Titration of commercial products

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Lab Report #1:

Titration of commercial products

Experiment #1:

Aim: To determine the concentration of acetic acid in vinegar.

Variables:

Independent: Volume/Amount of NaOH

Dependent: Volume/Amount of acetic acid produced (CH3COOH)

Controlled: Molarity of CH3COOH, Molarity of NaOH, pressure (not controlled, assumed), Temperature

Materials: Check photocopy.

Diagram:

Procedure: Check photocopy.

Data Collection:

Results for vinegar:

Data Processing:

The equation for the reaction of titration of acetic acid is the following:

CH3COOH (aq) + NaOH (aq)  Na+CH3COO-(aq) + H2O (l)

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Experimentally, it shows that a mole of acetic acid (CH3COOH) reacts with a mole of magnesium hydroxide (NaOH) (both in aqueous state) to form a mole of Na+CH3COO- and a mole of water. Since 1 dm3 gives one mole of NaOH, the amount used in this experiment (4.95ml) is equivalent to a total of 4.95x10-3 mol. Hence, it reacts with the same amount (in moles) of acetic acid. Then, provided the molar mass of acetic acid (60g mol-1), it can be deduced that 4.95x10-3 mol weights 0.297g:

(60g mol-1 * 4.95x10-3 mol = 0.297g)

And, if there is 0.297g per 5.0ml of vinegar, ...

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