[millerst]

Hello,

I've been doing this question for a while now and I can't seem to get the correct answer... 

The question is: Calculate the concentration of acetic acid in your vinegar sample. Use the average volume and concentration of sodium hydroxide, and the volume of vinegar.

CH3COOH + NaOH --> CH3COONa  +  H20

Average sodium hydroxide: 17.8 mL or 0.0178 L
Concentration of Sodium hydroxide: 0.25 mol/L
Volume of vingear: 5 mL or 0.005 L

C = (N)(V)
What do I do first? I know I need the amount of solute (in mol), in this case vinegar...but to do this I need to find the number of moles of sodium hydroxide?

N = C/V = 0.25 mol/L/0.0178L = 14.0 mol NaOH?

Since a 1:1 ratio, therefore the concentration of CH3COOH is 14.0 mol? I do not believe this is correct, seems a little large of an amount…plus it doesn’t compare close to the rest of the class. If someone could help me I’d appreciate it.

[enahs]

For every mole of Sodium Hydroxide reacted there is one mole of Acetic Acid. How many moles of Sodium Hydroxide was that? That is the number of moles of Acetic Acid. That many moles of Acetic Acid was in 5mL, so the concentration of that Acetic Acid was?

[allanf]

Actually N=C*V,

So the number of moles is: .25 mol/L * .0178L or about .0045 moles.

In your original post you divided the two instead of multiplying, that's where the error came from.

[millerst]

So if there is .0045 moles of Sodium Hydroxide then there is .0045 moles of Vinegar?

So then I can use C = N/V to find its concentration?

C = .0045 / 0.005   = 0.9

So the concentration of acetic acid is 0.9 mol/L?

[allanf]

Yes to all three.

[millerst]

What would I do to find the mass of acid in the volume of vinegar used (5 mL)?

I know I first need to find the molar mass of CH3COOH which is 60 g/mol, but after that I'm lost...

[enahs]

How many moles of acid was in that volume of solution? That times the molecular weight = mass of acid in solution.