Chemistry

Q1

In order to determine the concentration of acetic acid in a sample of vinegar, a student titrated a sample of vinegar with potassium hydroxide solution. Dada from four trials was collected and recorded in a data table.

Titiration of 10.0 mL of CH3COOH(aq) with 0.250 mol/L KOH (aq)
trial 1, 2 3 4
Fianl Reading(ml) 14.5 24.0 39.4 40.1
Intial Reading (mL) 0.5 11.5 27.0 27.8
Volume of KOH (mL) 14.0 12.5 12.4 12.3

Based on the data, it can be concluded that the concentration of vinegar is
A. 0.250 mol/L
B. 0.310 mol/L
C. 0.320 mol/L
D. 0.620 mol/L

Q2.
Use the following information to asnwer the question

Four flasks Containing 100 mL of HCL(aq)

Flask 1 = 2.5 mol/L
Flask 2= 0.20 mol
Flask 3= 200 ppm
Flask 4= 2.0g

What is the order of the flasks containg the highest to the lowest concentration of acid? Enter your answer as four digits.

_ _ _ _


Acetic acid is CH3COOH and it is the right end H that is the acidic one. I will call that H and let Ac stand for the remainder of the molecule.
HAc + KOH ==> H2O + KAc

mols KOH = L x M.
You have L and you have molarity.

Since the ratio of HAc to KOH is 1:1, then mols KOH also equals mols HAc.

Then, remembering the definition of molarity = M = mols/L, simply substitute mols and L is 10.0 mL (convert to L). You will need to calculate each of the four titrations and you will obtain four values. One of the values is not very close to the others; I don't know if you are to use statistics to see if that one is an outlier or not. I assume you will know what to do with the numbers.

Q2. You must convert each of the quantities to mols/L. The first one is done. The second one is 0.20 mol/100 mL = 0.20 mol/0.1 L = 2.0 M
The third one is 200 ppm. Convert to M.
The last one is 2.0 g. Convert to mols and divide by 0.1 L.
Post your work if you get stuck.

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  1. KOH + CH3COOH à KCOCH3 +H2O

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