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burtwood Offline
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Chemistry Help! Molar Concentrations! - May 27th 2009, 11:18 PM

I have to do a lab that has to be handed in soon so any help would be greatly appreciated. The question says: A sample of vinegar from the school cafeteria is diluted by a factor of 10 to make a 100.00 mL solution. The diluted solution is titrated with a standard sodium hydroxide solution using phenolphthalein is the indictator. What is the molar concentration of acetic acid in a sample of vinegar?

My oberservations are:
NaOH= 0.098 g/mol
NaOH= 60.00 mL

Vinegar = 50.00 mL

Trial 1
Final Buret Reading (mL) = 11.30
Inital Buret Reading (mL) = 2.20
Volume of HCl (aq) added (mL) = 9.10

Trial 2
Final Buret Reading (mL) = 20.30
Inital Buret Reading (mL) = 11.30
Volume of HCl (aq) added (mL) = 9.00

Trial 3
Final Buret Reading (mL) = 29.20
Inital Buret Reading (mL) = 20.30
Volume of HCl (aq) added (mL) = 9.10

Help me PLEASE!
   
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Re: Chemistry Help! Molar Concentrations! - May 28th 2009, 03:51 AM

Ok, here we go!

First of all, you know that vinegar is just a dilute solution of acetic acid, right?

Next, it's helpful to write a balanced chemical equation for the reaction of vinegar and sodium hydroxide.

CH3COOH(aq) + NaOH(aq) --> NaCH3COO + H20
An acid and a base react to form a salt (in this case, sodium acetate) and water. Just a normal neutralization reaction.

Then I'm afriad you've lost me. You added hydrochloric acid?

Usually in a titration lab, you have a set amount of solution A and you know the concentration of a solution B. The process of titration tells you how much of solution B you need to neutralize the given quantity of A. Using those three pieces of information (the quantity of A and the quantity and concentration of B) you can usually find the fourth, missing piece (usually the concentration of A). But here you seem to have three solutions. I'm not really sure what to do with that...
Are you sure it's volume of HCl added, not volume of CH3COOH?


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