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The distinctive odor of vinegar is due to acetic acid, CH3COOH, which reacts with sodium hydroxide according to: CH3COOH(aq)+NaOH(aq)−→−H2O(l)+NaCH3COO(aq) If 3.45 mL of vinegar needs 42.5 mL of 0.115 M 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?

Respuesta :

Answer:

85.2g

Explanation:

Given :

Ca = ? Cb = 0.115M Va = 3.45 Vb = 42.5ml Na = 1 Nb = 1

CaVa = CbVb

Ca = ( 0.115 × 42.5)÷3.45 = 1.42M

From the relation mass concentration = molar concentration × molar mass.

Molar mass of ethanoic acid = 12 + 3(1) + 12 + 2(16) + 1 = 60g/mol

Mass concentration of ethanoic acid = 60 × 1.42 = 85.2g