Experiment 9: Weak Acids and Buffers

Experiment 9 Weak Acids and Buffers GRADING RUBRIC EXPERIMENT 9 – LAB REPORT Worksheet (80 pts) Day & time of the lab: ___4/5/22 9 AM________________________ The lab report packet for this experiment should consist of the following in THIS ORDER: The lab report packet for this experiment should consist of the following: 1. (10 pts) IN-lab notebook pages 1.1. Experimental procedure 1.2. Recorded experimental values and the file name (as saved on your USB drive) 1.3. Observations if any were noted 2. (70 pts) The WORKSHEET 3.1 Tables with captions 3.2 Figures of graphs and spectra with captions 3.3 Complete (full sentence) answers to the discussion questions (DQ; Provide wellrounded answers composed with complete sentences for full credit. If you need to refer to any of the data or tables produced during the “Data Analysis” section, clearly indicate their location within your answers for the graders to find them readily.) 3.4 Data analysis Complete this worksheet and turn in your digital lab packet to Canvas. Feel free to adjust space in this document as needed. The tables provided here are only an example of these may look like, feel free to replace the entire tables with your own creations using Excel, etc All tables and figures (graphs) must be accompanied by captions which should be positioned above tables and below figures. TURN IN the digital copy to Canvas on the due date by 11:59 pm. All electronic material must be completed by the due date to avoid late penalties. (70 pts) WORKSHEET – (DATA ANALYSIS, DQs, Figures, Tables) 1 ● You MUST present all calculations in a typed form here using the equation function in Word 2 3 4 5 6 (10 pts) Include any/all relevant figures (titration curves and second derivative curves). Include appropriate captions. (1 pt each) Titration curves (5) are included and are properly labelled and with captions. Figure 1 Trial 1 Acetic Acid vs NaOH Titration Curve 12 10 pH 8 6 4 2 0 0 2 4 6 8 10 12 14 16 Volume (mL) Figure 1: A volume (mL) versus pH graph plotted to demonstrate the titration curve of Trial 1 of acetic acid titrated with NaOH. The equivalence point volume was 10.40 mL. Figure 2 Trial 2 Acetic Acid vs NaOH Titration Curve 12 10 pH 8 6 4 2 0 0 2 4 6 8 10 Volume (mL) 7 12 14 Figure 2: A volume (mL) versus pH graph plotted to demonstrate the titration curve of Trial 2 of acetic acid titrated with NaOH. The equivalence point volume was 8.05 mL. Figure 3 Trial 3 Acetic Acid vs NaOH Titration Curve 12 10 pH 8 6 4 2 0 0 2 4 6 8 10 12 14 Volume (mL) Figure 3: A volume (mL) versus pH graph plotted to demonstrate the titration curve of Trial 3 of acetic acid titrated with NaOH. The equivalence point volume was 9.53 mL. Figure 4 Buffered Acetic Acid vs. NaOH Titration Curve 14 12 pH 10 8 6 4 2 0 0 2 4 6 8 10 12 14 Volume (mL) Figure 4: A volume (mL) versus pH graph demonstrating a titration curve with a buffered acetic acid solution titrated against NaOH. The equivalence 8 point volume is 8.83 mL. Figure 5 pH Buffered Acetic Acid vs HCl Titration Curve 5 4.5 4 3.5 3 2.5 2 1.5 1 0.5 0 0 2 4 6 8 10 12 14 16 Volume (mL) Figure 5: A volume (mL) versus pH graph demonstrating a titration curve with a buffered acetic acid solution titrated against NaOH. The equivalence point volume is 10.90 mL. Figure 6 9 2nd Derivative Acetic Acid and NaOH Trial 1 10 8 2nd Derivative 6 4 2 0 -2 0 2 4 6 8 10 12 14 -4 -6 Volume (mL) Figure 6: A volume (mL) versus pH graph demonstrating a second derivative curve for Trial 1 of acetic acid being titrated against NaOH. The pH at half equivalence point is 4.54. Figure 7 2nd Derivative Acetic Acid and NaOH Trial 2 80 2nd Derivative 60 40 20 0 0 2 4 6 8 10 12 14 -20 -40 Volume (mL) Figure 7: A volume (mL) versus pH graph demonstrating a second derivative curve for Trial 2 of acetic acid being titrated against NaOH. The pH at half equivalence point is 4.63. Figure 8 10 2nd Derivative Acetic Acid and NaOH Trial 3 80 60 2nd Derivative 40 20 0 0 2 4 6 8 10 12 14 -20 -40 Volume (mL) Figure 8: A volume (mL) versus pH graph demonstrating a second derivative curve for Trial 3 of acetic acid being titrated against NaOH. The pH at half equivalence point is 4.50. Figure 9 2nd Derivative Buffered Acetic Acid and NaOH 2nd Derivative 0 2 4 6 8 10 12 14 Volume (mL) Figure 9: A volume (mL) versus pH graph demonstrating a second derivative curve for a buffered acetic acid being titrated against NaOH. The pH at equivalence point is 8.93. 11 Figure 10 Buffered Acetic Acid and HCl 2nd Derivative 0.6 2nd derivative 0.4 0.2 0 -0.2 0 2 4 6 8 10 12 14 16 -0.4 -0.6 -0.8 Volume (mL) Figure 10: A volume (mL) versus pH graph demonstrating a second derivative curve for a buffered acetic acid being titrated against HCl. The pH at equivalence point is 2.86. (1 pt each) captions. Second derivative curves (5) are included and are properly labelled and with Discussion Questions (Data Analysis calculations are detailed below) 1. (3 pts) Using the average pH at the half-equivalence point from the acetic acid titrations (Table 1), determine the pKa of the acetic acid solution. How was the pKa determined? Using the average pH at the half-equivalence point from the acetic acid titrations. The pKa was determined using the calculated pH values at the half-quivalence point in all the three trials and then taking the average of them. With that being said, averaging 4.54, 6.63, and 4.50 results in an average of 4.56. PART A: Preparation and Standardization of NaOH (6 pts) Table 1: NaOH volumes at the equivalence and half-equivalence points of each acetic acid titration, obtained from the second derivative curves. Trial NaOH volume (mL) at NaOH volume (mL) at pH at half-equivalence equivalence point half-equivalence point point 1 10.40 5.2 4.54 12 2 8.05 4.03 4.63 3 9.53 4.78 4.50 Average 9.33 4.67 4.56 2. (3 pts) Calculate the Ka of your acetic acid solution. Discuss this calculation. Based on the value of Ka, is acetic acid a strong acid or a weak acid? Why? The pKa value is simply the negative logarithm of the Ka value meaning that pKa = -logKa. Thus, if the pKa for the acetic solution is 4.56, raise both sides of the original relationship between Ka nd pKa to a power of 10 to get the Ka value. With that being said, the Ka value is 10- 4.56 which is the Ka value of a weak acid. This is because the larger the Ka the stronger the acid, but 10-4.56 is not a relatively large number meaning it is not a strong acid. However, if the Ka was 10-2 or larger, then it would be considered a strong acid. 𝑝𝐾𝑎 = −𝑙𝑜𝑔𝐾𝑎 4.56 = −𝑙𝑜𝑔𝐾𝑎 𝐾𝑎 = 10!".$% 3. (3 pts) Calculate the number of moles of acetic acid that were present in the original buffer solution prepared in Step 18 of the procedure. Assume no change in volume upon addition of sodium acetate trihydrate. Discuss this calculation. To calculate the number of moles of acetic acid that were present in the original buffer solution prepared in step 19, the liters of solution needed to be multiplied by the molarity that is known of the acetic acid stock solution. There 0.5M acetic acid solution which was then multiplied by 0.03 L of solution because 30 mL of solution was needed. After multiplying 0.5M by 0.03 L, there are 0.015 moles of acetic acid, assuming there is no change in volume upon the addition of sodium acetate trihydrate. (0.5𝑀)(. 03𝐿) = .015 𝑚𝑜𝑙 𝑎𝑐𝑒𝑡𝑖𝑐 𝑎𝑐𝑖𝑑 4. (3 pts) Calculate the number of moles of acetate that were present in the original buffer solution prepared in Step 18 of the procedure (assume the exact amount of sodium acetate trihydrate was obtained). The molecular mass of sodium acetate trihydrate is 136.08 g/mol. Discuss this calculation. The numbers of moles of acetate that were present in the original buffer solution of the procedure can be found by divide the exact number of sodium acetate trihydrate that was measured by the molar mass. Thus, 1.025 g of acetate was measured in the experiment, which is then divided by the molar mass, 136.08 g/mol, resulting in 0.0075 moles of acetate. 13 1.025 𝑔 𝑠𝑜𝑑𝑖𝑢𝑚 𝑎𝑐𝑒𝑡𝑎𝑡𝑒 = .0075 𝑚𝑜𝑙𝑒𝑠 𝑠𝑜𝑑𝑖𝑢𝑚 𝑎𝑐𝑒𝑡𝑎𝑡𝑒 136.08 𝑔/𝑚𝑜𝑙 5. (3 pts) Use the Henderson–Hasselbalch equation to calculate the theoretical pH of the buffer solution (prior to the addition of NaOH) using the values obtained in Discussion Questions 3 and 4 and the theoretical value of Ka for acetic acid, 1.8 ⋅ 10–5. Discuss this calculation. To calculate this value, the Henderson-Hasselbalch equation is used which states that the pH = pKa + log [A-]/[HA]. With that being said, the pKa for Ka of 1.8 x 1.8-5 is 4.74 because when the negative logarithm is taken 4.74 is the result. With that piece of information combined with the values from DQ 3 and 4 the theoretical pH can be calculated: pH = 4.74 + log(.0075/ .015). This is simplified to 4.74 + log(.0075) – log (.015) which is then simplified to 4.74 + log(.5) resulting in a theoretical pH of 4.44. . 0075 𝑚𝑜𝑙𝑒𝑠 𝑎𝑐𝑒𝑡𝑎𝑡𝑒 . 015 𝑚𝑜𝑙𝑒𝑠 𝑎𝑐𝑒𝑡𝑖𝑐 𝑎𝑐𝑖𝑑 4.74 + log(. 5) = 4.44 6. (2 pts) What was the measured pH of the buffer solution prior to the addition of NaOH? Use the theoretical pH calculated in Discussion Question 5 and the following equation to determine and discuss the % error of your results. theoretical pH − measured pH % error = J J × 100 % theoretical pH 𝑝𝐻 = 4.74 + log 4.44 − 4.71 J × 100 % 4.44 The measured pH of the buffered acetic acid solution that was measured was 4.71. Now using the formula given, the measured pH is subtracted from the theoretical over the theoretical multiplied by 100. The absolute value of the number is the percent error. This is a relatively low percent error meaning that the buffered solution is close to the theoretical pH. PART B: Buffered Acetic Acid vs. NaOH 6.08 = J (4 pts) Table 2: Relevant volumes and pH readings during the NaOH titration of the buffer. Titration Conditions NaOH added (mL) pH Initial 0 4.71 Buffering capacity reached 7.42 5.71 Equivalence point 8.83 8.93 14 7. (3 pts) During the addition of NaOH in Part B, which buffer species reacted with the NaOH? Provide the balanced formula unit equation for this reaction. The Acetic acid species reacted with the NaOH. NaCH COO(aq) + NaOH (aq) -> NaCH COO (aq) + H O (l) 3 3 2 8. (3 pts) Based on the volume of NaOH added in Step 9 of the Data Analysis, determine the buffering capacity of the buffered acetic acid solution. You may disregard the volume of “1 L” from the definition and refer to the volume used of the original buffer. Discuss the calculation and be sure to include the correct units. .00186 moles is the buffering capacity to the buffered acetic acid solution. This is determined by multiplying the volume when the buffering capacity was reached by the known molarity of the solution which is .25M. The volume of the buffering capacity when it was reached was 7.42 mL after the NaOH was added. Thus, when multiplying the molarity by the volume, moles is left, which is what the units of the buffering capacity is in. Furthermore, 7.42 mL needs to be converted into liters which is .00742 L which is then multiplied by .25 M, resulting in a buffering capacity of .00186 moles. (. 00742 𝐿)(. 25 𝑀) = .00186 𝑚𝑜𝑙𝑒𝑠 𝑏𝑢𝑓𝑓𝑒𝑟𝑒𝑑 𝑎𝑐𝑒𝑡𝑖𝑐 𝑎𝑐𝑖𝑑 9. (6 pts) At volumes below the volume determined in Step 9 of the Data analysis section, discuss the relative changes in pH following subsequent NaOH additions. What happened to the pH following subsequent additions of NaOH after the Step 9 volume was reached? Why? Below the volume determined in step 9 of the data analysis section, the changes in pH increased by approximately .06 every addition of NaOH. However, after the step 9 volume, the pH began increasing by much larger factors. This indicated that the equivalence point was near because the pH was increasing by more, conveying the fact that a dramatic change in pH was near. PART C: Buffered Acetic Acid vs. HCl (4 pts) Table 3: Relevant volumes and pH readings during the HCl titration of the buffer. Titration Conditions HCl added (mL) pH Initial 0 4.71 15 Buffering capacity reached 9.00 3.71 Equivalence point 10.90 2.86 10. (3 pts) During the addition of HCl in Part C, which buffer species reacted with the HCl? Provide the balanced formula unit equation for this reaction. Acetic acid was the buffer species that reacted with the HCl in Part C. NaCH COO(aq) +HCl(aq) -> CH COOH (aq) + NaCl(aq) 3 3 11. (3 pts) Based on the volume of HCl added in Step 14 of the Data Analysis, determine the buffering capacity of the buffered acetic acid solution. You may disregard the volume of “1 L” from the definition and refer to the volume used of the original buffer. Discuss the calculation and be sure to include the correct units. The buffering capacity of the buffered acetic acid solution is .0023 moles. This is determined by multiplying the volume when the buffering capacity was reached by the known molarity of the solution which is .25M. The volume of the buffering capacity when it was reached was 9.00 mL after the HCl was added. Thus, when multiplying the molarity by the volume, moles is left, which is what the units of the buffering capacity is in. Furthermore, 9.00 mL needs to be converted into liters which is .00742 L which is then multiplied by .25 M, resulting in a buffering capacity of .0023 moles. (. 009𝐿)(. 25𝑀) = .00225 𝑚𝑜𝑙𝑒𝑠 𝑏𝑢𝑓𝑓𝑒𝑟𝑒𝑑 𝑎𝑐𝑒𝑡𝑖𝑐 𝑎𝑐𝑖𝑑 12. (3 pts) At volumes below the determined volume for when buffering capacity was reached, discuss the relative changes in pH following subsequent titrant additions. What happened to the pH following subsequent additions of titrant after the buffering capacity volume was reached? At volumes below the determined volume for when the buffering capacity was reached, 9.00 mL, the pH would decrease on average about .04 after every addition of HCl. However, after the buffering capacity was reached, the pH began to decrease in larger increments after every addition of the HCl. This is because after the buffering capacity has been reached, the amount of conjugate base or acid has been diminished through neutralization. Thus, in this case, the pH began to increase, because the buffer solution has been depleted through neutralization with HCl. Therefore, the pH began to decrease because more HCl has been added, but the buffering capacity has been exceeded. 13. (3 pts) Compare the two buffering capacities obtained from Discussion questions 9 and 12. Theoretically, these two values should be identical. Why? Use as much detail as possible to support your explanation. 16 These two values should be identical from discussion 9 and 12 because the same buffer is being used. Thus, depending on the buffer and how concentrated it is, is what determines the buffer capacity. In this case, a 0.05 M buffered acetic acid was created, which was used in both Part B and C. Therefore, theoretically, the buffer would have same buffer capacity because the buffer capacity depends on the concentrations of the chemicals which were used to prepare the buffer. Moreover, because 1.21 g of acetate, 15mL of DI H2O, and 15 mL of .5M acetic acid was used to prepare the .25M acetic acid buffer, which was used in both sections, the theoretical buffer capacity would be the same because the same materials, compounds, and concentrations were used to prepare the buffer. 14. (3 pts) If the concentrations of both acetic acid and acetate were doubled (volume stays the same) when preparing the buffer, would you expect the buffering capacity to increase, decrease, or stay the same? Why? If both concentrations of the acetic acid and acetate were doubled, while the volume stays the same, the buffering capacity is expected to increase because the concentration of the buffer increasing. Moreover, buffering capacity is dependent on the concentration of the buffer and because the acetate and the acetic acid was increased while the volume stayed the same, the concentration of the buffer would increase, thereby, increasing the buffering capacity. Acetic acid is the weak acid, and its conjugate weak base is the sodium acetate. Thus, doubling both quantities increases the concentration of the buffer, as the volume is kept constant, which then increases the buffering capacity, allowing the pHs to increase or decrease, depending on if strong acid or base is added, at a slower rate. 15. (2 pts) Describe two observed/possible sources of error in this experiment and discuss how to prevent them. One source of error in this experiment is that there were two solutions NaOH were created within a trial because by accident, a NaOH solution was disposed. Therefore, the second time a NaOH was produced, the concentrations were not the same, resulting in different equivalence point volume ranges within a trial. This error should have been prevented, however, because of simple misreading of the lab manual, the solution was thrown away. Additionally, another error, is not closing the stopcock close to the equivalence point, meaning too much of the titrant was added, and the pH changed dramatically but there was too much titrant that was added. To prevent this error, after the buffering capacity is reached, meaning when the pH started to increase faster, the titrant should be added very carefully so that the equivalence point volume is accurate. 17