Chemistry question....HELP

[iLoveDivX]

The balanced equation is CH3COOH + NaOH ---> NaCH3CO2 + H2O

I know the moles of NaOH added is .0038 and I need to find out what the moles of CH3COOH (acetic acid) is. Anyone know?

This was a titration lab where I used 2.577 grams of vinegar and they need to know how much acetic acid is in vinegar.

 

[thEnEuRoMancER]

In your balanced equation one mole of NaOH reacts exactly with one mole of CH3COOH. This means your .0038 moles of NaOH reacted with .0038 moles of CH3COOH.

 

[Kelvrick]

I want to be the first to say I suck at chemistry. Good luck.

 

[iLoveDivX]

you sure neuromancer? so there's really no calculation involved since it's a 1:1 right? thanks.

 

[StevenYoo]

<< In your balanced equation one mole of NaOH reacts exactly with one mole of CH3COOH. This means your .0038 moles of NaOH reacted with .0038 moles of CH3COOH. >>



Agreed. If that's the balanced equation, then the 0.0038 moles of NaOH reacted with 0.0038 moles of the ethanoic acid. It's 0.0038 moles across the board.

-Steven

 

[iLoveDivX]

thanks a lot guys, really appreciate it.

 

[bolomite]

Wait a sec. How exactly was the titration carried out? What indicator was used? If you're saying that you added .0038 mol of NaOH to reach equivalence, then yes, there was also that amount of acetic acid present.

 

[iLoveDivX]

titration used phenolphthalein (blah, however it's spelled). but yeah, i added .0038 moles of NaOH to reach a change in color.

 

[bolomite]

OK, so now you can determine the mass of CH3COOH, then take your result and divide by 2.577 g, that result x100, to get the percentage of acetic acid in vinegar. I think store-bought vinegar is around 3-5% acetic acid.

 

[bolomite]

BTW, it comes out to 8.8% acetic acid in the vinegar.