What is the concentration of acetic acid in the 200 mL volumetric flask?

[nautica]

Add 200 mL of vinegar and oil dressing to a separatory funnel to separate the oil fromm the aqueous vinegar layer. When this is complete you have 100 mL of vinegar. You then put a 10 mL aliquot of this viegar into a 250 mL volumetric flask and dilute to the mark. Next, you take out a 50 mL aliquot from the 250 mL flask, add it to a 200 mL volumetric flask, and dilute to the mark. Knowing that vinegar is around 5% acetic acid (assuming 100 mL of vinegar has a mass of 100 grams, then there are roughly 5 grams of acetic acid in 100 mL solution), what is the concentration of acetic acid in units of molarity in the 200 mL volumetric flask?

I need some direction in setting up this problem.

Thanks
Nautica

btw - sorry for the long read

 

[chem_tr]

Hello, I think a standard analytical chemistry knowledge will help you about this question.

A 10 ml-aliquot diluted 25-fold is further diluted 4-fold. Therefore, a hundredfold concentration decrease should be present here. In the first solution, you state that 5 grams of acetic acid ($CH_3COOH$; M_w=24+4+32=60 gram/mol) is present, so you can find the initial and final molarity from there.

 

[nautica]

Is it that simple?

5g/60g * 0.01 L / 100 = 8 x 10^-6 M

 

[nautica]

^^^^^^^^^^^^^^^

 

[chem_tr]

Well, if you've done it correctly, then the solution is like that. It doesn't seem to be wrong; a very dilute acetic acid solution is likely in this case.

 

[thunderfvck]

Yes.

5 g/60 g/mol = 8.3 E-2 mol * 0.1 L = 8.3 E-3 mol/L

There's your c1.
v1 = 0.01 L
v2 = 0.25 L
Then it's just c2 = (c1v1)/v2

c2 = 3.3 E-4 mol/L

And then you take 50 ml's of this...

v1 = 0.05 L
c1 = 3.3 E-4 mol/L
v2 = 0.2 L

c2 = 8.3 E-5 mol/L

Our answers differ by a tenth because of the first step. In 100 g (100 ml) of acetic acid there are 5 grams. So I based my molarity on this. What you had done to get the E-6 is used the 10 ml instead of 100, which I don't think is correct since there are not 5 grams of CH3COOH in 10 ml, but in 100 ml only. So...I don't know. Make what you will of it.

 

[chem_tr]

Yes, you are right. A hundredfold concentration decrease is valid for only initial solutions. We take a 10-mL aliquot form the original 100 mL-one, so we should apply V₁*0,1 here. The second application needs to be 0,2 afterwards, since 50 mL of solution is taken from 250 mL. Thunderfvck, your approach seems to be okay, let's just check the numbers only.