In: Chemistry
Questions asked after completing quantitative volumetric analysis of pickling vinegar & NaOH (regular vinegar consists 5% acetic acid, CH3COOH. Piclking vinegar consists 7% of the acid).
1. Is there a trend in the values obtained in class? (there was a trend) Explain.
2. If regular vinegar costs $1.00/L & pickling vinegar costs $1.25/L, which is the best by? (i.e use a logical argument to determine which gives you the most acetic acid for your dollar)
3. Why do we only include the trials that agree withing 1% of each other? What is the difference between accuracy & precision for a series of measurements?
4. Identify 4 important sources of systematic (not human) error. Discuss why they are important.
5. If 0.25M potassium hydroxide solution was used as the titrant, instead of 0.1 M NaOH (aq), would your titration have use dmore or less itrant to reach the end point? Use a logical argument to determine how much more or less titrant would be required.
2. 5% CH3COOH means 5g in 100 g water, so 1L or 1000g of water will contain regular Vinegar = 50 g (Acetic Acid).
Where as 1L or 1000g of water, 7% pickling vinegar = 70 g acetic acid
So, per g price =$ 1/50 for 5% = $0.02 and $1.25/70 for 7% = $0.018
Thus 7% acetic acid or pickling vinegar is cheaper.
3. When we use significant figures then an implicit uncertainty is considered, the last digit is considered uncertain. For example, a result reported as 1.15 implies a minimum uncertainty of ±0.01 or 1% and the value lies in between 1.14 to 1.16.
The accuracy of a set of observations is the variance between the average of the measured values and the true value of the observed quantity. The precision of a set of measurements is a measure of the range of values found, that is, of the reproducibility of the measurements.
Accuracy and precision are related and this is illustrated by an example of firing a rifle at a target where the blue dots below represent hits on the target:
4.
5. From acidimetry and alkalimetry and law of equivalence, we know that
V1S1 = V2S2
In this case V2S2 is fixed and let us consider S1=0.1M NaOH,
Then considering, V1, volume of NaOH = V2S2/0.1 mL
If we consider S3 = 0.25 KOH and its volume = V3
Then V3 = V2S2/0.25
Thus the ratio V3/V1=0.1/0.25 =0.4 or V3 = 0.4xV1
The volume of the titrant of 0.25 M KOH will 0.4 times of the voulme of 0.1M NaOH solution.
1. Information not availabale.