[ccea]

I have been having problems solving acid base titrations. My professor hasn't gotten to this topic as yet but i have a lab on titration coming up, and i need help on how to solve this


A vinegar sample was analyzed and found to be 5.88 acetic acid by volume. The density of pure acetic acid is 1.049g/mL. What volume of 0.500M of NaOH would be required to titrate 10mL aliquot of this vinegar?

I think a useful formula would be D=m/V

[Borek]

A vinegar sample was analyzed and found to be 5.88 acetic acid by volume. The density of pure acetic acid is 1.049g/mL. What volume of 0.500M of NaOH would be required to titrate 10mL aliquot of this vinegar?

This is the most stupid question I have seen lately.

Volume concentration is one of the worst ways of describing solution composition. Check volume-volume percentage discussion and what is wrong with percentages.

But that's not all. Density of the solution rarely changes linearly with concentration, but sometimes dependence is close to linear. In case of acetic acid is not only non-linear, it has a maximum. Pure water has in 20 deg C temperature density of 0.9982 g/mL, pure acetic acid has density of 1.0497 g/mL, 78% w/w acetic acid has density of 1.0700 g/mL.

All that makes calculation of the concentration very difficult - and ambiguous.

If you mix 5.88 mL of acetic acid with 94.12 mL of water you will get 99.4123 mL of solution of 1.0339 M concentration. (Note that mixing 5.88+94.12 you may expect final volume to be 100 mL - it is not due to contraction).

If you mix 5.88 mL of acetic acid with 94.71 mL of water you will get 100 mL of solution of 1.0278 M concentration.

I wonder which result (1.0339 or 1.0278) will be considered correct.

I also wonder where you are supposed to use pure acetic acid density - IMHO it is of no use here.

To check these results please download CASC - trial version has built in density table for acetic acid. Solution mixer is a tool I used when calculating above results. It doesn't support volume/volume concentrations due to their ambiguity, but it can be used with some skill to do such calculations too. Contact me off forum if you need further details.

Oh, and next time don't post such questions in inorganic chemistry forum

[mike]

First write the equation for the reaction acetic acid plus NaOH.

If you have 10mL of vinegar and you know that 5.88% of it is acetic acid work out the volume of acetic acid.

Once you know the volume of acetic acid and the density of acetic acid you can work out the mass of acetic acid that you have. (yes your equation is useful).

Convert mass of acetic acid to number of moles of acetic acid (n=m/M).

Every 1 mole of acid reacts with one mole of base so the number of moles of acetic acid must equal the required number of moles of the base (NaOH).

Now from the number of moles of the base, and the concentration of the base (given in the question) calculate the volume of base requried (v=n/c).

Hope this helps  

[mike]

This is the most stupid question I have seen lately.  

Wow, not a very science-like thing to say Borek. I thought there were no "stupid" questions in science.

I think statements like this may be a little off putting to science students.

[Borek]

Wow, not a very science-like thing to say Borek. I thought there were no "stupid" questions in science.

I think statements like this may be a little off putting to science students.

Sorry if I wasn't clear enough. I was not referring to ccea, but to the teacher - author of the original question. There are questions that you should never ask your students as they generate more confusion and lead to the knowledge that is plain wrong.

If you have 10mL of vinegar and you know that 5.88% of it is acetic acid work out the volume of acetic acid.

Once you know the volume of acetic acid and the density of acetic acid you can work out the mass of acetic acid that you have. (yes your equation is useful).

This is wrong. As I have shown in my previous post using volume/volume % you will obtain wrong value for molar concentration of acetic acid in solution. Your approach contain hidden assumption that the mixing of solutions is a linear process. It is not!

The student learns from this question: "Wow, so whenever I have this kind of solution I can just multiply volume by % and I get correct result" - and s/he is already in big troubles, as its is not true and in some cases it will create error in % range.

Note that if the question was about 5.88% weight/weight solution of 1.0067 g/mL density everything will be OK and there will be no confusion possible!

From this point of view this is stupid question. And it is very sad conclusion

[mike]

Ok that's cool, you're right it is not the student but the teacher.

Do you think that maybe you are making this answer unnecessarily confusing. At what level of chemistry would you expect a student to go into as much detail as you have, solving simple problems with software. Seems like a bit of over-kill to me.

Can you give an answer with your method? and how different is it from my method?

Mike

[Borek]

Do you think that maybe you are making this answer unnecessarily confusing. At what level of chemistry would you expect a student to go into as much detail as you have, solving simple problems with software. Seems like a bit of over-kill to me.

That's not the answer that is confusing, that's the question. Note that if the questions will be given using weight/weight percentage there is no problem with calculations and the student - regardless of the level - will be not forced to make wrong assumptions.

Can you give an answer with your method? and how different is it from my method?

How do we define preparation of solution of given volume/volume percentage?

5.88 mL of acetic acid filled up to 100 mL? If so, both results will be the same 1.0278 M, so you may argue that it doesn't matter. In a way it doesn't matter, but you have gave the student false confidence that he controls the situation. Ask him about the concentration of water in the same solution and he is dead.

5.88 mL of acetic acid mixed with 94.12 mL of water? I will get 1.0339 M, you will have the same 1.0278 M as above. 0.6% difference. For other solutions it will be more (and remember that the student is convinced that his results are correct!)

[mike]

Yes this is true, the student will have made an assumption, although this is very common in science education.

So basically if the teacher had reworded the question (ie w/w rather than v/v) there would be no problem.

Well, thankyou very much for your knowledge (this is a great forum for sharing this type of information).

Cheers mate,

Mike

[Borek]

Check links from my first post for more details about problems with different concentrations.