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, it can be deduced that a mass of 5.94g of CH3COOH will be formed per 100ml:
0.276 → 5.0ml
X → 100ml
X → 0.276x (100/5.0) = 5.94g of acetic acid.
(Using the average values. The ‘actual’ volume formed for the trial #1 is 5.88g and for the trial #2 is 6.00g)
Errors & Improvements:
- No precise way to judge when the titration ends (with naked eye)
- Similar but not identical way of rinsing for the burette.
- Inaccurate way of measuring the volume of NaOH used.
- Rinsing provided by one subject
- More reverse titration applied
- Repeat experiment to deduce an average
Conclusion:
The value for the average amount of the acetic acid produced is 5.94g (~5.9g ± 0.1) per 100ml of vinegar, and therefore complies with the federal requirement of a minimum of 4g of acetic acid per 100ml of vinegar.
Experiment #2:
Aim: To determine the amount (as a percentage) of magnesium hydroxide in the milk of magnesia tablets.
Variables:
Independent: Volume/Amount of Hydrochloric acid – HCL
Dependent: Concentration of Mg(OH)2 found in the magnesia tablets
Controlled: Molarity and amount of HCl, Molarity and amount of NaOH, pressure (not controlled, assumed), Temperature
Materials: Check photocopy.
Diagram:
Procedure: Check photocopy.
Data Collection:
Data Processing:
The equation for the reaction of titration of magnesium hydroxide is the following:
2HCl (aq) + Mg(OH)2(s) → MgCl2(aq) + 2H2O(l)
In order to proceed with the equation, there must be present two moles of HCl for each mole of Mg(OH)2, hence the acid is added in excess. The actual experiment shows that for 4.83g of Magnesia tablets, a volume of 14.6 cm3(with a concentration of 1 mol dm-3) was needed to react completely with the Mg(OH)2, and provide a surplus (excess, neutralized later by the NaOH). The volume is equivalent to:
(1dm3 ~ 1mol, then 14.6 cm3 ~ 0.015mol)
0.015mol x 36.46gmol-1 =~ 0.55g
And, as two moles of HCl reacted with one mole of Magnesium hydroxide, an approximate amount of 0.008mol of Mg(OH)2 was used, that is,
2HCl → 1 Mg(OH)2
0.015mol / 2 = 0.008mol
0.008mol x 58g mol-1 = 0.46g
And as there is 0.46g of Mg(OH)2 / 4.83 of milk of magnesia, an approximate 9.5% (~10%) of milk of magnesia tablets are composed by Mg(OH)2 .
Errors & Improvements:
- No precise way to judge when the titration ends (with naked eye)
- Similar but not identical way of rinsing for the burette.
- Inaccurate way of measuring the mass of Milk of Magnesia
-
Not persistent stirring provided in order to dilute the Mg(OH)2
- Rinsing provided by one subject
- More reverse titration applied
- Repeat experiment to deduce an average
Conclusion: The experiment showed that an approximate 10% of the milk of magnesia tablets are made of Magnesium hydroxide (Mg(OH)2). It also showed that the value is close to the one established by Phillips, which is 8g/100ml.