The Ka and Molar Mass of a Monoprotic Weak Acid

The Ka and milling machinery Mass of a Monoprotic fragile Acid Chemistry Lab 152 prof James Giles November 7, 2012 Abstract The purpose of this look into was to determine the pKa, Ka, and molar flock of an unidentified acid (14). The pKa was put together 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 vast as to their strength. The difference between promiscuous and strong acids can be as much as 10 orders of magnitude.Strong acids decouple more completely than weak acids, heart and soul they produce higher concentrations of the conjugate nursing home anion (A-) and the hydronium cation (H30+) in etymon. HA(aq) + H20 (( A- + H3O+ With the fol small(a)ing formula the class to which an acid dissociates (Ka) can be compute and given a numerical value. Ka = A-H3O+ / HA Ka is the courtly way of measuring an acids strength. The purpose of this test was to determine the Ka of an unsung acid, alo ng with its pKa and molar mass. Experimental The knocked out(p)lander acid for this experiment was 14.The experiment began with the facility and standardization of NaOH resultant. It was calculated that 2. 00 grams of NaOH pellets were needed to tack together 0. 5 L of 0. 1 M NaOH radical. The solution was then standardized by conducting three titration trials. It was calculated that 0. 7148 grams of KHP were necessary to ware 35 mL of the 0. 1 M NaOH. Three takes of KHP were weighed approximating this number (Table 1). each sample was mixed with 40 mL of deionized pee and 2 drops of phenolphthalein in 3 Erlenmeyer flasks. Each flask was then titrated with the NaOH to a featherbrained pink end shoot.The messs of NaOH were record, 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 weigh ed out on an electronic balance to blanket(a) 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 molar concentration was immersed in the acid solution and an sign pH reading of 2. 61 was recorded.A burette 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 point the amount of NaOH added each clock was reduced. As the Table 2 shows, the pH rose significantly with the addition of footling NaOH over this interval. This information was plotted victimization chartical Analysis producing a titration arch graph of pH vs. NaOH (Graph 1). Additional calculations and graphs were produced to service of process identify the par 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 = inguens / plenty 0. 1 M Na OH = 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 streaks 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 bow 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 2 0. 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 in that respect is a high degree of symmetricalness among the 3 graphs and therefore a low degree of error in this experiment. tally 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 establish 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 well-nigh in the lower quarter of strength. This fancy 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 ScienceChemistry, tabular array Community College. The Ka and Molar Mass of a Monoprotic Weak Acid (handout).