Chemistry Homework

Question

A student titrates 2.025 g of vinegar with 0.121 M NaOH. It takes 14.01 mL of NaOH to reach endpoint. What is the mass percent acetic acid in the vinegar sample? Show your calculations below. 

Answer 1

The question asks for mass % of acetic acid. This means how many grams of acetic acid are in 100 g of solution.

Let's first find the moles of acetic acid (CH3COOH) present:

CH3COOH + NaOH ==> CH3COONa + H2O

mols NaOH used = 14.01 mls x 1 L/1000 mls x 0.121 mol/L = 0.001695 mols NaOH

mols CH3COOH present = 0.001695 b/c the reaction above is 1:1 mol ratio of NaOH : CH3COOH

Now, we find the mass of CH3COOH:

0.001695 mols CH3COOH x 60.05 g /mol = 0.1018 g CH3COOH

mass % = 0.1018 g / 2.025 g (x100%) = 5.03 % (3 sig. figs.)