The Ka and Molar Mass of a Monoprotic Weak Acid

The Ka and Molar Mass of a Monoprotic Weak Acid Chemistry Lab 152 Professor: James Giles November 7, 2012 Abstract: The purpose of this experiment was to determine the pKa, Ka, and molar mass of an unknown acid (#14). The pKa was found to be 3. 88, the Ka was found to be 1. 318 x 10 -4, and the molar mass was found to be 171. 9 g/mol. Introduction Acids differ considerable as to their strength. The difference between weak and strong acids can be as much as 10 orders of magnitude.Strong acids dissociate more completely than weak acids, meaning they produce higher concentrations of the conjugate base anion (A-) and the hydronium cation (H30+) in solution. HA(aq) + H20 (( A- + H3O+ With the following formula the degree to which an acid dissociates (Ka) can be calculated and given a numerical value. Ka = [A-][H3O+] / [HA] Ka is the conventional way of measuring an acid’s strength. The purpose of this experiment was to determine the Ka of an unknown acid, along with its pKa and mol ar mass. Experimental The unknown acid for this experiment was #14.The experiment began with the preparation and standardization of NaOH solution. It was calculated that 2. 00 grams of NaOH pellets were needed to prepare 0. 5 L of 0. 1 M NaOH solution. The solution was then standardized by conducting three titration trials. It was calculated that 0. 7148 grams of KHP were necessary to neutralize 35 mL of the 0. 1 M NaOH. Three samples of KHP were weighed approximating this number (Table 1). Each sample was mixed with 40 mL of deionized water and 2 drops of phenolphthalein in 3 Erlenmeyer flasks. Each flask was then titrated with the NaOH to a light pink endpoint.The volumes of NaOH were recorded, averaged, and the standardized. The molarity of the NaOH was found to be 0. 0981. Assuming a molar mass of 100 g/mol, it was calculated that 0. 3930 g of acid was needed to neutralize 40 mL of the standardized NaOH solution. This amount was weighed out on an electronic balance to full preci sion and added to a clean 250 mL beaker. The acid was first diluted with 10 mL of isopropanol and then 90 mL of water. A pH meter was immersed in the acid solution and an initial pH reading of 2. 61 was recorded.A buret filled with the NaOH solution was incrementally added to the acid solution and the changing pH values were recorded (Table 2). As the pH meter approached the equivalence point the amount of NaOH added each time was reduced. As the Table 2 shows, the pH rose significantly with the addition of little NaOH over this interval. This information was plotted using Graphical Analysis producing a titration curve graph of pH vs. NaOH (Graph 1). Additional calculations and graphs were produced to help identify the equivalence point: ? pH/? V vs. NaOH (Graph 2) and Vtotal x 10-ph vs. NaOH (Graph 3) Tables and CalculationsPreparation of 500 mL of 0. 1 M NaOH M = moles / volume 0. 1 M NaOH = moles NaOH / 0. 5 L H20 = 0. 05 moles NaOH 0. 05 moles NaOH x 39. 986 g/mol NaOH = 1. 99 g NaOH Preparation of KHP 0. 1 M NaOH = moles NaOH / 0. 035 mL NaOH = . 0035 moles NaOH 0. 0035 moles KHP x 204. 233 g/mole KHP = 0. 7148 g KHP Table 1: NaOH Titration Trials |Trial |KHP |NaOH (to titrate to endpoint) | | |(grams) |(mL) | |1 |0. 7159 |35. 75 | |2 |0. 7147 |35. 65 | |3 |0. 7149 |35. | | | | Avg. 35. 66 | Standardization of NaOH 0. 0035 moles NaOH / . 03566 mL NaOH = 0. 0981 M NaOH Table 2: pH vs. NaOH Values |NaOH |pH |NaOH |pH |NaOH |pH |NaOH |pH | |(mL) | |(mL) | |(mL) | |(mL) | | |0 |2. 61 |19. 2 |4. 54 |22. 15 |6. 56 |25. 4 |9. 74 | |2 |2. 94 |19. 4 |4. 58 |22. 2 |6. 2 |25. 9 |9. 82 | |4 |3. 18 |19. 6 |4. 61 |22. 25 |6. 87 |26. 4 |9. 96 | |5 |3. 3 |19. 8 |4. 65 |22. 3 |6. 98 |26. 9 |10. 02 | |6 |3. 4 |20 |4. 68 |22. 35 |7. 06 |27. 4 |10. 11 | |7 |3. 49 |20. 2 |4. 72 |22. 4 |7. 14 |28. 4 |10. 21 | |8 |3. 58 |20. 4 |4. 77 |22. 5 |7. 24 |29. 4 |10. 1 | |9 |3. 66 |20. 6 |4. 84 |22. 6 |7. 44 |31. 4 |10. 46 | |10 |3. 73 |20. 8 |4. 88 |22. 7 |7. 58 |33. 4 |10. 58 | |11 | 3. 88 |21 |4. 94 |22. 8 |7. 73 |35. 4 |10. 67 | |12 |3. 91 |21. 2 |5. 02 |22. 9 |7. 89 |36. 4 |10. 75 | |13 |3. 97 |21. 4 |5. 11 |23 |8. 03 |39. 4 |10. 87 | |14 |4. 04 |21. |5. 25 |23. 1 |8. 17 |42. 4 |10. 96 | |15 |4. 11 |21. 7 |5. 32 |23. 2 |8. 38 |44. 4 |11. 02 | |16 |4. 19 |21. 8 |5. 45 |23. 3 |8. 51 | | | |16. 5 |4. 24 |21. 85 |5. 52 |23. 4 |8. 65 | | | |17 |4. 29 |21. 9 |5. 62 |23. 6 |8. 92 | | | |17. 5 |4. 34 |21. 95 |5. 71 |23. 8 |9. 9 | | | |18 |4. 4 |22 |5. 86 |24. 1 |9. 27 | | | |18. 5 |4. 45 |22. 05 |6. 1 |24. 4 |9. 39 | | | |19 |4. 52 |22. 1 |6. 4 |24. 9 |9. 62 | | | Graph 1: pH vs. NaOH Titration Curve [pic] Estimated volume of NaOH at equivalence point based on titration curve: 22. 30 mL NaOH. Table 3: ? pH/? V vs. NaOH Values |NaOH |? pH/? V |NaOH |? pH/? V |NaOH |? pH/?V |NaOH |? pH/? V | |(mL) | |(mL) | |(mL) | |(mL) | | |2 |0. 12 |19. 2 |0. 2 |22. 1 |3. 2 |24. 4 |0. 46 | |4 |0. 12 |19. 4 |0. 15 |22. 15 |3. 2 |24. 9 |0. 24 | |5 |0. 1 |19. 6 |0. 2 |22. 2 |3 |25. 4 | 0. 16 | |6 |0. 09 |19. 8 |0. 15 |22. 25 |2. 2 |25. 9 |0. 28 | |7 |0. 9 |20 |0. 2 |22. 3 |1. 6 |26. 4 |0. 12 | |8 |0. 08 |20. 2 |0. 2 |22. 35 |1. 6 |26. 9 |0. 18 | |9 |0. 07 |20. 4 |0. 35 |22. 4 |1 |27. 4 |0. 1 | |10 |0. 15 |20. 6 |0. 2 |22. 5 |2 |28. 4 |0. 1 | |11 |0. 03 |20. 8 |0. 3 |22. 6 |1. 4 |29. 4 |0. 075 | |12 |0. 06 |21 |0. |22. 7 |1. 5 |31. 4 |0. 06 | |13 |0. 07 |21. 2 |0. 45 |22. 8 |1. 6 |33. 4 |0. 045 | |14 |0. 07 |21. 4 |0. 7 |22. 9 |0. 1 |35. 4 |0. 08 | |15 |0. 08 |21. 6 |0. 7 |23 |1. 4 |36. 4 |0. 04 | |16 |0. 1 |21. 7 |1. 3 |23. 1 |2. 1 |39. 4 |0. 03 | |16. 5 |0. 1 |21. 8 |1. 4 |23. 2 |1. |42. 4 |0. 03 | |17 |0. 1 |21. 85 |2 |23. 3 |1. 4 | | | |17. 5 |0. 12 |21. 9 |1. 8 |23. 4 |1. 35 | | | |18 |0. 1 |21. 95 |3 |23. 6 |0. 85 | | | |18. 5 |0. 14 |22 |4. 8 |23. 8 |0. 3 | | | |19 |0. 1 |22. 05 |6 |24. 1 |0. 4 | | |Graph 2: ? pH/? V vs. NaOH [pic] Estimated volume of NaOH at equivalence point based on ? pH/? V vs. NaOH graph: 22. 30 mL NaOH. Table 4: Vtotal x 10-ph vs. NaOH Values |NaOH |Vtotal x 10-ph |NaOH |Vtotal x 10-ph | |(mL) | |(mL) | | |19. 8 |0. 000443 |21. 6 |0. 000121 | |20 |0. 000417 |21. 7 |0. 000104 | |20. 2 |0. 000385 |21. 8 |7. 70E-05 | |20. 4 |0. 000346 |21. 85 |6. 60E-05 | |20. 6 |0. 000298 |21. 9 |5. 0E-05 | |20. 8 |0. 000274 |21. 95 |4. 30E-05 | |21 |0. 000241 |22 |3. 00E-05 | |21. 2 |0. 000202 |22. 05 |1. 80E-05 | |21. 4 |0. 000166 | | | Graph 3: Vtotal x 10-ph vs. NaOH [pic] Estimated volume NaOH at equivalence point based on Vtotal x 10-ph vs. NaOH graph: 22. 20 mL NaOH Calculating Ka of Unknown Acid pH at ? equivalence point volume: 3. 88 Ka = 10 -3. 88 = 1. 318 x 10 -4 Ka = 1. 318 x 10-4 Calculating the Molar Mass of the Unknown Acid 0. 0981 M NaOH = moles acid / . 02330 L NaOH = 0. 023 moles acid 0. 3930 g acid / 0. 0023 moles acid = 171. 9 g/mol Analysis of Error There is a high degree of agreement among the 3 graphs and therefore a low degree of error in this experiment. According to the Graphical Analysis program, Graphs 1 and 2 indicated that the total volume of NaOH at the equivalence point was 22. 30 mL. Graph 3 indicated a volume of 22. 20 mL, a difference of 0. 1 mL. Discussion Based upon the range of possible values for Ka, anywhere from 3. 2 x 109 for Hydroiodic acid (one of the strongest) to 5. 8 x 10-10 for Boric acid (one of the weakest), this experiment’s unknown acid solution (Ka = 1. 18 x 10-4) falls roughly in the lower quarter of strength. This estimate fits its titration curve. In general, strong acids quickly go from a very low pH to a very high pH, e. g. , 2 to 12, while weak acids quickly go from a lower pH to a higher pH, e. g. , 6 to 10. The unknown solution for this experiment jump from 5 to 10 pH, which is consistent with a Ka of 1. 318 x 10-4 and a weaker acid. References Darrell D. Ebbing and Steven D Gammon, General Chemistry, 9th ed. Cengage Learning: Ohio, 2009. Department of Physical Science—Chemistry, Mesa Community College. The Ka and Molar Mass of a Monop rotic Weak Acid (handout).