posted by Sarah on .
Commercial vinegar was titrated with NaOH solution to determine the content of acetic acid, HC2H3O2. For 20.0 milliliters of the vinegar 26.7 milliliters of 0.600-molar NaOH solution was required. What was the concentration of acetic acid in the vinegar if no other acid was present?
(A) 1.60 M
(B) 0.800 M
(C) 0.600 M
(D) 0.450 M
(E) 0.200 M
I know the answer is B, but how do you get this? These types of problems completely lose me. Is a rice table necessary?
Vinegar is acetic acid, which I will write as CH3COOH. It's the right end H that reacts with the NaOH.
CH3COOH + NaOH ==> CH3COONa + HOH.
SO, mols NaOH = L x M = 0.0267 L x 0.600 M = ?? mols NaOH.
The whole idea of titration is that the mols of the titrant will tell us the amount of the unknown material in the sample.
So we have ?? mols NaOH.
You can see we replaced the 1 H on CH3COOH with OH of the NaOH, so we have 1 mol CH3COOH per 1 mol NaOH. That means we must ALSO have ?? mols CH3COOH.
Since mols = M x L, you know mols CH3COOH, you know L CH3COOH from the problem, solve for M, the molarity.
It can't get much simpler than that; you may be making it too hard. # mols = M x L is all you need to know.
Okay. I got it now. Thank you so much! :)