Chemistry Acid-Base Titration

Chemistry Strong Acid and Weak bottom Titration Lab Cherno Okafor Mr. Huang SCH4U7 November 21st, 2012 information Collection and Processing tightness of the standard HCl radical 0. 1 M data Collection effort 1 Trial 2 Trial 3 Final HCl Buret recitation 0. 05 mL 38. 3 45 54. 5 Initial HCl Buret Reading 0. 05 mL 29. 9 38. 3 45 Volume of NaHCO3 use 0. 1 mL 9. 2 9. 5 9. 8 Qualitative Data I used the methyl orangeness index which was suitable for my titration because of its clear and distinct colouring material deepen from orange to a bright ruby at the endpoint * At the beginning of the titration afterward I added 3 drops of methyl orange into the fore (NaHCO3) and swirled, I began titrating the caustic (HCl) slowly, and initi onlyy in the methyl orange and base, there was a exact amount of red colour present, but then it quickly disappeared overdue to insufficient HCl (H+ ions)then I gradually kept titrating more than acid while swirling and there was even more red colour present, until finally I reached the endpoint when the orange-yellow colour had completely transformed into a red colour * Changes from an orange-yellow colour (slightly higher pH 4. 4) to a bright red colour (at low pH 3. 1) at the endpoint and point of compare * Baking Soda (NaHCO3) absorbed the odour caused by the rugged acid of HCl when I mixed the two bleach-like smell ProcessingIf the concentration of an acid or base is verbalised in molarity, then the slew of the solution multiplied by its concentration is equal to the moles of the acid or base. Therefore, the following relationship holds nVb x Cb = Va x Ca Where Vb = the spate of the base Cb = the concentration of the base Va = the volume of the acid Ca = the concentration of the acid n = the mole factor In the case of hydrochloric acid and Sodium Bicarbonate (Baking Soda), the mole ratio is one to one, thus the mole factor is 1. Therefore, the volume of atomic number 11 bicarbonate multiplied by its concen tration in molarity is equal to the moles sodium bicarbonate. The moles of sodium hydrated oxide are equal to the number of moles of hydrochloric acid in the reaction.The neutralisation reaction equation becomes HCl + NaHCO3 NaCl + H2O + CO3 Hence, Cb = Va x Ca / Vb. Trial 1 Calculation * first-class honours degree we need to find the change of volume of the acid used up in the titration Va = Vfinal Vintial Va = 38. 3 0. 05 29. 9 0. 05 Va = 8. 4 0. 1 mL Therefore, nVb x Cb = Va x Ca (1)(9. 2 0. 1)(Cb) = (8. 4 0. 1) (0. 1 0. 0005) Cb = (8. 4 0. 1) (0. 1 0. 0005) / (9. 2 0. 1) Cb = (8. 4 1. 19%) (0. 1 0. 5%) / (9. 2 1. 09%) Cb = (0. 84 1. 69%) / (9. 2 1. 09%) Cb = 0. 0913 2. 78% 0. 0913 0. 00254M is the concentration of the base for trial 1Theoretical Base Concentration = 0. 1 0. 0005 M data-based Base Concentration = 0. 0913 0. 00254 M Trial 2 Calculation * First find change of volume of the acid used up in the titration Va = Vfinal Vinitial Va = 45 0. 05 36 0. 05 Va = 9. 0 0. 1 mL Therefore, nVb x Cb = Va x Ca (1)(9. 5 0. 1)(Cb) = (9. 0 0. 1) (0. 1 0. 0005) Cb = (9. 0 0. 1) (0. 1 0. 0005) / (9. 5 0. 1) Cb = (9. 0 1. 1%) (0. 1 0. 5%) / (9. 5 1. 05%) Cb = (0. 9 1. 6%) / (9. 5 1. 05%) Cb = 0. 0947 2. 65% 0. 0947 0. 00251M is the concentration of the base for trial 2 Theoretical Base Concentration = 0. 1 0. 005 M Experimental Base Concentration = 0. 0947 0. 00251 M Trial 3 Calculation * First find change of volume of the acid used up in the titration Va = Vfinal Vinitial Va = 54. 5 0. 05 45 0. 05 Va = 9. 5 0. 1 mL Therefore, nVb x Cb = Va x Ca (1)(9. 8 0. 1)(Cb) = (9. 5 0. 1) (0. 1 0. 0005) Cb = (9. 5 0. 1) (0. 1 0. 0005) / (9. 8 0. 1) Cb = (9. 5 1. 05%) (0. 1 0. 5%) / (9. 8 1. 02%) Cb = (0. 95 1. 55%) / (9. 8 1. 02%) Cb = 0. 0969 2. 57% 0. 0969 0. 00250M is the concentration of the base for trial 3 Theoretical Base Concentration = 0. 1 0. 0005 MExperimental Base Concentration = 0. 0969 0. 00250 M * No w, I will comely all 3 trials Trial 1 0. 0913 2. 78% 2. 78% / century% x 0. 0913 = 0. 0913 0. 00254 M Trial 2 0. 0947 2. 65% 2. 65% / 100% x 0. 0947 = 0. 0947 0. 00251 M Trial 3 0. 0969 2. 57% 2. 57% / 100% x 0. 0969 = 0. 0969 0. 00250 M Therefore (0. 0913 + 0. 0947 + 0. 0969) (0. 00254 + 0. 00251 + 0. 00250) / 3 trials = (0. 2829 0. 00755) / 3 = (0. 0943 0. 00252) MAverage Concentration of base for the 3 trials * character Error = Theoretical Actual / Theoretical x 100% = (0. 1 0. 0005) (0. 0943 0. 00252) / (0. 1 0. 0005) x 100% = 0. 0057 0. 00302 / 0. 1 0. 0005) x 100% = (0. 57 0. 00352) x 100% = 5. 7% 0. 00352 end and Evaluation Conclusion In this titration lab, I used a strong acid HCl (hydrochloric acid) vs. a weak base NaHCO3 (sodium bicarbonate/baking soda). My intent was to find the concentration of the weak base after it has been titrated with the strong acid. The theoretical staple fibre solution had a concentration of 0. 1 0. 0005 M. In my experimen t, the value I obtained was 0. 0943 0. 00252 M, which is pretty close to 0. 1. I also had a very humble error percentage at just 5. 7% 0. 00352 error. My observational value was only off by 0. 0057 (0. 1- 0. 0943) with a total uncertainty of 0. 00402 (0. 005 + 0. 00352) from the theoretical value of the base concentration. Evaluation/Improvement Some of the most notable errors in my procedure to mention are the small quantities being used and hence the inaccuracy in measurements. Perhaps I could have arranged the titration to have big titres, which would reduce errors by using larger quantities such as a higher concentration for the standardized solution. In addition, there was also close to splattering/loss of the acidic solution being titrated into the canonic solution, as it came into contact with the edges and surface of the flask, which in turn, presumably initiated errors in volume measurements.Also, this means that not all of the acid that was added reacted efficiently with the basic solution mixed with methyl orange indicator. Moreover, there could have been impurities in the basic solution itself and as well as the indicator causing a higher culture than the theoretical value of concentration. The leakage that resulted from the stock cock may have caused the HCl to alter slightly in content because of the reaction with some of the chemicals in the external environment (air). There was also some eternal sleep that could have been left behind in the basic flask when I washed it with distilled water after the neutralization of each trial. Perhaps drying it could have made a difference instead of leaving it wet.Maybe some of the neutralized solution was left behind after I washed out the flask, and it mixed with the tiny water droplets also left behind in the flask. Before I started the next trial, it could have interfered with that titration and generate inaccuracy. Another error to mention is getting the exact endpoint during the titration. Th e indicator could have ranged from different shades of red (starting with orange) but I assumed that the moment it turned a standard red colour, it was civilisationed. In addition, I could mention that I may not have properly swirled the solutions before beginning the titration exhibit to make sure nothing (residue) settles at the bottom.This could have impacted the inaccurate colour change of the indicator in the neutralization and hence unknown standard colour. I also kept on adding drops when the solution was already a red colour towards the end. However, this may have either darkened or lightened the colour too much in an effort to change the precision of the indicator colour at the equivalence point or end point. Finally, at some moments, I was in a hurry to finish titrating, and so I may have flushed out the acid in large amounts. I realize that near the neutralization point, the acid must be released in drops. However, for the third trial, I did grade of flush out a large amount of the acid and therefore could have missed the neutralization point which could cause errors in results.