Vinegar contains acetic acid, CH_3COOH. You can determine the mass of acetic aci

Question

Vinegar contains acetic acid, CH_3COOH. You can determine the mass of acetic acid in a vinegar sample by titrating with sodium hydroxide of known concentration. The reaction that occurs is CH_3COOH(aq) + NaOH(aq) rightarrow CH_3COONa(aq) + H20(I) If you find that a 25.00 mL sample of vinegar requires 41.33 mL of a 0.953 M solution of NaOH for titration to the equivalence point, how many grams of acetic acid are there in the vinegar sample? What is the molar concentration of the acetic acid in the vinegar?

Explanation / Answer

V = 25 ml of Vinegar

V2 = 41.33 ml of NaOH

M2 = 0.953 M of NaOh to equivalence

find grams of acetic acid in sample

moles of acid = moles of base

Mol of base = M*V = 41.33*0.953 = 39.39 mmol of base

therefore, there are 39.39 mmol of acid

MW of Acetic Acid = 60 g/mol

mass = mol/MW = (39.39*10^-3) / 60 = 6.56*10^-4 g of Acetic Acid

b)

Find M of HAc

M = mol/V = mmol/ml = 39.39 mmol / 25ml = 1.576 M of HAc

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