Question

given a 5.0 g sample of a vinegar having a mass percentage of acetic acid equal to 9.1%(m/m), calculate what volume of NaOH solution of molarity 0.82M will be necessary to threats this vinegar sample.

Answer 1

As we know that vinegar is a consist of an acetic acid, water and other traces of compounds. So when we will add NaOH to vinegar the acid-base reaction will occur.

Since NaOH is a strong base and acetic acid is weak acid so Strong base - weak acid reaction.

NaOH will threat completely acetic acid completely when NaOH will neutralize an acetic acid and Ph will be less than 7.

we know that

mass of acetic acid = (9.1/100)*5 = .455g

So, mole of acetic acid = 0.455g/60g = 0.00758 and  

vinegar contains mostly by acetic acid and water

therefor remaining weight = mass of water = 5 - 0.455g = 4.545g

because vinegar is a liquid solution

therefor at neutralization point

M1*V1 = M2*V2 (M1 = molarity of NaOH, M2 = Molarity of acetic acid )

( V1 = volume of NaOH solution, V2 valume of acetic acid solution)

0.82*V1 = (0.00758/V2)*V2

=> V1 = 0.00758/0.82 = 0.009243L

V1 = 9.243ml = answer