Experiment 6 Acids and Bases: Analysis of Vinegar Objectives - Carry out an acid
ID: 1025904 • Letter: E
Question
Experiment 6 Acids and Bases: Analysis of Vinegar Objectives - Carry out an acid-base reaction - Learn how to do a titration -Determine the molarity and (w/w)% of acetic acid in vinegar Vinegar is a solution of 4-5% (w/w) acetic acid (HC,CO) in water. Federal standards require acetic acid in water are colorless but sometimes because of other additives like caramel or wine, some brands will have color to them. In this experiment, you will use non-colored vinegar, which is known as "white vinegar". This experiment is a quantitative analysis of the amount of acetic acid in vinegar. It will be carried out by the method of titration in which a measured quantity of vinegar is exactly neutralized by a known amount of a base (NaOH). Calculations using the quantities required for neutralization yield the concentration of the acetic acid in vinegar. a minimum acetic acid content of four percent in commercial brands. Solutions of According to Bronsted-Lowry definition, an acid is a substance that will release protons or H' ions. In the case of a strong mineral acid such as HCL, the dissociation of the acid is essentially complete in water. HCI (aq) + H2O (l) H30' (aq) Cl. (aq) + In the case of weak organic acid, such as acetic acid the dissociation in water is not nearly as complete. (Below the formula of acetic acid is written in the HA acid form, to show that only the first H is given out as a proton or H't) HCH3CO2 (aq) + H2O() H30' (aq) CH3C02. (aq) + This dissociation of acetic acid is characterized by an acid dissociation constant of 10% This means that only about one out of 100,000 molecules are actually dissociated when replaced in distilled water The technique of titration is used to determine the exact amount of acid or base present in a solution. To determine the concentration of an acid solution such as vinegar, a sample er flask. A base solution of a known concentration is added until the acid sample is completely neutralized. The concentration of the acid solution in terms of molarity and (w/w)% can then be as follows calculatedExplanation / Answer
Post-lab
1. Possible sources of error
If the volume of vinegar taken was higher or lower than actual value, the final value of acetic acid will also be higher or lower accordingly.
If air bubble is present in burette, the moles NaOH calculated would be higher, so moles acetic acid will also be higher. Final acetic acid molarity would thus also be greater than actual value.
If burette is rinsed with water instead of NaOH solution. The NaOH solution gets diluted, more is needed for neutralization. Moles of NaOH calculated would be higher and thus higher molarity of acetic acid would be calculated.
2. the moles of acetic acid in vinegar solution remains the same with or without the 25 ml DI water, therefore presence of water would not change the results.
3. If Mg(OH)2 a dibasic base was used instead of monobasic NaOH, the ratio of base to acetic acid would be 1 : 2.
reaction,
Mg(OH)2 + 2CH3COOH ---> Mg(CH3COO)2 + 2H2O
with NaOH the ratio of base to acid is 1 : 1.
NaOH + CH3COOH --> NaCH3COO + H2O
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