The reaction between Acetic acid and sodium hydroxide is given
by
CH3COOH + NaOH ⟶
CH3COONa + H2O
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1mole of CH3COOH = 1mole of NaOH
CH3COOH solution                                           Â
                 Â
NaOH solution
M1 = molarity of CH3COOH
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M2 = Molarity of NaOH=0.35M
V1 =Volume of CH3COOH =2 mL
            V2
= Volume of NaOH used=15.86mL
n1 =
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n2 = 1 (From equation)
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                              M1
=
                                              Â
M1 =
Molarity of Aceticacid in vinegar, M1 = 2.775 M
Strength of CH3COOH solution = Molarity x Molecular
mass
                                    Â
       = 2.775 X 60.06= 166.66
g/lit.
% of CH3COOH in 2.0 mL of Vinegar solution = Strength
x
                                                        Â
       = 166.66 X 0.002 X
100
% of CH3COOH
in 2.0 mL of Vinegar solution = 33.33%