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Chemistry: Strong Acid and Weak Base Titration Lab

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Introduction

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)ï then I gradually kept titrating more 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 equivalence Baking Soda (NaHCO3) absorbed the odour caused by the strong acid of HCl when I mixed the two: bleach-like smell Processing If 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. ...read more.

Middle

Cb = (0.9 ± 1.6%) / (9.5 ± 1.05%) Cb = 0.0947 ± 2.65% ï 0.0947 ± 0.00251M is the concentration of the base for trial 2 Theoretical Base Concentration = 0.1 ± 0.0005 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% ï 0.0969 ± 0.00250M is the concentration of the base for trial 3 Theoretical Base Concentration = 0.1 ± 0.0005 M Experimental Base Concentration = 0.0969 ± 0.00250 M Now, I will average all 3 trials: Trial 1: 0.0913 ± 2.78% ï 2.78% / 100% x 0.0913 = 0.0913 ± 0.00254 M Trial 2: 0.0947 ± 2.65% ï 2.65% / 100% x 0.0947 = 0.0947 ± 0.00251 M Trial 3: 0.0969 ± 2.57% ï 2.57% / 100% x 0.0969 = 0.0969 ± 0.00250 M Therefore: (0.0913 + 0.0947 + 0.0969) ...read more.

Conclusion

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. ...read more.

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