Abstract The identification of acids is very important in analytical chemistry. (1) Acids can be identified by their Ka and their molar mass. In this experiment an unknown acid is to be identified using a titration. Data from the titration will be used to construct a titration curve from which the mass and the Ka of the unknown acid can be determined. Upon completion it was determined that the unknown acid was benzoic acid. The experimental pKa was found to be 4.18 while the actual pKa is 4.20. The molar mass of the unknown was experimentally determined to be 125.25 g/mol, while the actual molar mass of Benzoic acid is 122 g/mol. William Fall

Introduction: A titration is an analytical technique that involves utilizing solutions of known concentrations to determine the concentration of an unknown solution. (3)Titrations are represented by titration curves. Titration curves provide valuable information as they allow for the concentration of the analyte to be determined. One very important point on the titration curve is the equivalence point, where the moles of acid equal the moles of base. At the equivalence point the known amount of base has exactly neutralized the acid. In this experiment, an unknown acid is to be titrated with a known quantity of base, previously standardized 0.97 M sodium hydroxide. The moles of base can be determined by multiplying the concentration by the volume. But how is an equivalence point determined? If a line is drawn through the part of the curve with the greatest slope; the center of that line is roughly the equivalence point.(2) Equivalence point can be determined more accurately by taking the first and second derivative of the titration curve. (2) Logger Pro software was used to make the first and second derivative of the titration curve for this experiment. For a first derivative graph, the equivalence point occurs where the change in pH over the change in volume is the greatest. (2) For the second derivative plot, the point at which the plot crosses “0” is the equivalence point. (2) William Fall

Introduction Another important point on titration curve is the midpoint. The midpoint point occurs at half the volume of the equivalence point. At the midpoint, the concentration of weak acid and its conjugate base are equal. The Henderson-Haselbach equation states that pH=pKa+log[base]/[acid] . When the acid and conjugate base are equal in concentration the ratio is one, the log of one is one. Therefore, at the half-equivalence point ph=pKa. (1) pKa is the negative log of the Ka of the acid. The Ka is the ionization constant, it describes the strength of the acid and identifies the acid since every acid has its own unique Ka.(1) This is important since the goal of the experiment is to determine the identity of an unknown acid. In order to calculate the molar mass, the acid sample must first be weighed before the titration. If the mass is known, then the mass of the acid divided by the moles of the acid at the equivalence point will equal the molar mass. At the equivalence point, the moles of base added equals moles acid. Multiplying the concentration of the base by the volume at the equivalence point will determine the moles of base, and thus moles of acid. By comparing the experimentally determined molar mass and Ka, the unknown acid can be identified. (1) William Fall (4)

Methodology First a pH probe was properly prepared by washing with tap then distilled water. Next the LabQuest was properly setup for the titration. A 100 ml beaker was tared and 0.100 grams of the solid unknown acid was massed. The acid was then dissolved in 5 ml of ethanol, and 45 ml of water was added to the beaker, as well as a stirring bar. The titration was carried out, a ring stand and pH probe clamp were setup . The pH probe was inserted into the solution and the previously standardized 0.97 M NaOH was added drop wise from a 1.0 ml hypodermic syringe. The volume of base added was recorded using the LabQuest. The titration was finished and all the materials were thoroughly cleaned. The image to the right is an example of a titration setup similar to the one conducted In this lab. (5) William Fall

Graphs Titration Curve This is the titration curve, upon determining the equivalence point the midpoint on the titration curve can be determined. Dividing the equivalence point volume by two will yield the midpoint volume. At the midpoint the pH= pKa. William Fall

Graphs First Derivative of Titration Curve This is the plot of the first derivative of the titration curve. The equivalence point is located where the change in pH over the change in volume is the greatest.(2) William Fall

Graphs Second Derivative of Titration Curve This is the graph of the second derivative of pH vs volume of base added. Where the plot exactly crosses “0” is the equivalence volume.(2) William Fall

Discussion To determine the identity of the unknown acid, the pKa and molar mass are needed. The titration will provide the data. Using the second derivative plot of the titration curve it was determined that 0.7984 mL of base was added to the beaker to reach the equivalence point. The mid-point must be determined because it provides the pKa of the unknown acid. At the mid-point the concentrations of weak acid and its conjugate base are equal. (1) The mid-point can be determined by dividing the equivalence volume by 2. It was determined that the mid-point volume was 0.3992 mL. At the midpoint, the Henderson- Haselbach equation, pH= pKa +log[base]/[acid[, can be used to determine the pH. Since the concentrations of weak acid and base are equal at the midpoint, pH= pKa. It was determined that the pKa of the unknown acid was 4.18. (6) William Fall

Discussion In order to determine the molar mass the number of moles of base needed to completely neutralize the acid must be determined. (1) The acid is neutralized completely at the equivalence point. (1) The moles of NaOH at the equivalence point can be determined by multiplying the concentration (0.97 M) by the volume at the equivalence point (7.984 x 10-4 L). ( 0.97 M) x (7.984 x 10-4 L) = 7.744 x 10-4 L The number of moles of base needed to completely neutralize the acid was 7.744 x 10-4 moles. To determine the molar mass of the acid, divide the initial mass of the acid in grams by the moles of base required to neutralize the acid. (1) 0.0970 g acid/ (7.744 x 10-4 molbase) = 125.5 g/mol  The molar mass of the unknown acid was determined to be 125.5 g/mol. Based on the pKa and molar mass the unknown acid was identified as Benzoic acid. The actual pKa of Benzoic acid is 4.20 and the actual molar mass is 122 g/mol. The data gathered in the experiment is most likely slightly off due to experimental error. Conducting more than one trial would be ideal as it would yield more data which could be used to take averages. (7) (8) William Fall

Conclusion In this experiment an unknown solid organic acid was to be identified. To identify the acid its pKa and molar mass were needed. A titration was performed with previously standardized 0.97 M NaOH. A titration curve was constructed using data from the titration. The second derivative of the titration curve was used to determine the experimental pKa of the acid, 4.18. The molar mass was calculated by dividing the initial mass of the acid by the moles of acid at the equivalence point. The experimental molar mass of the unknown acid was calculated to be 125 g/mol. The unknown acid was identified as Benzoic acid. The acid was identified based on its experimental molar mass and pKa, which were very similar to the actual molar mass and pKa of Benzoic acid. (122 g/mol and 4.20 respectively) The titration is very important in analytical chemistry as it provides the concentration of an unknown solution. (1) Titrations also provide information such as pKa and molar mass, that can lead to the identification of chemicals. (1) Every acid has its own pKa value, thus determining the pKa will identify the acid. This concept is used in forensic science, where pKa’s are used to identify acids at crime scenes. An unknown sample could be collected at a crime scene, at a forensics lab its pH could be tested. A titration could be carried out to determine the unknown’s concentration. If the substance was an acid, its pKa and molar mass could be obtained and the unknown would be identified, potentially helping in a criminal investigation. The titration is a powerful tool as it provides pivotal information with many different applications. William Fall

References 1- Langella, Marsilio. PWISTA.com The Identification Of A Solid Organic Acid 2- Langella, Marsilio . PWISTA.com Titration Notes . PDF 3- Perdue Univeristy. “Lab Techniques” What is a Titration? http://chemed.chem.purdue.edu/genchem/lab/techniques/titration/index.html 4- Image from expertsmind.com. pH Titration Curves. http://www.expertsmind.com/topic/measurement-of-ph/ph-titration-curves-910984.aspx 5- Image from prenhall online textbook. http://wps.prenhall.com/wps/media/objects/3312/3392202/blb1703.html 6-Image from University of Illinois. Titrations http://butane.chem.uiuc.edu/cyerkes/chem102ae_fa08/homepage/Chem102AEFa07/Lecture_Notes_102/Lecture27-102.htm 7- Image from Wikimedia http://commons.wikimedia.org/wiki/Category:Benzoic_acid#/media/File:Benzoic_acid_3D.png 8- Image from mdidea.com http://www.mdidea.com/products/proper/proper02703.html William Fall