Hey everyone, just had a quick one about some titration theory.
"A student conducts an experiment to determine the ethanoic acid concentration in a commercial
brand of vinegar. An outline of her procedure and her measurements are provided below.
230.0 mL deionised water added to a 250 mL volumetric flask
20.0 mL of vinegar added to the flask
20.00 mL aliquots added to flasks and titrated against 0.150 M NaOH."
The solutions said that an error in this experimental design was that "20 mL added to 230 mL of water will not necessarily give 250 mL. Some liquids are miscible in each other. All concentration calculations will be subsequently affected," and suggested that "It should be made up to the mark and not have 20 mL added to 230 mL."
Would miscibility not result in an error regardless if we're filling the volumetric flask up to the mark or adding 20 to 230? Cheers
Hi saransh,
I think the main error described here is that the vinegar to the volumetric flask before the deionised water.
The analyte (vinegar) should always be added first to ensure that the correct concentration is achieved. When deionised water is added directly to the vinegar already in the flask, up to the \(250 mL\) mark, you know that the volume totals as close to \(250 mL\) as possible, and that the \(n(vinegar)\) is as close to \(20 mL\) as possible.
If the deionised water is added to the volumetric flask first, then the separately measured \(20 mL\) of vinegar is added to the deionised water, you risk decreasing accuracy. As the uncertainty of the volumetric flask and the pipette used to measure the vinegar might result in inaccurate measurements, separately measuring the deionised water and vinegar and then adding them together combines the potential for error from two pieces of equipment. That is: when vinegar is added first, although inaccuracy can still come from the uncertainty of the \(n(vinegar)\) measurement, the total \(v\) wouldn't be affected as much.
If the deionised water is added first, both the vinegar and water would be more likely to be measured inaccurately, compounding the error. I'll list the possible errors from measurement in each case here:
Water after vinegarHigh vinegar: less water added to achieve \(v=250 mL\), inaccurately high \([CH_3COOH]\)
Low vinegar: more water added to acheive \(v=250 mL\), inaccurately low \([CH_3COOH]\)
Vinegar after waterLow water/low vinegar: \(v<250 mL\), inaccurately low \([CH_3COOH]\) calculated as \(v=250 mL\) will be used
Low water/high vinegar: \(v(solution)\approx\ 250 mL\), aliquots have higher \([CH_3COOH]\), high \([CH_3COOH]\) calculated
High water/low vinegar: \(v(solution)\approx\ 250 mL\), aliquots have lower \([CH_3COOH]\), low \([CH_3COOH]\) calculated
High water/high vinegar: \(v>250 mL\), high \([CH_3COOH]\) calculated as \(v=250 mL\) will be used
High water/accurate vinegar: low \([CH_3COOH]\) from \(v>250 mL\) and aliquot \([CH_3COOH]<\) actual
Low water/accurate vinegar: high \([CH_3COOH]\) from \(v<250 mL\) and aliquot \([CH_3COOH]>\) actual
Accurate water/high vinegar: high \([CH_3COOH]\) from \(v>250 mL\) and aliquot \([CH_3COOH]>\) actual
Accurate water/low vinegar: low \([CH_3COOH]\) from \(v<250mL\) and aliquot \([CH_3COOH]<\) actual
As you can see, measuring the deionised water and vinegar separately, then adding the water first creates many more potential errors than adding the vinegar first and topping up the volume with water accordingly. This makes vinegar first, water second a more accurate approach.
Other errorsI don't see an error that would result from the miscibility of liquids. If anything, I would've thought that the analyte and titrant being immiscible would pose a problem, as this suggests that they would not react with each other and sit in the conical flask as a heterogenous solution.
Some other errors you could discuss, in case you need more ideas:
- No mention of equipment rinsing or swirling the solution to dissolve particles
- Indicator is not added to the analyte prior to titration
- The phrasing used is 'aliquots added to flasks and titrated
against \(0.150M NaOH\)', which falsely implies that this is a back titration
- No mention of the titration process beyond 'titrate vinegar': should specify that three concordant titres must be achieved/results must be recorded (precision)
Anyways, that's what I think is going on in this question; it does seem a little strange. Hope I could help, let me know if there are any errors (see what I did there)