The distinctive odor of vinegar is due to acetic acid, HC2H3O2. Acetic acid reacts with sodium...

The distinctive odor of vinegar is due to acetic acid, HC2H3O2. Acetic acid reacts with sodium hydroxide in the following fashion: HC2H3O2(aq) + NaOH(aq)-> H2O(l) + NaC2H3O2(aq). If 2.52 mL of vinegar requires 34.9 mL of 0.1031 M (molarity) NaOH to reach the equivalence point in a titration, how many grams of acetic acid are in a 1.00 qt sample of this vinegar? (1 L = 1.0567 qt)

Expert Solution

HC2H3O2(aq) + NaOH(aq)-> H2O(l) + NaC2H3O2

no of moles of NaOH = molarity of NaOH * volume in L

= 0.1031*0.0349 = 0.003598 moles

From balanced equation 1 mole of CH3COOH react with 1 mole of NaOH

no of moles of NaOH = no of moles of CH3COOH

no of moles of CH3COOH = 0.003598 moles

2.52 ml vinegar contains 0.003598 moles

mass of CH3COOH = no of moles * G.M.Wt

= 0.003598*60 = 0.2159 gm

= 0.2159*1000/2.52 = 85.67g/L

1 US quart = 0.95L

= 85.67*0.95 =81.38gm acetic acid ina 1qt sample of vinegar

queen_honey_blossom answered 2 years ago

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