Determinating the percent acetic acid in vinegar using titration Essay
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1. As we add sodium hydroxide to vinegar with the phenolphthalein the solution instantly turned pink and then instantly turned colorless.
2. As we got closer to equating the moles of H+ Ions with OH- the pink would stop turning into colorless leaving a pale pink color.
In order to determine the mass of the acid, we have to calculate the moles of acetic acid used, and we can get that by multiplying the volume of sodium hydroxide used by its molarity that we got in the previous lab where we standardized a base.
In titration labs we assume that the number of moles of the base is the same as the acid’s:
1. 2. 3. Now we multiply the volume by the molarity:
1. 2. Got the molarity from the previous lab
3. Assuming that the moles of OH- = H+ we just multiply the moles by the molar mass to give the mass of acetic acid:
1. 2. 3. Now we divide the mass of acetic acid by the mass of the vinegar and then multiply it by a hundred to get the percentage.
For starters, the random error of the molarity of the previous lab is 4.77%, this will help us determine the random error of the other values we calculated in the previous part. Let us start by calculating the uncertainty in the volume:
1. 2. 3. Now we will calculate the uncertainty in the moles:
1. 2. 3. This uncertainty will be accompanied by the mass as well, because we multiplied the moles by a number that doesn’t have an uncertainty thus adding 0.
Now for the uncertainty of the final result:
1. 2. This uncertainty is confusing to put next to our result since it is in percentage too, since this 5.80% of 4.70%, I turned value to an absolute value for convenience:
So the final result will be:
According to my sources, the theoretical value of percent acetic acid is between 4% and 6%, I used the average of the values which is 5%. Now we can calculate the final percent error:
1. 2. 3.
Applying the calculations for trail 1 and trail 2, we get the following data table:
Volume used (mL ± 0.10 mL)
Moles of acetic acid
Mass of acetic acid
Percent acetic acid in vinegar
0.00290 moles (± 5.53%)
0.174 g (± 5.53%)
4.70% (± 0.273%)
0.194 g (±5.54%)
4.92% (± 0.280%)
0.173 g (± 5.53%)
4.60% (± 0.267%)
I took the average of the values and got:
With a percent error of:
1. 2. 3.
We conclude that the percent acetic acid in vinegar is 4.74% with an uncertainty of 0.273, as for the total percent error, we got 5.20% below the acceptable value which is 5%. The random error is in range with the acceptable value which means that no systematic error have taken place in this lab. I think that we’ve got good values considering the bumps we had along the road that will be discussed in the evaluation.
In this lab, we have made some assumptions that may have affected our final result. First, we found out that sodium hydroxide can react with carbon dioxide and we realized that after we conducted the experiment in an open system (where carbon dioxide is in the air), this makes sense because when sodium hydroxide reacts with carbon dioxide, it reduces the concentration thus giving us less moles thus less mass for acetic acid which leads us to a lower percentage than the accepted value and that’s what we got. We can easily fix this by sealing or insolating the reaction by simply covering the burette with a cork and by sealing the beaker that we titrated in with a plastic wrap but leaving a hole for letting the base through.
Another thing to consider is the fact the phenolphthalein might have affected the lab. Since this is a weak acid strong base titration, the end point is above 7 pH. The range at which phenolphthalein turns from colorless to pink is from 8.2-10.0. This means that the endpoint was not reached because the phenolphthalein turned pink the lower ranges. This will decrease in the value of volume used thus the moles of acetic acid, which leads to the decrease of the percent acetic acid. We can fix that by doing more trials but with different indicators each time, this will help us determine or verify out endpoints. This makes sense because in trial 2, the value of volume of base used was higher than all trials, and in that trial the solution was not paler than the others, it was violent pink, and we got a better value than the other two trials.