Question #7d6ae

1 Answer
Nov 10, 2016

The given sample of vinegar is 5.1% acetic acid by mass.

This means 100g acetic acid contains 5.1g acetic acid.

It is assumed tha density of vinegar is=1" g/mL"

So we can say

100mL or 0.1L Vinegar contains 5.1g acetic acid.

So the strength of vinegar solution is =5.1xx10=51gL^-1

Now molar mass of CH_3COOH

=2*12+2*16+4*1=60" g/mol"

So strength of the acetic acid solution

S_1=(51g/L)/(60g/"mol")=0.85M

Volume of acetic acid solution

V_1=5mL

Let this solution is nutralised by NaOH solution of strength S_2=0.22M and of Volume V_2mL

The balanced equation of nutralisation reaction is

CH_3COOH +NaOH->CH_3COONa+ H_2O

This equation reveals that the stochiometric ratio of reactants is 1:1

So this follows the relation

S_2xxV_2=S_1xxV_1

=>0.22xxV_2=0.85xx5

=>V_2=(0.85xx5)/0.22=19.3mL