Question

Speciality vinegars will have different acetic acid concentrations. A student analyzed a speciality vinegar by pipetting 10.0 mL of the vinegar into a flask. This sample was titrated with 0.128M NaOH. The titration required 20.6 mL of the base to reach the endpoint. What is the percent by mass of acetic acid in the speciality vinegar? Use the correct number of significant figures. Do not include a unit or percent sign when you type in your answer or it will be counted wrong.

Answer 1

the reaction is given by

CH3COOH + NaOH ---> CH3COONa + H20

we can see that

moles of CH3OOH present = moles of NaOH reacted

also

moles = molarity x volume (L)

so

moles of NaOH added = 20.6 x 10-3 x 0.128

moles of NaOH added = 2.6368 x 10-3

so

moles of CH3COOH present = 2.6368 x 10-3

now

mass = moles x molar mass

so

mass of CH3COOH = 2.6368 x 10-3 x 60

mass of CH3COOH = 0.158208 g

now

given

10 ml of vinegar

assuming density of vinegar as 1 g / ml

so

mass of solution = 10 g

now

% mass = mass of Ch3COOH x 100 / mass of solution

% mass = 0.158208 x 100 / 10

% mass = 1.58208

so

percent by mass of acetic acid is 1.58208