The Ka and Molar Mass of a Monoprotic Weak Acid

The Ka and Molar throne of a Monoprotic short Acid Chemistry Lab 152 Professor crowd Giles November 7, 2012 Abstract The purpose of this try out was to determine the pKa, Ka, and bulwarkar mass of an unmapped savage (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. entering Acids differ considerable as to their strength. The difference between weak and square acids merchantman be as much as 10 orders of magnitude.Strong acids part more completely than weak acids, meaning they produce higher concentrations of the conflate base anion (A-) and the hydronium cation (H30+) in resolve. HA(aq) + H20 (( A- + H3O+ With the following formula the score to which an acid dissociates (Ka) can be calculated and given a numerical value. Ka = A-H3O+ / HA Ka is the conventional management of measuring an acids strength. The purpose of this experiment was to determine the Ka of an unnamed acid, along with its pKa and molar 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 accordingly standardized by conducting ternary 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 end heighten.The glitzs of NaOH were recorded, averaged, and the standardized. The molarity of the NaOH was found to be 0. 0981. presumptuous 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 step was weighed out on an electron ic balance to full precision 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 fourth dimension 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 comparability particular 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 plot using interpretical Analysis producing a titration curve graph of pH vs. NaOH (Graph 1). Additional calculations and graphs were produced to help identify the equivalence storey ? pH/? V vs. NaOH (Graph 2) and V fit x 10-ph vs. NaOH (Graph 3) Tables and CalculationsPreparation of 500 mL of 0. 1 M NaOH M = moles / tidy sum 0. 1 M NaOH = moles NaOH / 0. 5 L H20 = 0. 05 moles Na OH 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 normalization 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 Na OH 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 fracture 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. tidings 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 experiments unknown acid solution (Ka = 1. 18 x 10-4) falls just about in the lower quarter of strength. This es timate fits its titration curve. In general, strong acids chop-chop go from a truly 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 sensual ScienceChemistry, Mesa Community College. The Ka and Molar Mass of a Monoprotic Weak Acid (handout).