A student wanted to determine the molar concentration of acetic acid in a commer

Question

A student wanted to determine the molar concentration of acetic acid in a commercial vinegar sample. He used a volumetric pipet to deliver 3.00 mL of vinegar to an Erlenmeyer flask, added about 30 mL of water and 3 drops of phenolphthalein indicator. He performed the titration and obtained the following data:

Flask ; Initial Volume (0.0997 M NaOH) mL ; Final Volume (0.0997 M NaOH) mL
1 ; 50.00 ; 25.10
2 ; 25.10 ; 0.35
3 ; 50.00 ; 25.00

a. Determine the molarity of acetic acid in the vinegar sample.

This is what I have (And I think it's right?!) :
mean value of NaOH = 24.88ml
moles of CH3COOH = (24.88)(0.0997) = 2.48

But I need help on part B!..
b. Convert molar concentration to % Acetic acid

Explanation / Answer

let molarity of acetic acid be yM if added 3.00 ml means millimoles = 3y final volume = 33ml after 30 ml water added. [CH3COOH] = 3y/33M = 0.09yM mean value of NaOH used = 24.88ml Using, millimoles of NaOH = 24.88 x 0.0997 = 2.48 one mole of NaOH reacts with 1 mole of CH3COOH, so moles of CH3COOH = 2.48 or, 3y = 2.48 or, y = 0.83M

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