a) A back titration is implemented in indigestion and constipation remedy analysis using a milk sample. A magnesia milk sample weighing 3.964 g and containing magnesium hydroxide, Mg(OH)2, was added to 60.00 mL of 0.350 M hydrochloric acid (HCl) in a conical flask. After the reaction was completed, the excess HCl in the conical flask required 30.45 mL of 0.216 M sodium hydroxide (NaOH) for titration to achieve the end point. Calculate the percentage (w/w) of the active ingredient Mg(OH)2 in the sample.
The reaction: Mg(OH)2 + 2HCl → MgCl2 + CO2 + H2O
b) Distinguish between end point and equivalence point.
Answer (1)
To solve this problem, we'll calculate the amount of magnesium hydroxide, Mg(OH)_2, present in the milk sample through a back titration method.
Step-by-step Solution:
Determine the moles of HCl initially added:
We have 60.00 mL of 0.350 M HCl. The moles of HCl added can be calculated using:
Moles of HCl = Volume (L) × Molarity (mol/L) = 0.060 × 0.350 = 0.021 mol
Calculate the moles of excess HCl that reacted with NaOH:
We use 30.45 mL of 0.216 M NaOH to neutralize the excess HCl. The moles of NaOH used are:
Moles of NaOH = Volume (L) × Molarity (mol/L) = 0.03045 × 0.216 = 0.00658 mol
Since the reaction between HCl and NaOH is one-to-one, the moles of excess HCl are also 0.00658 mol.
Calculate the initial moles of HCl that reacted with Mg(OH)_2:
The moles of initial HCl that reacted with Mg(OH)_2 can be found by subtracting the moles of excess HCl from the initial moles of HCl:
\text{Moles of HCl reacted with Mg(OH)_2} = 0.021 - 0.00658 = 0.01442 \text{ mol}
Determine the moles of Mg(OH)_2:
According to the reaction Mg(OH) 2 + 2 HCl → MgCl 2 + 2 H 2 O , 1 mole of Mg(OH)_2 reacts with 2 moles of HCl.
Therefore, the moles of Mg(OH)_2 present is:
\text{Moles of Mg(OH)_2} = \frac{0.01442}{2} = 0.00721 \text{ mol}
Convert the moles of Mg(OH)_2 to mass:
Using the molar mass of Mg(OH)_2 (which is approximately 58.32 g/mol):
\text{Mass of Mg(OH)_2} = \text{Moles} \times \text{Molar Mass} = 0.00721 \times 58.32 = 0.42059 \text{ g}
Calculate the percentage (w/w) of Mg(OH)_2 in the sample:
\text{Percentage of Mg(OH)_2} = \left( \frac{0.42059}{3.964} \right) \times 100\% \approx 10.61\%
b) Distinction between the End Point and Equivalence Point
End Point: This is the point in a titration where the indicator changes color. It signifies the completion of the reaction as seen by the human eye. The end point is used as an approximation to the equivalence point.
Equivalence Point: This is the theoretical point at which the number of moles of titrant is stoichiometrically equivalent to the number of moles of analyte in the solution. It marks the completion of the chemical reaction as per the stoichiometry of the reaction, and it is ideally where the end point should occur.
