mols NaOH used = M x L = estimated 0.0028 but you need a more accurate answer.
Convert mols NaOH to mols acetic acid and that is 1:1; therefore, mols acetic acid (vinegar) = estimated 0.0028.
g acetic acid = mols x molar mass = approx 0.0028 x about 60 = about 0.17g.
So you had about 0.17g vinegar in 50 mL or about 0.34g in 100 mL and that is % w/v or about 0.34%.
Note: commercial vinegar is about 3-5% w/v.
a 50ml aliquot sample of vinegar is titrated with a 0.078 M NaOH sol'n. The phenolpthalein end point was reached after the addition of 35.6 ml.
Equation:
HC2H3O2 + NaOH _____> NaC2H3O2 + H20
Calculate the % by mass/vol. of acetic acid in the sample.
1 answer
1 answer