Standardization of a Primary Standard & Determination of Concentration by Acid-Base Titration

Standardization of a Primary Standard & Determination of Concentration by Acid-Base Titration It is often necessary to test a solution of unknown concentration with a solution of a known, precise concentration. The process of determining the unknown s concentration is called standardization. Solutions of sodium hydroxide are virtually impossible to prepare to a precise molar concentration because the substance is hygroscopic. In fact, solid NaOH absorbs so much moisture from the air that a measured sample of the compound is never 100% NaOH. On the other hand, the acid salt potassium hydrogen phthalate, KHC 8 H 4 O 4, can be measured out in precise mass amounts once it has been dried in an oven. It reacts with NaOH in a simple 1:1 stoichiometric ratio, thus making it an ideal substance to use to standardize a solution of NaOH. At the equivalence point, moles acid = moles base = moles salt formed. Since the ratio of reactants is 1:1, the neutralization equation, M A V A =M B V B, also applies since it is just another statement of moles acid = moles base = moles salt formed at the equivalence point. OBJECTIVES In this experiment, you will Prepare an aqueous solution of sodium hydroxide to a target molar concentration. Determine the exact concentration of your NaOH solution by titrating it with a solution of potassium hydrogen phthalate, abbreviated KHP, of precise molar concentration. Figure 1 MATERIALS Data Collection Mechanism solid sodium hydroxide, NaOH ph Sensor solid potassium hydrogen phthalate, KHP ring stand distilled water 250 ml beaker plastic weighing dish or weighing paper 250 ml Erlenmeyer flask balance (± 0.01 g) 50 ml and 100 ml graduated cylinders utility clamp magnetic stirrer and stirring bar buret and buret clamp

100 ml graduated cylinder Part I: STANDARDIZATION OF PRIMARY STANDARD PROCEDURE 1. Measure out 100 ml of distilled water into a 250 ml Erlenmeyer flask. 2. Set up the data collection system. a. Connect the ph Sensor to the interface. b. Start the data collection program. c. Set up data collection for Events with Entry mode. 3. Measure out the mass of NaOH that is needed to prepare a 0.100 M solution and add it to the flask of distilled water. Swirl the flask to dissolve the solid. CAUTION: Sodium hydroxide solution is caustic. Avoid spilling it on your skin or clothing. 4. Measure out the mass of KHP that will neutralize 25 ml of 0.100 M NaOH solution. Dissolve the analyte, the KHP, in about 50 ml of distilled water in a 250 ml beaker. 5. Set up a ring stand, buret clamp, and buret to conduct a titration (see Figure 1). Place a utility clamp on the ring stand to hold the ph Sensor in place during the titration. 6. Rinse and fill the buret with the titrant, the NaOH solution. Place a magnetic stirring bar in the beaker (if available) and place the beaker of KHP solution on a magnetic stirrer under the buret. Connect the ph Sensor to the utility clamp so that the tip of the sensor in immersed in the KHP solution but does not interfere with the movement of the magnetic stirring bar. 7. Conduct the titration carefully. When you have completed the titration, dispose of the reaction mixture as directed. Before conducting a second titration record the results of your first titration, either save the electronic data, or record your data in a table or chart. 8. Repeat the titration with a second KHP solution. Analyze the titration results and record the equivalence point. Save the results of the second titration. 9. Use your titration data to determine the equivalence point, which is the largest increase in ph upon the addition of a very small amount of NaOH solution. A good method of determining the precise equivalence point of the titration is to take the second derivative of the ph-volume data, a plot of 2 ph/ vol 2.

You may use another method to analyze titration data, called a Gran Plot. Proposed in the early 1950s by G. Gran, this method uses the reciprocal of ΔpH of the titration data (where ΔpH = ph value previous ph value). The graph of volume of 1/ΔpH vs. titrant volume resembles a V-shaped plot. The inflection point of this plot is the equivalence point volume of the titration. To use this method, you will first need to create a new calculated column, 1/ΔpH. You can do this in your calculator or with computer graphing software such as Logger Pro. On your resulting plot of 1/ΔpH vs. volume (see Figure 2), interpolate to find the intersection of two best-fit regression lines. The precise volume where the two linear fits intersect will be the equivalence point volume. In the sample graph shown here, the equivalence point is 10.58 ml. Figure 2

DATA TABLE Trial Equivalence point 1 2 PART I: PRE-LAB QUESTIONS 1. Calculate the mass of sodium hydroxide needed to prepare 100 ml of a 0.100 M solution. 2. Calculate the mass of KHP needed to react completely with 25 ml of a 0.100 M NaOH solution. Consider the reaction equation to be as shown below. HP (aq) + OH (aq) H 2 O( ) + P 2 (aq) 3. Calculate the molarity of a solution of sodium hydroxide if 23.64 ml of this solution is needed to neutralize 0.5632 g of KHP. 4. It is found that 24.68 ml of 0.1165 M NaOH is needed to titrate 0.2931 g of an unknown monoprotic acid to the equivalence point. Calculate the molar mass of the acid. 5. The following data was collected for the titration of 0.145 g of a weak monoprotic acid with 0.100 M NaOH as the titrant: Volume of NaOH added, ml 0.00 5.00 10.00 12.50 15.00 20.00 24.00 24.90 25.00 26.00 30.00 ph 2.88 4.15 5.58 4.76 4.93 5.36 6.14 7.15 8.73 11.29 11.96 (a) Use computer graphing software to graph ph vs. volume of NaOH. (b) Analyze your graph. What is volume of NaOH required to reach the equivalence point? (c) What is the ph at the equivalence point? (d) Calculate the molar mass of the weak monoprotic acid? (e) Give the K a and pk a value of the acid. Justify your answer.

PART I: POST-LAB QUESTIONS AND DATA ANALYSIS 1. Graph your data. 2. Calculate the molar amount of KHP used to neutralize the NaOH solution. 3. Calculate the molar concentration of the NaOH solution that you prepared. 4. Compare the actual molarity of your NaOH solution with your goal of 0.100 M. 5. A student fails to wash the weighing paper when transferring the KHP sample into the beaker. What effect does this error have on the calculated molarity of the NaOH solution? Mathematically justify your answer. 6. A student failed to notice an air bubble trapped in the tip of the buret during their experiment. What effect does this error have on the calculated molarity of the NaOH solution? Mathematically justify your answer.

Acid-Base Titration A titration is a process used to determine the volume of a solution that is needed to react with a given amount of another substance. In this experiment, your goal is to determine the molar concentration of two acid solutions by conducting titrations with the standardized NaOH solution you mixed in the previous lab exercise. You will be testing a strong acid, HCl, solution and a weak acid, HC 2 H 3 O 2, solution. The reaction equations are shown below in net ionic form. H + (aq) + OH (aq) H 2 O(l) HC 2 H 3 O 2 (aq) + OH (aq) H 2 O(l) + C 2 H 3 O 2 (aq) The stoichiometry of the two reactions is identical; thus, your calculations will be straightforward. However, you will observe a significant difference in how the two acid solutions react with NaOH. In this experiment, you will monitor ph as you titrate. The region of most rapid ph change will then be used to determine the equivalence point. The volume of NaOH titrant used at the equivalence point will be used to determine the molarity of the HCl solution. OBJECTIVES In this experiment, you will Accurately conduct acid-base titrations. Determine the equivalence point of a strong acid-strong base titration. Determine the equivalence point of a weak acid-strong base titration. Calculate the molar concentrations of two acid solutions. MATERIALS Data Collection Mechanism YOUR 0.100 M standardized NaOH solution ph Sensor hydrochloric acid, HCl, solution, unknown concentration ring stand acetic acid, HC 2 H 3 O 2, solution, unknown concentration 250 ml beaker distilled water magnetic stirrer and stirring bar 50 ml buret and buret clamp 25 ml graduated cylinder utility clamp 50 ml graduated cylinder

PART II: ACID-BASE TITRATION PROCEDURE 1. Obtain and wear goggles. 2. Add 50 ml of distilled water to a 250 ml beaker. Obtain about 10 ml of a hydrochloric acid solution of unknown concentration from the dispensing buret your instructor has set up. Record the exact volume of HCl that you obtained. CAUTION: Handle the hydrochloric acid with care. It can cause painful burns if it comes in contact with the skin. 3. Place the beaker on a magnetic stirrer and add a stirring bar. If no magnetic stirrer is available, stir the reaction mixture with a stirring rod during the titration. 4. Set up the data collection system. a. Connect a ph Sensor to the interface. b. Start the data collection program. c. Set up data collection for Events with Entry mode. 5. Rinse and fill the buret with your titrant, the standardized NaOH solution of known concentration (about 0.100 M). CAUTION: Sodium hydroxide solution is caustic. Avoid spilling it on your skin or clothing. 6. Use a utility clamp to suspend the ph Sensor on the ring stand (see Figure 1). Position the ph Sensor so that its tip is immersed in the HCl solution but is not struck by the stirring bar. Gently stir the beaker of acid solution. 7. Conduct the titration carefully. Listed below is a suggested method of running a titration. a. Record an initial ph reading, before adding any 0.100 M NaOH solution. b. Add NaOH in small increments that raise the ph of the mixture by about 0.15 units at a time, until you reach a ph of about 3.5. c. At this point, add NaOH drop by drop. Continue in this manner until you have recorded data through the equivalence point. d. When the ph of the mixture is about 10, add larger increments of NaOH that raise the ph by about 0.15 units, or 1 ml of NaOH at a time, until the ph remains constant. 8. When you have finished the titration, dispose of the reaction mixture as directed. Rinse the ph Sensor with distilled water in preparation for a second titration. Before conducting the second titration, record your data electronically, or in a table or chart. 9. Use your titration data to determine the equivalence point, which is the largest increase in ph upon the addition of a very small amount of NaOH solution. 10. Print and save the graph and data set of ph vs. volume for the HCl trials. 11. Repeat the necessary steps to test the acetic acid solution and analyze the titration data. Print and save the graph and data set for the acetic acid titrations.

PART II: DATA TABLE HCl Trial Volume HCl Equivalence point CH 3COOH Trial Volume CH 3COOH Equivalence point 1 1 2 2 PART II: POST LAB QUESTIONS AND DATA ANALYSIS 1. Calculate the molar amounts of NaOH used in the reaction with the HCl solution and with the HC 2 H 3 O 2 solution. 2. Calculate the molar concentration (molarity) of the HCl solution and the HC 2 H 3 O 2 solution. 3. Compare the actual molar concentrations of your two acid solutions with your calculated molarities. Were the calculated molarities of your acid solutions within a reasonable range (about 5%) of the actual values? If not, suggest reasons for the inaccuracy. 4. The equivalence points of the two titration curves were not in the same ph range. Explain.

TEACHER INFORMATION 1. This experiment conforms to the guidelines for the fifth and sixth laboratory experiments listed in the College Board AP Chemistry guide (the Acorn book). 2. Potassium hydrogen phthalate (KHP), commonly used to standardize NaOH solutions, has the chemical formula, KHC 8 H 4 O 4, and an accepted molar mass of 204.22 g/mol. This solid should be dried in an oven prior to having students mass their sample. If you must dry it ahead of time, then store it in a desiccator prior to use. 3. If you are pressed for time, you can make a larger batch of the 0.100 M solution for standardization ahead of time by adding 4 grams of solid NaOH to 1.0 L of DI water. 4. Prepare the 0.1M HCl by adding 8.6 ml of concentrated HCl to sufficient water to make 1 L of solution. 5. Prepare the 0.1M CH 3 COOH solution by adding 5.7 ml of glacial acetic acid to sufficient water to make 1 L of solution. You may substitute store bought vinegar as well. 6. The stored calibration information for the ph sensor works well for this experiment. However, you should examine your probes for mold growth and consult the directions that came with your model for proper storage solutions. If they are Vernier sensors, then sensor information can be found at http://www2.vernier.com/booklets/ph-bta.pdf 7. If using Drop Counters, for best results, instruct your students to perform a new calibration on their Drop Counters. If time does not permit a new calibration, a substitute value of 28 drops/ml may be used. This value will provide good results. If students will not be calibrating their Drop Counters, instruct them to follow the procedure of manually entering a calibration value and to enter a value of 28 for the drops/ml ratio. 8. If using LoggerPro software, it is possible to zoom in on the equivalence point in the 2nd derivative vs. volume plot, and then perform a linear fit on the data near the equivalence point. Once a linear fit is established, it is possible to interpolate along that fit until the 2nd derivative value is approximately 0, corresponding to the equivalence point volume of NaOH titrant. HAZARD ALERTS

Hydrochloric Acid: Highly toxic by ingestion or inhalation; severely corrosive to skin and eyes. Hazard Code: A Extremely hazardous. Sodium Hydroxide: Corrosive solid; skin burns are possible; much heat evolves when added to water; very dangerous to eyes; wear face and eye protection when using this substance. Wear gloves. Hazard Code: B Hazardous. Acetic acid: Corrosive to skin and tissue; moderate fire risk (flash point: 39 C); moderately toxic by ingestion and inhalation. Hazard Code: A Extremely hazardous. The hazard information reference is: Flinn Scientific, Inc., Chemical and Biological Catalog Reference Manual, P.O. Box 219, Batavia, IL 60510, (800) 452-1261, www.flinnsci.com Answers to Part I PRE-LAB QUESTIONS 1. Calculate the mass of sodium hydroxide needed to prepare 100 ml of a 0.100 M solution. ( liters) M # moles = MV 0.100 0.100L 0.0100 moles NaOH required 40.0 g 0.0100 mol NaOH 0.400 g NaOH required mol 2. Calculate the mass of KHP needed to react completely with 25 ml of a 0.100 M NaOH solution. Consider the reaction equation to be as shown below. There is a 1:1 molar ratio of NaOH : KHP NaOH KPH ( liters) HP (aq) + OH (aq) H 2 O( ) + P 2 (aq) M # moles = # moles = MV 0.100 0.025L 0.0025 moles KHP required 204.22 g 0.0025 mol KPH 0.5106 g KHP required mol 3. Calculate the molarity of a solution of sodium hydroxide if 23.64 ml of this solution is needed to neutralize 0.5632 g of KHP. There is a 1:1 molar ratio of NaOH : KHP 1 mol # moles KPH = 0.5632 g KPH 0.002758 mol KHP = # moles 204.22 g mol 0.002758 molnaoh M 0.1167 M L.02364 L NaOH 4. It is found that 24.68 ml of 0.1165 M NaOH is needed to titrate 0.2931 g of an unknown monoprotic acid to the equivalence point. Calculate the molar mass of the acid.

grams Molar mass = mol # moles MV 0.1165 0.02468 ml 0.002875 mol NaOH = # moles M NaOH liters acid There is a 1:1 molar ratio of NaOH : monoprotic acid 0.2931 g Molar mass acid = 101.9g/mol 0.002875 mol 5. The following data was collected for the titration of 0.145 g of a weak monoprotic acid with 0.100 M NaOH as the titrant: Volume of NaOH added, ml 0.00 5.00 10.00 12.50 15.00 20.00 24.00 24.90 25.00 26.00 30.00 ph 2.88 4.15 5.58 4.76 4.93 5.36 6.14 7.15 8.73 11.29 11.96 (a) Use computer graphing software to graph ph vs. volume of NaOH. The following graph was produced in LoggerPro 3.6 with the connect points option selected. (The dashed red lines have been drawn on the graph as part of the answers to the remaining questions pertaining to this graph.) (b) Equivalence Point Volume NaOH 25mL (c) Equivalence Point ph 8.5 (d) ½ Equivalence Point ph 4.75 (d) ½ Equivalence Point Volume NaOH 12.5 ml (b) Analyze your graph. What is volume of NaOH required to reach the equivalence point?

Students should draw the best line they can passing through the geometric midpoint of the vertical region of the titration curve and read where it intersects the x-axis. The reported value should be close to 25 ml. Answers will vary based on student interpretations of data from the graph.

(c) What is the ph at the equivalence point? Students should draw the best line they can passing through the geometric midpoint of the horizontal region of the titration curve and read where it intersects the x-axis. The reported value should be close to a ph of 8.5. Answers will vary based on student interpretations of data from the graph. (d) Calculate the molar mass of the weak monoprotic acid? Answers will vary based on student interpretations of data from the graph. There is a 1:1 molar ratio of NaOH : weak monoprotic acid NaOH liters M # moles = MV 0.100 0.02500 L 0.0025 mol NaOH 0.145 g weak acid MM weak monoprotic acid 58.0 g/mol 0.0025 mol weak acid (e) Give the K a and pk a value of the acid. Justify your answer. Answers will vary based on student interpretations of data from the graph. For weak acids, the ph = pk a at the half-equivalence point. The graph shows an equivalence point when 25mL of NaOH has been added, therefore the half-equivalence point volume is, well half of that, so 12.5 ml. Students should draw a straight vertical line that intersects the titration curve s x-axis at 12.5 ml and then draw a straight horizontal line at that intersection point so that they can easily read the ph from the y-axis. Therefore, the ph at the half-equivalence point is approximately equal to 4.75. From that we estimate that the K a value is in the neighborhood of 10 5. Students need practice with estimation skills! The actual calculation follows: ph pk a at half-equivalence point 4.75 pk 10 K 1.8 10 a 4.75 5 a PART I: SAMPLE DATA TABLE Trial Equivalence point 1 25.80 ml 2 25.60 ml

Answers to PART I: POST-LAB QUESTIONS AND DATA ANALYSIS 1. Graph your data. Graphs will vary. 2. Calculate the number of moles of KHP used to neutralize the NaOH solution. This quantity was calculated as Question 1 in the Pre-Lab Quesitons. There is a 1:1 molar ratio of NaOH : KHP NaOH KPH ( liters) M # moles = # moles = MV 0.100 0.025L 0.0025 moles KHP required 204.22 g 0.0025 mol KPH 0.5106 g KHP required mol 3. Calculate the molar concentration of the NaOH solution that you prepared. Answers will vary. If a student accurately measured the 0.5106 g KHP calculated above then: 0.0025 mol Trial 1: NaOH = = 0.09690 0.02580 L M Trial 2: 4. Compare the actual molarity of your NaOH solution with your goal of 0.10 M. 0.0025 mol NaOH = = 0.09766 M 0.02560 L Answers will vary depending upon the accuracy of your balance and the storage conditions and age of your solid NaOH. Point out to students that since NaOH is hygroscopic, the mass they initially measured contained water, therefore less actual NaOH is in solution than the mass recorded, thus it is not uncommon for students to have molarities less than 0.10 M. 5. A student fails to rinse the weighing paper into the beaker when transferring the KHP sample. What effect does this error have on the calculated molarity of the NaOH solution? Mathematically justify your answer. The mass of the KHP is reported as too high since some residue did not make it into the beaker. Therefore, the number of moles of KHP is too high, thus the number of moles of NaOH in the numerator of the molarity calculation makes the reported molar concentration of NaOH too high. 6. A student failed to notice an air bubble that passes through the tip of the buret during their experiment. What effect does this error have on the calculated molarity of the NaOH solution? Mathematically justify your answer. The volume of the NaOH is reported as too low since some of the volume passing through the buret was air. Since the volume term is in the denominator of the molarity calculation, the molarity is reported as too high.

PART II: SAMPLE DATA HCl Trial Volume HCl Equivalence point CH 3COOH Trial Volume CH 3COOH Equivalence point 1 5.00 0.0941 1 5.00 0.0941 2 5.00 0.0941 2 5.00 0.0941 Answers to PART II: POST LAB QUESTIONS AND DATA ANALYSIS 1. Calculate the molar amounts of NaOH used in the reaction with the HCl solution and with the HC 2 H 3 O 2 solution. For the sample data: mol 4 Trial 1: mol NaOH = M BVB 0.0941 0.00555 L = 5.22 10 mol L mol 4 Trial 2: mol NaOH = M BVB 0.0941 0.00553 L = 5.20 10 mol L mol 4 Trial 1: mol CH3COOH = M BVB 0.0941 0.00564 L = 5.31 10 mol L mol 4 Trial 2: mol CH3COOH = M BVB 0.0941 0.00566 L = 5.33 10 mol L 2. Calculate the molar concentration (molarity) of the HCl solution and the HC 2 H 3 O 2 solution. HCl Molarities: 4 5.22 10 mol Trial 1: 0.104 M HCl 0.00500 L 4 5.20 10 mol Trial 2: 0.104 M HCl 0.00500 L HC 2 H 3 O 2 Molarities: 4 5.31 10 mol Trial 1: 0.106 M CH3COOH 0.00500 L 4 5.33 10 mol Trial 2: 0.107 M CH3COOH 0.00500 L 3. Compare the actual molar concentrations of your two acid solutions with your calculated molarities. Were the calculated molarities of your acid solutions within a reasonable range (about 5%) of the actual values? If not, suggest reasons for the inaccuracy. All of the trials agreed nicely with the calculated molarities. This is mostly due to the use of either a ph meter, ph probe or drop counter which eliminates the need for an indicator. It also eliminates students over titrating the equivalence point by missing the end point (color change).

4. The equivalence points of the two titration curves were not in the same ph range. Explain. For the titration of NaOH and HCl, the equivalence point should be at ph = 7.00 since at the equivalence point moles acid = moles base = moles salt formed. All of the acid and base have reacted to form the salt, which in this case is NaCl and a neutral salt. For the titration of NaOH and HC 2 H 3 O 2, the equivalence point should be at a ph greater than 7.00. Again, at the equivalence point moles acid = moles base = moles salt formed and all of the acid and base have reacted to form sodium acetate, but this is not a neutral salt. This salt is basic as evidenced by the hydrolysis reaction below: C 2 H 3 O 2 + H 2 O H C 2 H 3 O 2 + OH Therefore, the resulting solution is basic.