Commercial vinegar contains a small percent
/nsg 9 .x3sjtdage of ethanoic acid. A student wishes to determine the concentration of ethanoic acid in vinegar. She titrates a solution of ethanoic acid against a
j gts/nd s3.9xstandard solution of potassium hydroxide of concentration $8.0\times10^{-4}$mol$dm^{-3}$.The equation for the reaction is $CH_3$$COOH_{(aq)}$+$KOH_{(aq)}$$\rightarrow$$CH_3$$COOK_{(aq)}$+$H_2O$(l)
A volume of 25.0$cm^3$ of vinegar was diluted with distilled water and made up to a volume of 250.0$cm^3$ in a volumetric flask. A burette was filled with the diluted solution.
25.0$cm^3$ of potassium hydroxide solution was added to a conical flask and three drops of phenolphthalein indicator added.
1.Ethanoic acid is considered a weak acid. Define the term ‘weak acid’.
2.The results of the titration are shown in the table below:
Trial | Volume of Titre$(cm^3)$ |
1 | 21.0 |
2 | 21.2 |
3 | 20.8 |
1.Using section 22 (Section 18- 2025 Syllabus) of the IB data booklet, identify the color at the endpoint.
2.Calculate the concentration of dilute ethanoic acid.
$\times10^{-4}mol\:dm^{-3}$
3.Calculate the concentration of ethanoic acid in vinegar.
$\times10^{-3}mol\:dm^{-3}$
Important note: The section numbers mentioned in the video solution are specific to the 2024 syllabus.
You can refer to the question stem to find the section numbers in the 2025 syllabus.