Homework Help: chemistry

Posted by Gweedo8 on Friday, June 12, 2009 at 1:51pm.

I am trying to answer a bunch of lab questions about a lab I did and am stuck on finding the morality of the diluted vinegar solution? Can anyone help me? (I hope I gave enough info)

Here where the lab procedures that I already went through:

Transfer 10mls (Vc)of vinegar to a 100mL volumetric flask (Vd). Fill the volumetric flask to the line with water. Using a pipet, transfer 10mLs of the diluted vinegar solution to a 125mL volumetric flask. Add 25mLs of distilled water and 2 drops of phenolphthalein indicator solution. Obtain a buret and fill it with 50mLs of sodium hydroxide

When the solution turns from colorless to pink, record how much (by difference) NaOH was added. (This occurred at 26.10 mLs)

Here are the steps that I am given to help solve my Homework:

-Using the morality of NaOH determined above (This was the first part of the Homework and I already determined that the Morality of NaOH was 31.90 mLs or .0319 L) and the volume of NaOH used, find the moles of NaOH. (I took .02610/.0307=.85016)

-Convert the moles of NaOH to moles of acetic acid using the balanced equation. (The moles should be the same for both .85016)

-Divide the moles of acetic acid by the volume of solution that contained those moles of acetic acid. (This is where I am stuck)

-Use the dilution formula to find the concentrated morality of acetic acid in vinegar.

chemistry - GK, Friday, June 12, 2009 at 8:56pm
Was the purpose of the experiment to find the concentration of vinegar or the concentration of NaOH? To make any sense out of all this we must know the concentration (molarity) in moles/liter of either the acid (vinegar) or the basic (NaOH) solution. I can't find it in the information given.
chemistry - Gweedo8, Sunday, June 14, 2009 at 10:02am
The last question on my assignment is to find the % of acetic acid in the vinegar. But their was another previous lab that we did do determine the molarity of NaOH which was .0307. So I started from the point where I took 10 mLs of diluted solution and 2 drops of indicator then slowly started adding NaOH. Which I eventually added 26.10 mLs before it turned pink. So then to get the moles of NaOH I took .0261L NaOH*.0307 M NaOH=0.0008 mol of NaOH then converted those moles to an equivalent mols of Acetic acid from the balanced equation. Then I took the moles of acetic acid and divided by .035 which equaled .02289.

Does that look like it all works out?
chemistry - GK, Sunday, June 14, 2009 at 8:37pm
1) (0.0307mol/L)(0.0261) = 0.00080127 moles NaOH
moles NaOH = moles Acetic Acid = 0.00080127 mol
0.00080127 mol ac. acid / 0.01000 L = 0.080127 mol/L acetic acid (vinegar).
2) Since the titrated vinegar was diluted 10 times, the original vinegar was 10 times more concentrated: 0.80127 mol/L
3) 1 mole of CH3COOH (acetic acid) = 60.05256 g/mol
(0.80127 mol/L)(60.05256 g/mol) = 48.12 g/L
48.12 g/L --> 48.12 g/1000 mL --> 4.812 g/100 mL
4.812 g/100 mL is equivalent to ?________ %
chemistry - Gweedo8, Monday, June 15, 2009 at 6:48pm
I understand how you obtained the grams of acetic acid but how do you find the percent, I am so of acetic acid in vinegar?

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