dark pisk endpoint A commercial vinegar is analyned for the percent acetic acid
ID: 544308 • Letter: D
Question
dark pisk endpoint A commercial vinegar is analyned for the percent acetic acid gresn. The ds for Tnial J Complete the table to determine tdhe percent. (See Report Sheet) Record caued significant figures in the table tekow ii values with the correct nurmsber ot A. Preparation of Vinegar Sample 1. Mass of vinegar () Part B.5 B. Analysis of Vinegar Sample 1. Buret reading, initial (mL.) 2. Buret reading. final (mL) 3. Volume of NaOH used (mL) 4. Molar concentration of NaOH - 3.20 Part B.6 solution (mo/L) 5. Moles of NaOH added (mol Show calcalation -Part B.7 6. Moles of CH,COOH in vinegar (mol) 7. Mass of CH,COOH in vinegar (e) 8. Percent by mass of CH,COOH Show calculation. Show calculation Part B.8 in vinegar (%) Show calculation 5. b. Fr Trials 2 and 3, the percent CHCOOH in vinegar was 5.01% and 4.66% respectively. a. What is the average percent of CH,COOH in the vinegar sample? Data Analysis, B. b What are the standard deviation and the relative standard deviation (%RSD) for the percent of CH 00H in the vinegar sample? Data Analysis, C and D.Explanation / Answer
You have a solution of 0.1M NaOH and 25 ml from it, the first thing to do is to calculate the number of moles that this solution contains
moles = Molarity * Volum = 0.1 * 0.025 = 0.0025 moles of NaOH, you will need 0.0025 moles of acetic acid to neutralize this, let´s turn this to grams, we need the molecular weight of acetic acid which is 60g/gmol
reaction is
CH3COOH + NaOH===== CH3COONa + H2O
moles = grams / molecular weight
grams = 0.0025 * 60 = 0.15 grams of acetic acid needed
From the statement we know that there is 1 gram for every mililiter 1 g/ml, this means that there will 1 Kg of solution in 1 liter of solution.
1 Liter = 1000 ml = 1000 total grams, 1000 total grams by 5% is 50 grams of acetic acid.
in 100 ml there will be 100 total grams, 5% of this is 5 grams of acetic acid
in 10 ml there will be 10 total grams, 5% of this is 0.5 grams of acetic acid
If we continue doing this analysis we will find that we need 3 ml of solution
3 ml = 3 total grams by 5% = 0.15 grams of acetic acid
we need 3 ml of acetic acid solution.
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