Posted by Jake on Wednesday, March 4, 2009 at 10:23pm.
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.
Chemistry - DrBob222, Wednesday, March 4, 2009 at 11:23pm
Determine moles acetic acid. moles NaOH = M NaOH x L NaOH.
Moles acetic acid = moles NaOH.
M = # moles/L of solution.
Chemistry - Jake, Wednesday, March 4, 2009 at 11:36pm
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?
Chemistry - Jake, Wednesday, March 4, 2009 at 11:37pm
"The molarity of the NaOH solution is 0.160." The number in the calculations is correct.
Chemistry - DrBob222, Thursday, March 5, 2009 at 12:59pm
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:
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