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 1
Theoretical 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 – 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 ± 0.0005 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% → 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% → 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) M→Average 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%
= (0.057 ± 0.00352) x 100% = 5.7% ± 0.00352
Conclusion 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 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 (0.0005 + 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.
by
cherno (student)
