posted by Jake on .
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.
Determine moles acetic acid. moles NaOH = M NaOH x L NaOH.
Moles acetic acid = moles NaOH.
M = # moles/L of solution.
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?
"The molarity of the NaOH solution is 0.160." The number in the calculations is correct.
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: