Help~!! Vinegar analysis

Question

1. Assuming the density of a 5�etic acid solution is 1.0 g/ml, determine the volume of the acetic acid solution necessary to neutralize 25.0 ml of 0.10 M NaOH.

2. A 31.42ml volume of 0.108 M NaOH is required to reach the phenolphthalein endpoint in the titration of a 4.441g sample of vineger. Calcuate the percent acetic acid in the vinegar.

Answer 1

1. Assuming the density of a 5etic acid solution is 1.0 g/ml, determine the volume of the acetic acid solution necessary to neutralize 25.0 ml of 0.10 M NaOH.

find moles of NaOH :

0.025 litres @ 0.10 moles/litre = 0.0025 moles NaOH

since it takes 1 mole of acetic acid to neutralize 1 mole of NaOH, we need 0.0025 moles of acetic acid, but how many grams is that,... use molar mass:

0.0025 grams @ 60 grams / mole = 0.15 grams of acetic acid

use the % by weight to find the weight of solution that has 0.015 grams of acid:

0.15 g acid @ 100g solution / 5 grams acid = 3.0 grams of solution is required

use density to find thw volume of 3.0 grams of solution:

3.0 g @ 1.g/ml = 3.0 ml of solution

Your answer is: 3.0 mls of vinegar

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2. A 31.42ml volume of 0.108 M NaOH is required to reach the phenolphthalein endpoint in the titration of a 4.441g sample of vineger. Calcuate the percent acetic acid in the vinegar.

find moles of NaOH:

0.03142 litres @ 0.108 mol/litre = 0.00339 moles NaOH

0.00339 moles NaOH--> 1:1 ratio --> 0.00339 moles acetic acid

find grams of acid:

0.00339 moles @ 60.0 g / mol = 0.2036 grams of acid

find %:

0.2036 g acid / 4.441 g sample (100) = 4.58% acid

your answer 4.58% acid