Titration Experiment with a solution of sodium hydroxide against a solution
of potassium hydrogen phthalate
Name: Pieter Ploem
Lab Partner Names: Andrea Scotti, Liam Padolina
Date: 10/4/2022
Research Question/Aim
What is the concentration of a solution of sodium hydroxide by titration against a standard solution
of potassium hydrogen phthalate?
Reaction Equation
H+A- (aq) + Na+ OH- (aq)  Na+A- (aq) + H2O (l)
Material & Method (See attached instruction sheet)
Raw Data (Quantitative)
Table displaying used equipment for measurement and their uncertainties
Weighing Scale (g)
± 0.01
Pipette Solution (cm3)
± 0.04
Burette Solution (cm3)
± 0.005
Volumetric Flask (cm3)
± 0.3
Eye Dropper Pipette (Drops)
No uncertainty
Table displaying the mass of Potassium Hydrogen Phthalate, the mass of the Weighing Bottle,
and the mass of Potassium Hydrogen Phthalate + the mass of the Weighing Bottle
Potassium Hydrogen Phthalate (g)
5.28
Weighing Bottle (g)
0.56
Potassium Hydrogen Phthalate + Weighing Bottle (g) 5.84
Table displaying the initial and final burette readings containing (NaOH) in a series of trials
Trials
Experimental
1
2
3
4
5
Trial
Final
24.45
49.00
24.35
48.20
23.80
47.65
Readings (cm3) Initial
0.00
24.45
0.10
24.35
0.00
23.80
Burette
Raw Data (Qualitative)
+
-
Table displaying the extent to which the solution containing the products Na A (aq) + H2O (l)
was pink-colored based on pure observation for a series of trials
Trials
Experimental
1
2
3
4
5
Pink
Pink
Very light
Very light
Very light
pink
pink
pink
Trial
Extent of
Bright Pink
Pink Color
Processed Data
Table displaying the initial and final burette readings containing (NaOH) including the difference
between the final and initial burette readings in a series of trials
Trials
Experimental
1
2
3
4
5
Trial
Final
24.45
49.00
24.35
48.20
23.80
47.65
Readings (cm3) Initial
0.00
24.45
0.10
24.35
0.00
23.80
Difference between
24.45
24.55
24.25
23.85
23.80
23.85
Burette
Final and Initial (cm3)
Trials 3, 4, and 5 will be used for accurate evaluation for this experiment as the differences in these
values vary the least amount in comparison to the other trials. Additionally to this, trials 3, 4, and
5 all displayed a very light pink color wherefore these trials were very close to the “end-point”
where the reaction was just complete. Due to these factors, once could declare that these trials are
hence the most accurate and reliable trials for evaluation.
Table displaying the final and initial burette readings of NaOH, the difference between the final
and initial burette readings of NaOH, and the mean value of the differences for trials 3, 4, and 5.
Trials
3
4
5
Final
48.20
23.80
47.65
Readings (cm3) Initial
24.35
0.00
23.80
Difference between
23.85
23.80
23.85
Burette
Final and Initial (cm3)
Mean Value of the
23.83
Differences (cm3)
With this data, the molarity of Sodium Hydroxide (NaOH) will be calculated
Given
-
Mass of Potassium Hydrogen Phthalate – 5.28g
-
Volume of Potassium Hydrogen Phthalate used for reaction – 25 cm3
-
Mean value for the volume of Sodium Hydroxide used for reaction – 23.83 cm3
-
Balanced Chemical Equation – HA(aq) + Na+OH-(aq) Na+A-(aq) + H2O(l)
Unspecified/Unknown
-
Molarity of Potassium Hydrogen Phthalate
-
Exact Molarity of Sodium Hydroxide
First, the Molarity of HA needs to be calculated. In order to do this, the number of moles in 5.28g
of HA must be calculated. The molar mass of HA is 204.22 gmol-1. The calculations to get the
number of moles can be seen in the following:
5.28𝑔
204.22𝑔/𝑚𝑜𝑙
≈ 0.026 𝑚𝑜𝑙𝑠
Since this number of moles was applied to 250 cm3 or 0.250 dm3 of distilled water, we can
determine the molarity of the HA solution. The calculations are seen in the following:
0.026 𝑚𝑜𝑙𝑠
0.250 𝐿
0.103𝑀
Now that the Molarity of HA is, determined, we can deduce the molarity of NaOH. Since only 25
cm3 of the 0.103M solution of HA was used in the reaction with the NaOH solution, we must
calculate the number of moles of HA present in the 25 cm3 of 0.103M solution. The calculations
can be seen in the following:
𝑛
= 0.103𝑀
0.025 𝐿
𝑛 = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠
0.025 𝐿 ∗ 0.103𝑀 = 𝑛
𝑛 = 0.0026 𝑚𝑜𝑙𝑠
Due to the equation, we know that HA and NaOH have a one-to-one ratio, therefore this means
that if HA has 0.0026 mols in a 25 cm3 solution, NaOH has 0.0026 moles in the calculated 23.83
cm3 solution. Knowing this, the molarity of NaOH will be calculated.
0.0026 𝑚𝑜𝑙𝑠
0.02383 𝐿
𝟎. 𝟏𝟎𝟖𝟓𝑴
Therefore, it can be concluded that the Molarity of the NaOH solution is 0.1085M or 0.1085 mol
dm-1.
The Uncertainty of the Molarity of NaOH
In order to find the uncertainty of the molarity of NaOH, the measured values with their uncertainty
that were included in the calculations that led to finding the value for the molarity of NaOH, need
to be analyzed and evaluated.
The first step in the calculation was to find the number of moles in 5.28g of HA which was done
by dividing the mass of HA “5.28g” by the molar mass of HA. 5.28 grams has an uncertainty
which can be calculated by adding the uncertainty of the recorded mass of HA + Weighing bottle
(±0.01g) with the uncertainty of the recorded mass of the Weighing bottle (±0.01g). This can be
calculated in this way since the mass of HA + Weighing bottle was subtracted by the mass of the
Weighing bottle to find the mass of HA. When adding or subtracting variables of the same type of
measurement (mass, length, etc.), we simply add the absolute uncertainties of these variables
together. Therefore, the uncertainty of the mass of HA is ±0.02g. Using the uncertainty of the mass
of 5.28g, we can find the percentage uncertainty of the number of moles in 5.28g of HA. This is
shown in the following calculations:
(
∆𝑢
) ∗ 100
𝑚
∆𝑢 = 𝐴𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑢𝑛𝑐𝑒𝑟𝑡𝑎𝑖𝑛𝑡𝑦
𝑚 = 𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑚𝑒𝑎𝑠𝑢𝑟𝑒𝑚𝑒𝑛𝑡
0.02
(
) ∗ 100
5.28
≈ 𝟎. 𝟑𝟖%
The next step in calculating the molarity of NaOH was to find the Molarity of HA. This was done
by dividing the calculated number of moles in 5.28g of HA by 250 cm3 of distilled water solution.
In order to find the uncertainty of the molarity of HA, the percentage uncertainty of the number of
mols in 5.28g of HA needs to be added to the percentage uncertainty of the 250 cm3 volume since
there variables are divided by each other. The 250 cm3 volume had an uncertainty of ±0.3 cm3
since this solution was measured in a volumetric flask with an uncertainty of ±0.3 cm3. The
percentage uncertainty of the 250 cm3 volume is 0.12%. Since the same calculation process to find
the percentage uncertainty was already shown above, the calculations for percentage uncertainties
won’t be shown moving forward. The sum of the percentage uncertainty of the number of mols in
5.28g of HA and the volume of 250 cm3 is ±0.50%. Therefore, the percentage uncertainty of the
molarity of the HA solution is ±0.50%.
The next step in calculating the molarity of NaOH was to find the number of moles in a 25 cm3
HA solution with the calculated 0.103M. This was done by multiplying the molarity of the solution
by 0.025 L. The volume of 25 cm3 has an uncertainty of ±0.04 cm3. In order to find the uncertainty
of the number of moles in this solution, the percentage uncertainties of both the Molarity of the
solution and the 25 cm3 must be added because they are multiplied by each other. The percentage
uncertainty for the 25 cm3 volume is ±0.16%. The sum of the percentage uncertainty of the
molarity of the HA solution and the percentage uncertainty of the 25 cm3 solution is approximately
±0.66%. Therefore, the percentage uncertainty of the number of moles in a 25 cm3 0.103M solution
of HA is ±0.66%.
The final step in calculating the molarity of NaOH was to divide the calculated number of mols in
a 25 cm3 0.103M solution of HA by the average volume of NaOH needed to exactly and fully
complete the reaction as this calculation would yield the precise molarity of NaOH. To find the
final percentage uncertainty of the molarity of NaOH, the percentage uncertainty of the number of
mols in a 25 cm3 0.103M solution of HA must be added to the percentage uncertainty of the
average volume of NaOH needed to exactly complete the reaction since these measurements are
being divided by one another. The uncertainty of the average volume of NaOH needed to exactly
complete the reaction has an uncertainty that needs to be calculated since this value is an average.
This average was taken from 3 different measurements of NaOH from the burette that has an
uncertainty of ±0.005 cm3. To find the average, the three burette readings are added together and
then divided by three. Since three measurements of the same absolute uncertainty were added
together, the possibility of an uncertain value increases by three. Knowing this, the absolute
uncertainty of the average volume of NaOH needed to exactly complete the reaction can be
calculated by adding the absolute uncertainties of the three measurements together. This sum is
±0.015 cm3. The percentage uncertainty is then ±0.06%. The sum of both the percentage
uncertainties of the average volume of NaOH needed to exactly complete the reaction and the
calculated number of mols in a 25 cm3 0.103M solution of HA is ±0.72%. Therefore, the
percentage uncertainty of the molarity of NaOH is ±0.72%.
Finally, the absolute uncertainty of the molarity of NaOH is shown in the following calculation:
∆𝑢 = 𝑚 ∗ 𝐹𝑢
𝐹𝑢 = 𝐹𝑟𝑎𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑢𝑛𝑐𝑒𝑟𝑡𝑎𝑖𝑛𝑡𝑦
∆𝑢 = 0.1085 ∗ 0.0072
±𝟎. 𝟎𝟎𝟎𝟕𝟖𝑴
The molarity of NaOH with its absolute uncertainty is 0.1085M ±0.00078M.
Evaluation
For this experiment, the aim was to answer the research question “What is the concentration of a
solution of sodium hydroxide by titration against a standard solution of potassium hydrogen
phthalate?”. By following the procedure of this experiment, I was able to deduce a quite accurate
value for the molarity of a NaOH solution. This value being 0.1085M ±0.00078M. The degree to
which this value is accurate can be statisticized when comparing my value for the molarity of
NaOH (0.1085M) with the accepted value for the molarity of NaOH (0.1038M) in the form of the
percent error. The calculations needed to find the percent error are the following:
𝑃𝑒𝑟𝑐𝑒𝑛𝑡 𝐸𝑟𝑟𝑜𝑟 = |
0.1085𝑀 − 0.1038𝑀
| ∗ 100
0.1038𝑀
𝑃𝑒𝑟𝑐𝑒𝑛𝑡 𝐸𝑟𝑟𝑜𝑟 = 4.53%
4.53% is a low percent error, especially for an experiment that requires many steps and requires
multiple measurements, hence greater uncertainties. The reason why the value for molarity is quite
accurate is due to a few factors. The first factor was the great success I had achieved when
gathering data. There were no severe incidents, other than one small incident, that would disrupt
the experiment drastically and hence affect the credibility of the data. I also made sure to do
multiple trials and select three very good pieces of data that had similar values and very light pink
solutions containing the products of the reaction. The very light pink solution helped verify that
the end-point was just reached and not bypassed significantly, thus once again adding to the
credibility of the data. I also made sure to gather all the data on the same day hence reducing the
potential random and systematic errors caused due to the changes in the surrounding environment.
However, there were certain factors that potentially negatively did affect the accuracy of
the resulting value. One of the factors is that we had different group members verify the
measurements. This occurred most frequently when observing the readings from the burette
wherefore the way we read and note down the measurements varied each time as it was based on
the personal judgment of multiple people. As mentioned previously, there was a small incident
that likely affected the final value for the molarity of NaOH. This was when my group members
and I had collected the HA in the weighing bottle and had already done the measuring of its mass.
The weighing bottle was placed down a bit too aggressively on the table where small amounts of
HA fell out of the container. We had tried to the best of our ability to put back the escaped bits of
HA back into the weighing bottle; however, it is almost certain that there were minute amounts
still lost. Due to this, the mass of the HA was to a certain extent reduced, which reduces the number
of moles in HA, which reduces the molarity of the HA solution and thus also reduces the molarity
of the NaOH solution. This is a valid incident to reflect on, even though it is still quite small, since
my final value for the molarity of NaOH was over the accepted value and if I had avoided this
incident, the molarity of NaOH would have been reduced and potentially closer to the accepted
value.
In general, the method and procedure were quite effective and reliable for gathering
accurate data. The method’s strong points are its choice of measuring devices as the great majority
of the measuring devices were capable of presenting very accurate measurements, especially the
25 cm3 pipette. The method also allowed for very efficient data gathering once everything was
prepared. However, the main methodological issues that I encountered were its lack of explanation
for disposing of the chemicals after the experiment was completed. The method doesn’t
specifically clarify that the solution with the products should be put away into a waste beaker.
Additionally, to this, it doesn’t clarify if the chemicals can be poured down the sink, or whether
they need to be disposed of through the EHS hazardous waste program. Next time the procedure
should provide an MSDS to clarify hazards more clearly to reduce the risk of injury.
To extend this investigation, we could observe the effects of changing the temperature of
the NaOH solution on how it affects the calculated molarity of NaOH through titration. According
to Vedantu.com, a highly rated live online tutoring company, “When temperature increases the
volume of the solution also increases and this reduces the molarity of the substance” (“What Is the
Effect of Temperature on Molarity Class 12 Chemistry CBSE”). This would extend my knowledge
and my understanding of the different factors that affect molarity and how to improve my ability
of being able to apply titration to find different molarities.
Works Cited
“What Is the Effect of Temperature on Molarity Class 12 Chemistry CBSE.” Vedantu.com, 6
July 2021, www.vedantu.com/question-answer/effect-of-temperature-on-molarity-class12-chemistry-cbse-60e51bbb22400a17eb302a9e.