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 motivation behind this trial was to decide the pKa, Ka, and molar mass of an obscure corrosive (#14). The pKa was seen as 3. 88, the Ka was seen as 1. 318 x 10 - 4, and the molar mass was seen as 171. 9 g/mol. Acquaintance Acids contrast impressive likewise with their quality. The contrast among feeble and solid acids can be as much as 10 sets of magnitude.Strong acids separate more totally than powerless acids, which means they produce higher centralizations of the conjugate base anion (A-) and the hydronium cation (H30+) in arrangement. HA(aq) + H20 (( A-+ H3O+ With the accompanying recipe how much a corrosive separates (Ka) can be determined and given a numerical worth. Ka = [A-][H3O+]/[HA] Ka is the traditional method of estimating an acid’s quality. The reason for this test was to decide the Ka of an obscure corrosive, alongside its pKa and molar mass. E xploratory The obscure corrosive for this trial was #14.The test started with the planning and normalization of NaOH arrangement. It was determined that 2. 00 grams of NaOH pellets were expected to plan 0. 5 L of 0. 1 M NaOH arrangement. The arrangement was then normalized by directing three titration preliminaries. It was determined that 0. 7148 grams of KHP were important to kill 35 mL of the 0. 1 M NaOH. Three examples of KHP were weighed approximating this number (Table 1). Each example was blended in with 40 mL of deionized water and 2 drops of phenolphthalein in 3 Erlenmeyer carafes. Every jar was then titrated with the NaOH to a light pink endpoint.The volumes of NaOH were recorded, found the middle value of, and the normalized. The molarity of the NaOH was seen as 0. 0981. Accepting a molar mass of 100 g/mol, it was determined that 0. 3930 g of corrosive was expected to kill 40 mL of the normalized NaOH arrangement. This sum was weighed out on an electronic parity to full ac curacy and added to a clean 250 mL measuring utencil. The corrosive was first weakened with 10 mL of isopropanol and afterward 90 mL of water. A pH meter was submerged in the corrosive arrangement and an underlying pH perusing of 2. 61 was recorded.A buret loaded up with the NaOH arrangement was steadily added to the corrosive arrangement and the changing pH esteems were recorded (Table 2). As the pH meter moved toward the comparability point the measure of NaOH included each time was diminished. As the Table 2 shows, the pH rose fundamentally with the expansion of little NaOH over this interim. This data was plotted utilizing Graphical Analysis creating a titration bend chart of pH versus NaOH (Graph 1). Extra computations and charts were created to help distinguish the proportionality point: ? pH/? V versus NaOH (Graph 2) and Vtotal x 10-ph versus 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 versus 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. 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 versus NaOH Titration Curve [pic] Estimated volume of NaOH at proportionality point dependent on titration bend: 22. 30 mL NaOH. Table 3: ? pH/? V versus 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 versus NaOH [pic] Estimated volume of NaOH at identicalness point dependent on ? pH/? V versus NaOH chart: 22. 30 mL NaOH. Table 4: V total x 10-ph versus 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 versus NaOH [pic] Estimated volume NaOH at equality point dependent on Vtotal x 10-ph versus NaOH diagram: 22. 20 mL NaOH Calculating Ka of Unknown Acid pH at ? identicalness 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 corrosive/. 02330 L NaOH = 0. 023 moles corrosive 0. 3930 g corrosive/0. 0023 moles corrosive = 171. 9 g/mol Analysis of Error There is a high level of understanding among the 3 charts and along these lines a low level of blunder in this inves tigation. As indicated by the Graphical Analysis program, Graphs 1 and 2 demonstrated that the all out volume of NaOH at the proportionality point was 22. 30 mL. Diagram 3 showed a volume of 22. 20 mL, a distinction of 0. 1 mL. Conversation Based upon the scope of potential qualities for Ka, somewhere in the range of 3. 2 x 109 for Hydroiodic corrosive (one of the most grounded) to 5. 8 x 10-10 for Boric corrosive (one of the most fragile), this experiment’s obscure corrosive arrangement (Ka = 1. 18 x 10-4) falls generally in the lower quarter of solidarity. This gauge accommodates its titration bend. By and large, solid acids rapidly go from an exceptionally low pH to an extremely high pH, e. g. , 2 to 12, while feeble acids rapidly go from a lower pH to a higher pH, e. g. , 6 to 10. The obscure answer for this examination bounce from 5 to 10 pH, which is steady with a Ka of 1. 318 x 10-4 and a more fragile corrosive. References Darrell D. Ebbing and Steven D Gammon, General Chemistry, ninth ed. Cengage Learning: Ohio, 2009. Branch of Physical Scienceâ€Chemistry, Mesa Community College. The Ka and Molar Mass of a Monoprotic Weak Acid (freebee).