Vinegar Analysis

Experiment: Analysis of Vinegar

Introduction

This is another quantitative analysis experiment. Look at the label of many items. It tells you how much of key substances is there. Vinegar should have the right amount of acetic acid in it for pickling and other household tasks. Tell students-READ THE LABEL. As experienced science students, they can know perform an analysis to determine the acetic acid content of household vinegar. This is representative of how all commercial materials end up with reported contents or analysis on label.

General Information

The basic principle behind titration is that one solution is added to another to result in a chemical reaction.

The process is set so exactly enough of one material is added to completely react with the other according to the equation for the reaction.

Here is an example. I use Ag+ + Cl- ч AgCl as an example of determining amount of chloride ion in, say, sea water. We want to determine the % Cl- in sea water. Remember

% = Part/Whole x 100%. In this case %Cl- = mass Cl-/Mass Solution x 100%

At an exact stoichometric amount of Ag+ , moles Ag+ = moles Cl- (see equation).

IF we know amount of Ag+, we know the amount of chloride.

Example: To just react with chloride in 25.0 mL of sea water solution in the flask, 20.5 mL of 0.1024 M Ag+ solution is added. (You've seen molarity, M, all semester, but it is very important now. That means MOLES of reagent per LITER of solution. So, the silver solution being added has 0.1024 moles in 1 L.) We want to determine number of moles on 20.5 mL, or 0.0205 L.

Steps: determine moles Ag+-> moles Cl- --> mass Cl-, then from mass Cl- and mass solution, % can be calculated.

#moles Ag+ , n = M x V = 0.1025 moles/L x 0.0205 L = 0.00210 moles Ag+ added

Relate to moles of Cl-. See above 0.00210 moles Ag+ x 1 mole Cl-/ 1 mole Ag+ = 0.00210 mole Cl-

Mass Cl- = 0.00210 mole Cl- x 35.45 g Cl-/ mole Cl- = 0.0744 g Cl-

To determine % by mass we need mass of solution

MassSoln = Volume x Density (Recall D = M/V)

Mass = 25.0 mL x 1.00 g/mL = 25.0 g

Then calculate % Cl- = 0.0744 g Cl-/25.0 g solution x 100% = 0.298%