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The Concentration Of Acetic Acid

Essay add: 22-10-2015, 20:34   /   Views: 182

Aim: To determine the concentration of acetic acid by titrating against a standard solution of NaOH (0.1M). Using this to find the equivalence point, half the equivalence point and the neutralizing pH that will give the PKA value of acetic acid.

Data collection:

Table 1.1: The change in pH of 10 ml of acetic acid (unknown concentration) when sodium hydroxide (0.1M) was titrated over 3 trials.pH ± 0.01Volume of NaOH added (ml) ± 0.1mlTrial 1Trial 2Trial 31.03.932.152.922.03.722.764.203.04.553.474.054.04.373.694.485.04.633.734.406.05.083.854.607.05.483.654.598.05.643.944.349.05.483.864.4210.05.574.124.5411.05.764.334.8512.06.634.515.2813.09.064.775.3214.011.345.915.3815.011.616.275.4516.012.168.216.1117.012.438.786.3418.012.3910.136.5819.012.4410.867.1320.012.6611.2810.2621.012.6511.1511.2622.012.6911.2911.5323.012.5911.4211.8024.012.6111.3212.2225.012.6211.6812.56Qualitative observations: The solution turned slightly milky initially, but then turned colourless after about 12-13 ml of NaOH was added.

Data processing:

Graph 1.1: The volume of NaOH titrated into 10 ml acetic acid vs. the pH for trial 1.Using the above graph, we can extrapolate the midpoint of the steep rise (11ml to 15 ml). The midpoint comes to 13 ml of NaOH (0.1 M) being titrated into the acetic acid and the neutralisation pH comes to about 9.06 (verified from table 1.1).ph = 8.21, volume = 16 mlGraph 1.2: The volume of NaOH titrated into 10 ml acetic acid vs. the pH for trial 2.Similarly we can do the same with trial 2.ph = 7.13, volume = 19 mlGraph 1.3: The volume of NaOH titrated into 10 ml acetic acid vs. the pH for trial 3.Similarly we can do the same for trial 3.Table 1.2: Summary of results of the neutralizing volume of NaOH (0.1M) and the pH of neutralization for trials 1,2 and 3.Volume of NaOH (ml) ± 0.1mlpH ± 0.01Trial 1139.06Trial 2168.21Trial 3197.13Average (±0.3 ml, 0.03)168.13Thus, from table 1.2 we can see that 8.13 is the equivalence point for this reaction and therefore half the equivalence point would be 4.065.Calculations and error propagation:Average volume:(Trial 1 + Trial 2 + Trial 3)/3= (12 + 16 + 19)/3= 48/3 = 16 mlError:ΔAvg. = (ΔTrial 1 + ΔTrial 2 + ΔTrial 3) x Avg.Trial 1

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