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.

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.

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?

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 ?________ %

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?

To find the molarity of the diluted vinegar solution, you need to follow these steps:

1. Calculate the moles of NaOH used: Since you determined that the volume of NaOH used is 26.10 mL (or 0.02610 L) and the molarity of NaOH is 31.90 mL (or 0.0319 L), you can calculate the moles of NaOH by multiplying the volume (in liters) by the molarity. In this case, it would be 0.02610 L * 0.0319 mol/L = 0.00083169 mol NaOH.

2. Convert the moles of NaOH to moles of acetic acid: Since the reaction between NaOH and acetic acid is a 1:1 ratio according to the balanced equation, the moles of NaOH will be equal to the moles of acetic acid. Therefore, the moles of acetic acid in the diluted vinegar solution would also be 0.00083169 mol.

3. Divide the moles of acetic acid by the volume of solution that contained those moles of acetic acid: From the lab procedure, you transferred 10 mL of the diluted vinegar solution to a 125 mL volumetric flask, so the volume of the solution containing the moles of acetic acid would be 10 mL (or 0.010 L). Divide the moles of acetic acid (0.00083169 mol) by the volume of the solution (0.010 L) to get the molarity: 0.00083169 mol / 0.010 L = 0.08317 M.

4. Use the dilution formula to find the concentrated molarity of acetic acid in vinegar: Since you initially diluted 10 mL of vinegar to a 100 mL volumetric flask, the diluted vinegar solution has a 10-fold dilution factor. Multiply the molarity of the diluted vinegar solution (0.08317 M) by the dilution factor (10) to obtain the concentrated molarity of acetic acid in vinegar: 0.08317 M * 10 = 0.8317 M.

Therefore, the molarity of the diluted vinegar solution is 0.08317 M, and the concentrated molarity of acetic acid in vinegar is 0.8317 M.