[psychoman]

A vinegar sample was analyzed and found to be 5.88% acetic acid by volume. The density of pure acetic acid is 1.046 g/ml. What volume of 0.200 M NaOH would be required to titrate a 2 ml aliquot of this vinegar??   Can anyone help me?. I have never heard of titration .=)

[Borek]

http://www.titrations.info/

[psychoman]

ok, so I attempted to answer the question and got an answer of 9.26 mL of 0.200 M NaOH.....can anyone confirm this answer(Is it right???) pls...

[Borek]

Show how you got there.

[psychoman]

first off I found the number of moles for acetic acid which I got as 0.116 mol. Next I determined the volume of NaOH by converting 2 mL of acetic acid into litres (0.002 L) then multiplied that by the number of moles of acetic acid, and finally multiplied that value (0.232 mol) by the molar mass of NaOH.....wait a minute!!!!!!!! I think I found the mass not the volume, s**t.   NOW WAT DO I DO?

[Borek]

first off I found the number of moles for acetic acid which I got as 0.116 mol

How.

Note, that once you know number of moles of acetic acid, you are almost there - thats what you need to calculate stoichiometric amount of base.

[psychoman]

I used the given info about 0.002 L of acetic acid, then I found the molar mass of acetic acid from the periodic table and calculated the mol of acetic acid. So how do i go about finishing the problem

[lamnesia]

hey psycho. Well i think i know where you got the question from and i'm also working on the same question for my lab.

I did the question and got an answer but i'm not sure if it is correct.. heres how i did it, i'll write it step by step.

1. Equation- CH3COOH + NaOH = CH3COONa + H2O
2. n=C x V
    since, n (NaOH) = n (CH3COOH)
    therefore, C(NaOH) x V(NaOH) = C(CH3COOH) x V(CH3COOH)
3. V(CH3COOH) = 0.0588 x 2mL = 0.1176mL
    Mass(CH3COOH) = 0.1176mL x 1.049g/mL = 0.1233624g
    Molar Weight of CH3COOH is 60.062g/mol
    C(CH3COOH) = 0.1233624g / 60.062g/mol = 0.00205392 mol/mL

 (0.1176mL x 0.00205392mol/mL) / 0.0002mol/mL = 1.2 mL


my answer seems abit off, thats why i'm asking for help.

[Borek]

I used the given info about 0.002 L of acetic acid, then I found the molar mass of acetic acid from the periodic table and calculated the mol of acetic acid.

Sorry, but the way you have presented your approach doesn't say a word about what you really did - and the amount of acid you have calculated is wrong.

[Borek]

C(NaOH) x V(NaOH) = C(CH3COOH) x V(CH3COOH)

Correct in general - note that product on the right is just a number of moles

C(CH3COOH) = 0.1233624g / 60.062g/mol = 0.00205392 mol/mL

Partially correct. Where does mL in the result comes from?

That is - it is as correct as it can be given type of information you were shown, that's not correct in gereneral, but that's not your fault. Whoever asked the question should think twice.

(0.1176mL x 0.00205392mol/mL) / 0.0002mol/mL = 1.2 mL

That's where you have gone awry, but that's effect of earlier mistake. 0.00205392 is not concentration.

[psychoman]

I'm sorry but we never used these formulas in high school as we never got to them and even learned...aside from density(back in grade 9 only =)...)   Pls can u tell us how to answer it and wat the final answer should be?

[Borek]

Some basic things:

http://www.chembuddy.com/?left=concentration&right=toc

http://www.chembuddy.com/?left=balancing-stoichiometry&right=toc

But if you need to go deeper (that is moles/masses) you will have to look somewhere else.

[lucas89]

May I ask why you're doing this problem? Is it out of your own self-interest? Because I find it quite weird that any teacher would assign a problem involving a titration without first showing their students the very most basic chemistry formulae.

[JGK]

I also find it unusual that someone at undergraduate chemistry level has never heard of titration