Essay: Identification and Synthesis of an Unknown Acid

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  • Identification and Synthesis of an Unknown Acid
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This report describes an experiment that was used to assist in the identification of an unknown acid. These experiments were conducted due to an unknown acid discovered by MSU Office of Environmental Health and Safety. A list of 26 possible acids, their molar mass, and pKa values were given by MSU EHS1. To identify the acid, it was titrated with a NaOH solution. To ensure that the final calculations of the titration was correct, the first part of the experiment was to standardize the NaOH solution that was given. This was done by titrating NaOH with a solution of KHP and water. This allowed the concentration of NaOH present in the NaOH solution to be identified in order to be used in the final calculation of the molar mass of the unknown acid2. Next, the second part of the experiment was performed to successfully identify the unknown acid. This was done by titrating a specific amount of the unknown acid in water with the NaOH solution. A pH probe was used with logger pro to track the pH of the solution as more NaOH was added to the acid solution, which allowed a titration curve to be created. This combined allowed both the pKa of the acid and the molar mass of the acid to be identified from the volume of NaOH used and the graph created by logger pro3. This allowed use to identify if the acid was monoprotic, diprotic, or triprotic and compare the molar mass and the pKa of the unknown acid with the list of possible known acids to successfully confirm the identity of the acid.


Part 1

Standardization of NaOH

The base that is used to neutralize the unknown acid is NaOH. In order to use NaOH to titrate the unknown acid, it is important to find out the concentration of NaOH in the solution. This was done by titrating NaOH with KHP which is an acid similar to that of the unknown acid.


.8 g KHP
Weigh boat and scale
100 mL of distilled water
Phenolphthalein (indicator)
150 mL beaker
2 250 mL Erlenmeyer flask
Burette Stand

To begin the process of standardizing NaOH, the necessary materials and chemicals should be gathered first. Measure out 0.8 g of KHP on a weigh boat on a scale. Collect 50 mL of distilled water in a 150 mL beaker. Rinse out the Erlenmeyer flasks and burette with NaOH to prevent contamination. Pour the .8 g of KHP into the Erlenmeyer flask and add the 50 mL of distilled water after. Swirl the flask for approximately 1 minute or until the KHP is fully dissolved. Add 4 drops of Phenolphthalein indicator to the solution. Set the burettes in the burette stand with the Erlenmeyer flasks positioned directly below it. Record the starting volume of the NaOH. Slowly and Continually, add NaOH to the KHP solution until a color change occurs. If the color appears too dark, repeat the titration.2

Use the following equation and steps to calculate the concentration of NaOH after it is successfully titrated2

Step 1 g of KHP(1 mole KHP/204.23 g KHP)=mol KHP
Step 2 moles KHP/L H2O used= M KHP
Step 3 moles KHP(1 mol NaOH/1 mol KHP)=moles NaOH
Step 4 moles NaOH/ L NaOH used= M NaOH

Part 2

Identification of Unknown Acid
.2 g of unknown acid 523
.207 M Solution of NaOH
30 mL distilled water
pH probe
250 mL Erlenmeyer flask
burette and burette stand
Logger Pro computer software
3 beakers of any size
buffer solution with a pH of 4
buffer solution with a pH of 7
Stir bar

Calibrating pH probe

In order to determine the pKa of the acid and its equivalence point, a pH probe must be used with logger pro to build a titration curve. First the pH probe has to be calibrated in order to determine the pH of the solution. To begin, connect the pH probe to the computer and open logger pro. Detach the storage buffer solution from the probe. Clean the the probe thoroughly with distilled water. Calibrated the probe using two buffer solutions. Pour a small amount of each buffer solution in separate beakers, one with a pH of 4 and one with a pH of 7. It should be enough of each solution to cover the tip of the probe. Start the calibration by first clicking Experiment > Calibrate > pH in logger pro. Place the probe into the 4-pH buffer solution and type the pH of the solution into the box marked Read 1. Clean probe thoroughly again with distilled water. This should be done each time the probe is placed in a new solution. Place the probe placed into the 7-pH buffer solution and type the pH of the solution into the box marker Read 2. Clean the probe again thoroughly with distilled water.3

Titration of Unknown Acid

The same method that was used to standardize NaOH is used to titrate the unknown acid. First, measure out .2 g of the unknown acid on a scale and pour it into a 250 mL Erlenmeyer flask. Collect 30 mL of distilled water in a beaker and pour it into the Erlenmeyer flask. Swirl the flask until the acid is completely dissolve. Clean a burette with the NaOH solution to avoid contamination. Set up the burette in the burette stand and fill the burette up with NaOH. Position the flask directly under the burette. Add 2 drops of phenolphthalein indicator into the solution. Record the starting volume by looking at the meniscus of the solution in the burette. Slowly begin adding NaOH into the solution and record the pH after every .5 mL of NaOH that is added to the solution.3


Standardization of NaOH Starting Volume of NaOH Ending Volume of NaOH Results
Trial 1 12.55 mL
Required additional NaOH
.2 mL
23.32 mL
8.72 mL Volume of NaOH5 used:19.29 mL or .019 L
Solution turned light pink
Trial 2 1.90 mL 25.52 mL Volume of NaOH4 used: 23.52 mL or .024 L
Solution turned dark pink

The solution should look like the image below when the titration reaches its equivalence point.10

Titration of unknown Acid
Titration Results
Trial 1 The solution turned dark pink5
Trial 2 The solution turned dark pink4
Trial 3 The solution turned dark pink also5
Trial 4 Starting Volume of NaOH-0 mL5
End Volume of NaOH-12.1 mL
The solution turned light pink at the equivalence point of the reaction

Titration Curve Graph to the Equivalence Point 3

Due to the curves in the graph and where the equivalence point occurs, the acid was identified as monoprotic.

Standardization of NaOH from Trial 12
.8g KHP(1 mol KHP/204.23 g KHP)= .004 mol KHP
.004 moles KHP/.05 L H2O = .08 M KHP
.004 moles KHP(1 mole NaOH/1 mol KHP)= .004 moles NaOH
.004 moles NaOH/.01929 L NaOH=.207 M NaOH

Calculation of the Molar Mass and pKa of the unknown acid4
0.004 mol NaOH / 0.019 L x 0.0124 L NaOH = .00261 mol
0.2 g unknown acid / .00261 mol = 76.6 g / mol (Molar Mass of unknown)
12.4 mL of NaOH required for equivalence point to be met.
12.4 / 2 = 6.2 mL half of equivalence point
At 6.2 mL of NaOH, the pH was between 3.87 – 4.12
Therefore, the pKa of the unknown is between 3.87-4.12


In order to successfully titrate the unknown acid, the NaOH solution required standardization to figure out the concentration of NaOH present in the solution. NaOH has to be standardized because its concentration cannot be found in an environment that contains air or water. NaOH is hygroscopic, which means that it absorbs water from the air and it also interacts with the carbon dioxide in the air to form sodium carbonate. Therefore, the exact concentration of NaOH cannot be determine so it has to be titrated with an acid to find the exact concentration of NaOH.9 Titrating the NaOH in the KHP solution required the same number of moles of NaOH to be added as the number of moles of the KHP solution. In one mole of KHP, there is one mole of hydrogen ions that can interact with one mole of sodium hydroxide, which contains one mole of hydroxide ions. At the equivalence point of the reaction, the amount of KHP and hydroxide ions have been mixed. This makes the solution more basic, due to the concentration of hydroxide ions and the formation of a KP ion, which is a weak base, contributing to basicity of the solution.9 When phenolphthalein is added to the solution a slight color change occurs when the equivalence point is reached. This is due to the color changing property of the indicator, which changes color when the hydrogen ion concentration changes. Also, the indicator changes color because its ion is colored. As the acidic solution becomes more basic the molecule loses its hydrogen ions, which causes phenolphthalein to give off its pink color. When ionization is complete the indicator turns the solution pink.7
In the second trial of the NaOH standardization, the solution became too dark. This is due to the concentration of the hydroxide ions being greater than the hydrogen ions in the solution, which meant that too much NaOH was added to the solution, making it too basic. In the titration of the unknown acid, as more NaOH was added, the solution slowly began to turn a slight pink color, but it didn’t stay pink. This was an indication that the solution was getting closer to the equivalence point, but it hadn’t reached the equivalence point.2 At the equivalence point, there was .00261 moles of NaOH present in the solution and .00261 moles of the unknown acid present in the solution. In one mole of NaOH there is one mole of the pH probe was used to collect the pH of the solution as .5 more mL of NaOH was added to the solution. As the amount of NaOH in the solution increased, the pH of the solution began to increase as the acid was being neutralized by the strong base. The concentration of hydroxide and hydrogen ions were inching closer to becoming the same. When the equivalence point was reached of the reaction between the unknown acid and NaOH, the pH of the solution was around 8.3 making it more basic and causing the ionized phenolphthalein to change color7. This means that the amount of obdurate hydroxide ions in the solution was greater than the concentration of hydrogen ions in the solution. This is also due to the NaOH interacting with the acid and forming a conjugate base.

As show in the graph the equivalence point of the reaction is when around 12.4 mL of NaOH was added to the acidic solution. At one-half of this volume of NaOH, which would be the half equivalence point, enough of NaOH was added to neutralize half of the acid. Since half of the acid reacted to form a conjugate base, the concentrations of the conjugate base and acid at the half-equivalence point was the same. Therefore, at the half-equivalence point, the pH is equal to the pKa8. The pKa identifies the strength of the acid. The lower the pKa is, the stronger the acid is.6 At the half equivalence of the reaction between NaOH and the unknown acid, the pKa was determined to be between 3.87 and 4.1, which demonstrated that the acid is a relatively strong acid, meaning that it almost completely disassociates in pure water. The curves in the titration curve graph show the pH rising. If the acid was diprotic or triprotic, there would be more curves in the graph due to multiple equivalence points.


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