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. Maybe Idk
What would be the ph of a 0.20M CH3COOH solution if no salt were present?
2 answers
2.27