given a 5.0 g sample of a vine

given a 5.0 g sample of a vinegar having a masspercentage of acetic acid equal to 9.1%(m/m), calculate what volumeof NaOH solution of molarity 0.82M will be necessary to threatsthis vinegar sample.

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

Since NaOH is a strong base and acetic acid is weak acid soStrong base – weak acid reaction.

NaOH will threat completely acetic acid completely when NaOHwill 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 = 0.00758and

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 aceticacid )

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

0.82*V1 = (0.00758/V2)*V2

=> V1 = 0.00758/0.82 = 0.009243L

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