Acid-Base Titration

[dinosaur93]

Just for clarification, correct me if im wrong and help me answer the following question to re-inforce our understanding..


The standard solution (which could be obtained from our primary standards) has a known concentration with an unknown volume.

while the other solution is of accurately known volume. - Placed in a burette in which the titre is obtained?

The standard solution however, has its concentration derived from the formula n = m/M_r and c =nV where V is 250cm³ of the volumetric flask. The standard solution if placed into a 20cm³ pipette (aliquot) and further transferred into a conical flask where a suitable indicator is added.



Questions:

1. Why is the aliquot placed into a conical flask instead of a beaker?

2. What happens when I still have some water left in the following apparatus. What will happen to my result? (eg. concentration too high, diluted, etc.)

a. Conical flask
b. burette
c. pipette

3. What if my burrette is dirty and is not throughly cleanse with the solution to be used for prior to the titration process?

4. Why does the titre have to be concordant, what is the main purpose of it?

5. Is poor standards of calibration a systematic or random error, Why?

6. What is the main advantage of acid-base titration compared to gravimetric, spectroscopy, chromatography? What makes it unique?

7. Does the acid always have to go to the burrette and base in the pipette?

8. Oops, A student suddenly put more than 3 drops of suitable indicator in the conical flask, How will this effect the result? (eg. higher concentration, lower concentration, etc)

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[Somye]

Hmmm I thought that generally your standard solution goes in the burette, but anyway..

1) Not entirely sure..
 
2a) Would not play an effect on the result, as the number of mol in the conical flask is fixed as there is a known volume and concentration
b) the unknown solution would be diluted and hence lead to an decreased calculation of concentration
c) The known solution would be diluted, and hence an accurate number of mol would not be present in the conical flask. This would lead to an increase in calculations as there is less of the known solution to react with

3) Refer to Q2b

4) Titres must be concordant in order to get an accurate result, as large errors may be present in non concordant titres

5)Hmmm.. not entirely sure, but I think it's a systematic error, as it's a flaw with design/instrumentation and is the same every time?

6) Cheap, easy, can be done in a basic laboratory.  Easier to have errors in gravimetric analysis I think, and its quantitative rather than qualitative, which may be better suitable for the purpose of the experiment

7) No, as I said earlier, it was my understanding that the known solution goes in burette, and the unknown in the conical flask.. But regardless, it doesn't matter, all you would need is a different indicator

8. Shouldn't make any significant effect on results

I hope this helped!

[Shenz0r]

b) the unknown solution would be diluted and hence lead to an decreased calculation of concentration

If you left water in the burette, it would lead to an increased calculation of concentration in the unknown solution. Because water was left in the burette, the standard solution will become diluted and hence more volume of the standard solution will be needed to titrate with the aliquot. You already know the concentration, if you multiply by more volume, you will get a greater number of mol at the equivalence point and hence a higher calculated concentration for your unknown solution.

[Somye]

b) the unknown solution would be diluted and hence lead to an decreased calculation of concentration

If you left water in the burette, it would lead to an increased calculation of concentration in the unknown solution. Because water was left in the burette, the standard solution will become diluted and hence more volume of the standard solution will be needed to titrate with the aliquot. You already know the concentration, if you multiply by more volume, you will get a greater number of mol at the equivalence point and hence a higher calculated concentration for your unknown solution.

Yeah, you would normally be right, but in this situation, the standard solution is in the conical flask for some reason. Hence, if more volume is required to neutralize a higher number of mol, it would lead to a decrease in the concentration of the unknown solution...

Hopefully that makes sense..

[Shenz0r]

b) the unknown solution would be diluted and hence lead to an decreased calculation of concentration

If you left water in the burette, it would lead to an increased calculation of concentration in the unknown solution. Because water was left in the burette, the standard solution will become diluted and hence more volume of the standard solution will be needed to titrate with the aliquot. You already know the concentration, if you multiply by more volume, you will get a greater number of mol at the equivalence point and hence a higher calculated concentration for your unknown solution.

Yeah, you would normally be right, but in this situation, the standard solution is in the conical flask for some reason. Hence, if more volume is required to neutralize a higher number of mol, it would lead to a decrease in the concentration of the unknown solution...

Hopefully that makes sense..

Oh yeah, missed that bit. Yeah, if you use more volume to titrate using the unknown solution, the calculated concentration will be decreased.

It's kinda like how a weak character takes 10,000 hits to kill a boss while a more powerful one can one-hit KO

[Somye]

Haha video game analogy.... Nice :p

Yeah, it is a bit weird though, why is the standard solution in the conical flask..