Problem:
Do manufacturers provide true or false concentrations of acetic acid in vinegars
Purpose:
The whole idea of this lab /experiment is to find out whether the percentage label of acetic acid in vinegar is true or false, which is provided by the manufacturers.

The purpose of this lab is to calculate the acetic acid content on various brands of vinegar and evaluate the manufacturer’s claim based on the percentage of acetic acid in the solution of vinegar.

Materials:
Refer to page 310 in Addison Wesley Chemistry 11
Experimental Design Chart
Brands
Cost per Litre (L) Manufacturer’s Claim Regarding Percent on Acetic Acid in Vinegar
Selection White Vinegar
$1.15
5%
Allen’s Cleaning Vinegar
$1.39
10%
Allen’s Pickling Vinegar
$1.01
7%

Sample A Vinegar
Sample B Vinegar
Sample C Vinegar
Volume of Sample (mL)
20.00mL
20.00mL
20.00
Density of Sample (g.mL-1)
1.1
1.096
1.0985
Mass of Sample (g)
22.0
21.92
21.97
Volume Percent of Acetic Acid in vinegar
4.375%
10.75%
6.5
Percent on vinegar label by company
5%
10%
7%

Analyzing and Interpreting
Balanced Equation of Vinegar

NaOH + CH3COOH  NACH3COO + H2O

1)
Sample A
0.0175 L of Sodium hydroxide (NaOH) was used for sample A
0.05 mol/L Concentration of Sodium hydroxide (NaOH) was used for Sample A

nNaOH= Volume X Concentration = (o.o5 mol/1 L)(0.0175L) = 0.000875 mol nCh3COOH = (0.000875 mol of NaOH) (1 mol of CH3COOH/1 mol of NaOH) = 0.000875 mol of Ch3COOH
Concentration of acetic acid =0.000875 mol/0.0200L = 0.04375 mol/L
CH3COOH Volume Percent(%) = (0.04375) (100%) = 4.375%

B sample
0.043 L of Sodium hydroxide (NaOH) was used for sample B
0.05 mol/L Concentration of Sodium hydroxide (NaOH) was used for Sample B nNaOH= Volume X Concentration = (0.05 mol/1L)(0.043L) = 0.00215 mol nCH3COOH = (0.00215 mol of NaOH)(1 mol of CH3COOH/1 mol of NaOH) = 0.00215mol of CH3COOH
Concentration acetic acid = ( 0.00215 mol/0.0200L) = 0.1075 mol/L
CH3COOH Volume Percent(%) = (0.1075)( 100%) = 10.75%

C sample
0.05 mol/L Concentration of Sodium hydroxide (NaOH) was used for Sample C
0.026 L of Sodium hydroxide (NaOH) was used for sample C

nNaOH= Volume X Concentration =0.05 mol/1L) (0.026L) = 0.0013 mol nCh3COOH = ( 0.0013 mol NaOH)(1 mol CH3COOH/1 mol NaOH)() = 0.0013 mol Ch3COOH
Concentration of acetic acid =(0.0013 mol/0.200L) = 0.065 mol/L
CH3COOH Volume Percent (%) = (0.065)(100%) = 6.5 %
2)a.
Sample A Vinegar
Sample B Vinegar
Sample C Vinegar
0.04375mol/L
0.1075mol/L
0.065mol/L

b)
Sample A Vinegar