Determining the percentage composition of Vinegar.

Question

In this lab, we were titrating CH₃COOH with NaOH. we used an average of 33.433 mL of 0.7 molarity NaOH to neutralize 25mL of CH₃COOH.

The questions are:

calculate the moles of NaOH from the average volume used.

Calculate the moles of acetic acid present originally.

Calculate the molarity of the acetic acid in the solution.

Calculate the mass of acetic acid in 1.00 L of solution.

Calculate the percentage of the acetic acid in the vinegar.

If you can answer any of these it would really help me in finishing the lab report with how you've done it so I can understand it. Thank you for the help

Answer 1

An important thing to know is that it takes one mole of NaOH to neutralize one mole of acetic acid (Aa).

We can use the equation Molarity=Moles/Volume, where for NaOH, this is 0.7. Since 33.433mL was used, we rearrange the equation to get moles; Moles = Molarity*Volume = 0.7*33.433*10^-3 = 0.0234 moles of NaOH. I multiplied but 10^-3 because the units for volume need to be in liters, and multiplying a volume in milliliters by 10^-3 is the conversion. Now that we have the moles of NaOH, we know that (because NaOH:Aa is 1:1) that there are also 0.0234 moles of Aa in the 25mL described in the question. Now to calculate molarity it is very simple, we just use the original equation; Molarity = Moles/Volume = 0.0234/0.025 = 0.936M. Again, the volume is multiplied by 10^-3 to convert to liters. By definition, Molarity is equivalent to moles/Liter, so in 1 liter of solution, you would have 0.936 moles of Aa. Lastly, the percentage of Aa in the vinegar is just the mass of Aa divided by mass of solution. Another way is dividing the moles of Aa by the moles of water, but this method is not used as often. You just need the mass of the 25mL solution to figure the last one out. Good luck!