Concentration of acetic acid in vinegar?


Posted on : 03-04-2014 | By : My Study Coach | In : Improve Your Concentration
1 Star2 Stars3 Stars4 Stars5 Stars (No Ratings Yet)
Loading ... Loading ...

Hey, we recently did a titration lab. Im kind of stuck on one question…

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

following are the values
average volume of NaOH added to vinegar to titrate it = 8.55mL
concentration of NaOH = 1 mol/L
volume of Vinegar = 10 mL

Now i have no idea where to go from here. can anyone help me?

Related posts:

  1. How to calculate the molar concentration of acetic acid in a sample of vinegar?
  2. What was the molar concentration of the original phosphoric acid solution?
  3. how to calculate concentration?
  4. How do you determine the Concentration of a Base at an Endpoint of a titration?
  5. finding the concentration?!?

Share this :

  • Stumble upon
  • twitter

Comments (1)

Vinegar is a weak MONOBASIC acid: CH3COOH.

The equation for the neutralization is:

From the equation, 1 mole of CH3COOH is neutralized completely by 1 mole of NaOH.

Hence, in your experiment, the number of mole of NaOH used is equal to the number of mole of CH3COOH.

Here are the informations you get from your experiment:

NaOH: 1 mol/L and 8.55mL
CH3COOH: 10 mL

The number of mole of A = (concentration of A) X (Volume of A)

(1) X (8.55/1000) = ( concentration of CH3COOH ) X (10/1000)
or (1) X (8.55) = ( concentration of CH3COOH ) X (10)

The concentration of CH3COOH = 0.855 mol/L

Post a comment

Powered by Yahoo! Answers

Powered by WP Robot

Remind Me Later

Sign Up for Your FREE Study Skills Video Course here