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PROCEDURE PART I: DILUTING THE VINEGAR SOLUTION The vinegar solution must be diluted by a factor of 5 to be suitable for titr
PART IV: TITRATING THE SAMPLES 1. Place the flask containing the measured vinegar sample with the phenolphthalein under the b
VALUE Volume original vinegar solution Volume diluted vinegar solution Dilution Factor 5 mL Molarity of NaOH 10.214 M TRIAL 1
EXPERIMENT 11 Post-Lab Questions DETERMINATION OF THE ACID CONTENT IN VINEGAR NAME DATE 1. Why doesnt it matter if water is
EXPERIMENT 11 DETERMINATION OF THE ACID CONTENT IN VINEGAR PURPOSE The purpose of this experiment is to determine the percent
After you reach the endpoint of the titration you can determine the number of moles of NaOH used from the volume of titrant n

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Answer #1

1) When the amount of vinegar is measured, we can calculate the strength using the data. No matter what amount of water we add , it won't change the amount of base needed to standardise the sample solution taken for titration. The base reacts in one to one molar ratio which means it deals with the no of moles present in the solution. The amount of water added won't change the amount of mole present.

2 ) We use pipette to measure a fixed amount of sample solution to be titrated. If it already contains some amount of water, the the actual amount of sample would be less and the concentration will be lower than usual.

3 ) The concentration of acetic acid in vinegar is too high . If the volumetric flask contains a little amount of water , then it won't change the concentration too diversely. Though, the actual concentration will be lower as the mark of volumetric flask will reach early if it contains some amount of water.

4 ) Let , the concentration of the solution is X moles per liter.

The molar mass of acetic acid is 60 g per mole. Density is 1.0052 g/mL.

So, using the formula ( grams of sample present/ grams of solution) * 100 = % (m/m)

(x * 60*100 )/(1.0052*1000) = 1.55

x = ( 1.55 * 1000 * 1.0052)/ 6000 = 0.25 M

The concentration is 0.25 M

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