1. Assume the vinegar is 5.00 % acetic acid by mass and the density of vinegar i
ID: 1034529 • Letter: 1
Question
1. Assume the vinegar is 5.00 % acetic acid by mass and the density of vinegar is 1.00 g/mL. Assume you have a 0.2000 M NaOH solution. Calculate the volume (in mL) of NaOH needed to titrate 5.00 mL of vinegar. Show your work and report your answer to correct significant figures.
2. In your own words, write a step-by-step experimental procedure for determining the % acetic acid in vinegar. The procedure you write must include sufficient detail so that anyone reading it will know exactly what to do in the lab
Explanation / Answer
Given
5 % acetic acid by mass
Given
volume of vineagar = 5 ml
mass of vinegar = 5 ml * 1 g/ml = 5 g
mass of acetic acid = (0.05 g / 1g of vinegar ) * 5 g of vinegar = 0.25 g of acetic acid
Molar mass of acetic acid = 60 g/mol
No. of moles = 0.25 g / 60 g/mol = 0.00417 moles
CH3COOH (aq) + NaOH (aq) --> CH3COONa (aq) + H2O (l)
0.00417 moles of acetic acid require 0.00417 moles of NaOH
Given Molarity = 0.2 mol/L
Volume = No. of moles / Molarity = 0.00417 moles / 0.2 mol/L = 0.0208 L = 20.8 ml
Answer 20.8 ml of NaOH
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