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Question The monoprotic acidic substance in vinegar is acetic acid (H3CCO2H). When 6.00g of a certain brand of vinegar was titrated with 0.0807 M NaOH, 37.56mL of NaOH was required. What is the mass percentage of acetic acid in the vinegar? Answer First the equation for the neutralization reaction: CH3COOH(aq) + NaOH(aq) ---> CH3COONa(aq) + H2O(l) The equation wasn't really necessary if you just look at the reactants and the fact that you have a monoprotic acid and a base that reacts one to one with monoprotic acids. Still, it is always a good step to take at the beginning of a problem, and one that I personally often make mistakes on. Let us begin by determining exactly how much sodium hydroxide was required to titrate the acetic acid. To do that we take the molarity (mol/L) and multiply it by the volume of this solution used to titrate it (in liters). 0.0807 mol / L x 0.03756 L = 0.003031092 mol NaOH Because it is a one to one reaction your mols sodium hydroxide will be equal to your mols of acetic acid. So take the molar amount and multiply by the molecular weight of acetic acid to get the weight of acetic acid present: 0.003031092 mol Acetic Acid x 60.05 g / mol = 0.1820170746 g Acetic Acid Finally, with the grams of acetic acid in hand, you can figure out the mass percentage (wt vs wt) of your solution by dividing your mass of acetic acid by the total mass of the solution and multiplying by one hundred. 0.1820170746 g Acetic Acid / 6.00 g solution = 0.03034 x 100 = 3.03 % |
