Chemistry Acid-Base Titration

Chemistry: Strong Acid and Weak Base Titration Lab Cherno Okafor Mr. Huang SCH4U7 November 21st, 2012 Data Collection and Processing Concentration of the standard HCl solution: 0. 1 M Data Collection: | Trial 1| Trial 2| Trial 3| Final HCl Buret Reading  ± 0. 05 mL | 38. 3| 45| 54. 5| Initial HCl Buret Reading  ± 0. 05 mL | 29. 9| 38. 3| 45| Volume of NaHCO3 used  ± 0. 1 mL | 9. 2| 9. 5| 9. 8| Qualitative Data: I used the methyl orange indicator which was suitable for my titration because of its clear and distinct colour change from orange to a bright red at the endpoint * At the beginning of the titration after I added 3 drops of methyl orange into the base (NaHCO3) and swirled, I began titrating the acid (HCl) slowly, and initially in the methyl orange and base, there was a tiny amount of red colour present, but then it quickly disappeared due to insufficient HCl (H+ ions)then I gradually kept titrating more acid while swirling and there was even more red colour present, unti l 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 equivalence * Baking Soda (NaHCO3) absorbed the odour caused by the strong acid of HCl when I mixed the two: bleach-like smell ProcessingIf the concentration of an acid or base is expressed in molarity, then the volume 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 volume 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 sodium bicarbonate multiplied by its concentration in molarity is equal to the moles sodium bicarbonate. The moles of sodium hydroxide are equal to the number of moles of hydrochloric acid in the reaction.The neutralization equation becomes: HCl + NaHCO3 NaCl + H2O + CO3 Hence, Cb = Va x Ca / Vb. Trial 1 Calculation: * First 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 Experimental 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 â⠂¬â€œ 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 * Now, I will average all 3 trials: Trial 1: 0. 0913  ± 2. 78% 2. 78% / 100% 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 * Percentage 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 Conclusion and Evaluation: Conclusion: In this titration lab, I used a s trong 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 basic solution had a concentration of 0. 1  ± 0. 0005 M. In my experiment, the value I obtained was 0. 0943  ± 0. 00252 M, which is pretty close to 0. 1. I also had a very small error percentage at just 5. 7%  ± 0. 00352 error. My experimental 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 bigger titres, which would reduce errors by using larger quantities such as a higher concentration for the standardized solution. In addition, there was also some splattering/loss of the acidic solution being titrated into the basic 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 reading 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 residue 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 provide inaccuracy. Another error to mention is getting the exact endpoint during the titration. The 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 finished. In addition, I could mention that I may not have properly swirled the solutions before beginning the titration process 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 sort of flush out a large amount of the acid and therefore could have missed the neutralization point which could cause errors in results.