Calculate the molar concentration of acetic acid (CH3COOH) in a 5.00-mL sample of vinegar (density is 1.00 g/mL) if it is titrated with 25.00 mL of NaOH.

Answer 1

Determine moles acetic acid. moles NaOH = M NaOH x L NaOH.
Moles acetic acid = moles NaOH.

M = # moles/L of solution.

Answer 2

Thank you very much.

The molarity of the NaOH solution is 0.162.

So I did the following:

25.00 mL NaOH*(1 L / 1000 mL)*(0.160 mol NaOH / 1 L NaOH)*(1 mol CH3COOH / 1 mol NaOH) = 0.004 mol CH3COOH

(0.004 mol CH3COOH / 5.00 mL) * (1000 mL / 1 L) = 0.8 M CH3COOH

Now I'm supposed to determine the mass percent of the acetic acid in the vinegar sample. I assume this is why I was given the density...

how do I use the concentration of acetic acid and the density of vinegar to determine the mass percent?

Answer 3

typo...
"The molarity of the NaOH solution is 0.160." The number in the calculations is correct.

Answer 4

moles CH3COOH x molar mass acetic acid = grams CH3COOH.
(g CH3COOH/100 g soln)*100 = mass percent. which you can do one of two ways.
I have 0.24 g/5 g so how much is in 100?
That will be 0.24 x 100/5 = 0.24 x 20 = 4.8 which is 4.8%.
The other way, I think, is simpler:
(0.24/5)*100 =