[jonnamn]

Determine the concentration of acetic acid in vinegar using a 0.110 M NaOH standard solution

I added 39.2 mL of NaOH to a flask that held 5 mL vinegar, 3 mL phenolphthalein, and 42 mL of distilled water when it turned pink.

I found the moles of acetic acid that were added = 0.892 M

But now I have to find the percent concentration of the acetic acid, and it's supposed to be between 4.0% and 5.5%

Can anyone help?

[Arkcon]

What is the formula to compute molarity?  What is the formula to compute % concentration?  How are they the same?  How are they different?  How can you use the answer to one formula to find the answer to the unknown?

Sorry to ask a bunch of questions, but that's how we help on this forum.  We give you hints, so you can get started, and figure it out for yourself.  You'll be better off, at at exam time, if you get used to conversions like this on your own.  They won't give you the same exact problem, but something similar.

[jonnamn]

The formula to figure out molarity that I used was:

39.2 mL NaOH * (0.110 mol NaOH/1000 mL) * (1/0.005 L) = 0.892 M

I don't know the equation to figure out the percentage concentration.  That's what I need help with.

[Borek]

What is the definition of the percentage concentration then?

[jonnamn]

The way the problem was stated is:

How many moles of NaOH were added? (I found that it was 0.862 M)

This value is the same number of moles of acid that were in the flask. Since we added 5mL of the original vinegar to the flask, we can find the concentration of acid in vinegar using your number of moles and 5 mL.

What is the concentration of acid in the vinegar?

[Borek]

It is obvious what the problem is, the same experiment is repeated in thousands of schools every year.

Please state the definition of percent concentration, we will start from there to find a way to calculate it.