a 2.00 mL sample of vinegar is titrated with 15.86 mL of 0.350 M NaOH to a pheno

Question

a 2.00 mL sample of vinegar is titrated with 15.86 mL of 0.350 M NaOH to a phenolphthalein end point
A) calculate the molar its of acetic acid in the vinegar solution
B) calculate the % of acetic acid in the vinegar

a 2.00 mL sample of vinegar is titrated with 15.86 mL of 0.350 M NaOH to a phenolphthalein end point
A) calculate the molar its of acetic acid in the vinegar solution
B) calculate the % of acetic acid in the vinegar

a 2.00 mL sample of vinegar is titrated with 15.86 mL of 0.350 M NaOH to a phenolphthalein end point
A) calculate the molar its of acetic acid in the vinegar solution
B) calculate the % of acetic acid in the vinegar

Explanation / Answer

The reaction between Acetic acid and sodium hydroxide is given by

CH3COOH + NaOH CH3COONa + H2O

       1mole of CH3COOH = 1mole of NaOH

CH3COOH solution                                                                NaOH solution

M1 = molarity of CH3COOH ?                            M2 = Molarity of NaOH=0.35M

V1 =Volume of CH3COOH =2 mL             V2 = Volume of NaOH used=15.86mL

n1 = 1                                                                           n2 = 1 (From equation)

                                                   

                                                        M1 =

                                                M1 =

Molarity of Aceticacid in vinegar, M1 = 2.775 M

Strength of CH3COOH solution = Molarity x Molecular mass

                                              = 2.775 X 60.06= 166.66 g/lit.

% of CH3COOH in 2.0 mL of Vinegar solution = Strength x

                                                                  = 166.66 X 0.002 X 100

% of CH3COOH in 2.0 mL of Vinegar solution = 33.33%

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