts of titrant added near the equivalence
point that the endpoint can be considered to have the same amount of titrant added as the
equivalence point. The endpoint is easily attainable because of the use of indicators, which are
chemicals that change color at the endpoint, thus fulfilling the function of their name. These are
specifically chosen so that their endpoint lies within a small range of the equivalence point of the
reaction, thus allowing the moles of titrant needed to neutralize a solution to be accurately
determined.
Prior to using a titrant to analyze the concentration of a solution, the titrant must be
standardized, meaning that the exact concentration of the titrant itself must be determined. In the
case of NaOH, the first titrant used, a 100% pure sample of known concentration of KHP was
Experiment 10
Using the same deductions as with the standardization, the molarity of each of the two
acetic acid solutions was determined. Then, by using the price and volume of the original bottles
of vinegar, the number of moles of acetic acid per dollar were calculated for each bottle. While
the percentages of acetic acid in each vinegar were very similar to each other, the apple cider
vinegar brand was undoubtedly the better buy, primarily because the volume (and thus total
number of moles by calculation) was much larger for a smaller price than the gourmet vinegar.
For the second set of solutions, which were mixtures of sodium carbonate and sodium
bicarbonate, an acidic titrant had to be prepared. This is because the carbonate and bicarbonate
ions can act as bases, which means that to neutralize them, an acidic titrant was needed. A stock
solution of HCl was prepared by dilution and standardized in a similar fashion to the NaOH
stock solution. However, the primary standard used was a known amount of pure Na 2
CO
3
. There
is also another key difference to this second standardization: KHP only needed one OH
molecule for every KHP molecule to become neutralized, but each Na 2
CO
3
molecule must
receive two protons to become fully neutralized. Because of this, two endpoints are associated
with Na 2
CO
3
, one that occurs when the Na 2
CO
3
has accepted the first proton and one when it has
accepted the second and become neutralized. The net equation for this reaction is as follows:
CO
3
2 -
(aq) + 2H
(aq) à H 2
CO
3
(aq)
The first endpoint that occurs happens around a pH of 8, so phenolthalein was used again. The
second ending occurred in the more acidic range of a pH around 3.5 to 5.5. This is around the
range of the bromcresol green indicator.
A second difference and complication in titrating the carbonate is that a byproduct of CO 2
is formed during the titration. This dissolved CO 2
actually acts as a buffer for the solution, thus
making it difficult to neutralize, since the buffer works to resist the change in the pH. To solve
Experiment 10
this problem, the solution was boiled once the endpoint was soon to be reached. With the CO 2
boiled out, the endpoint could be reached normally and without the resistance of a buffer.
With the standardized HCl titrant, an unknown concentration of sodium bicarbonate and
sodium carbonate could be analyzed to determine the concentrations of both the bicarbonate and
carbonate in the solution. To do this, the solution needed to be titrated twice, each time under
different conditions. The first titration was used to determine the total alkalinity of the solution,
which was determined based on the total amount of HCl consumed to neutralize the solution.
HCO
3
(aq) + H
(aq) à H 2
CO
3
(aq)
CO
3
2 -
(aq) + 2H
à H 2
CO
3
(aq)
This equation shows that 1 mole of HCl is consumed for each mole of bicarbonate and 2
moles are consumed for each mole of carbonate. However, unless either the bicarbonate amount
of the carbonate amount is known, nothing more than the total sum of the amounts of
bicarbonate and carbonate can be deduced. This is why the second titration was necessary; it
used the NaOH solution to react with any bicarbonate to produce only carbonate ions, which was
then precipitated out with CaCl 2
. This method cleverly allows one to analyze only the amount of
bicarbonate within the solution, regardless of how much carbonate is present. By adding NaOH
and CaCl 2
, the solution was left with excess OH
ions, and by titrating with HCl, this excess
amount could be determined. Since the total amount of OH
added was known, the exact amount
that reacted with bicarbonate could be calculated, and by the 1:1 stoichiometric ratio of the
chemical reaction, the amount of bicarbonate in the mixture could as well be determined.
OH
(aq) + HCO 3
(aq) à CO 3
2 -
(aq) +H 2
O(l)
With this value, the equation relating the HCl consumed in the titration for total alkalinity
could now to be solved for the amount of carbonate, since two of the three pieces were known.
Experiment 10
Weight: 0.0050 mol × 36.997 g/mol NaOH = 0.20 g NaOH
50% solution of NaOH à
3 .J3 b )*+&
3 .t3 b d"#lkg"X
Therefore, 0.4 grams of solution is needed.
Density of 50% NaOH at 25 °C: 1.52170 g/mL
3 .t3 b
2 .ZJ
x
yz
= 0.26 mL of 50% NaOH
- What is a primary standard? Can KOH be used as a primary standard? Explain.
A primary standard is the solid chemical that is used to standardize the secondary
standard, the titrant solution. This solid is weighed, dissolved, and brought to a known
volume, and through its titration, calculations can be conducted and the secondary
standard can be accurately standardized. Primary standards work best when they have a
high molecular weight, do not absorb molecules or react with the air, and can be stored at
high levels of purity. Under these criteria, KOH would not be a good primary standard;
its molecular weight is only 56 g/mol, and it reacts with and absorbs CO 2
readily from the
air. When titrating, it is necessary that the only reactions occurring are those between the
titrant and analyte; if KOH was used, other reactions would be occurring.
- An aqueous sample is analyzed and found to contain 4 percent NaOH (4 g in 96 mL H 2
O)
a) Calculate the concentration of the solution in molarity (assume the density of water is
1.00 g/mL and the volume of the solution is 100 mL)
t b )*+&
7,.,,0 b/6"#
= 0.100 mol NaOH
t b )*+&
- 2 ( d"#lkg"X
233 6"#
2 (
= 1.00 M concentration
b) If this solution is neutralized with 1 M sulfuric acid, how many mL of sulfuric acid
will be needed?
H
2
SO
4
produces 2 moles of H
ions for every 1 mole of H 2
SO
4
present; there is 1
mol of OH produced for every 1 mole of NaOH. Since the molar concentrations
are the same, 100 mL of 1 M sulfuric acid has the same number of moles of solute
as 100 mL of 1 M NaOH. Thus, the amount of sulfuric acid needed to neutralize
the solution is half of the amount of NaOH present:
50 mL needed to neutralize 100 mL of 1 M NaOH
c) What is the normality of the NaOH solution?
Experiment 10
Normality = molarity × equivalents (the number of moles of reactive
atoms/molecules) à 1 M NaOH; 1 mol OH (reactive molecule) produced.
Normality = 1 × 1 = 1 N
Post-Lab Questions:
- What precautions did you take in setting up a micro or a macro buret?
In this lab, the macro buret was carefully attached to the buret stand so that it would not fall
out during the titration. One of the most important safety precautions was taking the buret out of
the clamp and using a funnel when pouring strong acid or base into the buret. The buret itself
was also held while operating the stopcock during the titration to ensure that the buret would not
topple out of the clamp.
- What amount of oxalic acid (H 2
C
2
O
4
• 2H
2
O, MW = 126.0 g/mol) must be used to
prepare 10 mL of 0.130 M oxalic acid solution? This acid has two replaceable acidic
protons. What would be the normality of the acid?
- 273 6"#
2333 6(
´ 10 mL = 0.0013 mol oxalic acid needed
0.0013 mol ´ 126.0 g/mol = 0.16 g oxalic acid
0.130 M ´ 2 = 0.26 N
Since oxalic acid can donate two protons, the molarity is multiplied by a factor of
2 to obtain the normality of this solution.
- In you take 50 mL of 5 percent acetic acid, how many millimoles of acetic acid have you
taken?
50 mL solution ´ 1.005 g/mL = 50.25 g of solution
0.050 ´ 50.25 g solution = 2.5125 g acetic acid
J.Z2JZ b *fWkgf *fgh
/3.3Z b/6"#
23
Y
66"#
- 333 6"#
= 41.84 mmol = 42 mmol
