If %FFA (as Palmitic acid) is 5.22% for a 5 g oil sample. What is the volume of 0.1N NaOH required to titrate the oil sample? If the volume of 0.1N NaOH required to titrate blank i... If %FFA (as Palmitic acid) is 5.22% for a 5 g oil sample. What is the volume of 0.1N NaOH required to titrate the oil sample? If the volume of 0.1N NaOH required to titrate blank is 0.2 mL and the molar mass of Palmitic acid is 256.43 g/mol.
Understand the Problem
The question is asking to calculate the volume of a 0.1N NaOH solution required to titrate an oil sample given the percentage of free fatty acids (as Palmitic acid) and the mass of the sample. We'll use the provided percentage, mass, and molar mass to find the amount of Palmitic acid present and then determine the volume of NaOH needed based on stoichiometry.
Answer
The volume of 0.1N NaOH required is $9.97 \text{ mL}$.
Answer for screen readers
The volume of 0.1N NaOH required to titrate the oil sample is approximately $9.97 \text{ mL}$.
Steps to Solve
- Calculate the mass of Palmitic acid present
First, we need to find the mass of Palmitic acid in the oil sample.
Given:
- Percentage of free fatty acids (FFA) = 5.22%
- Mass of oil sample = 5 g
The mass of Palmitic acid can be calculated as: $$ \text{Mass of Palmitic acid} = \frac{\text{Percentage of FFA}}{100} \times \text{Mass of oil sample} $$ Substituting the values: $$ \text{Mass of Palmitic acid} = \frac{5.22}{100} \times 5 = 0.261 \text{ g} $$
- Convert mass of Palmitic acid to moles
Next, we convert the mass of Palmitic acid to moles using its molar mass.
Given:
- Molar mass of Palmitic acid = 256.43 g/mol
Using the formula: $$ \text{Moles of Palmitic acid} = \frac{\text{Mass of Palmitic acid}}{\text{Molar mass of Palmitic acid}} $$ Substituting the values: $$ \text{Moles of Palmitic acid} = \frac{0.261}{256.43} \approx 0.001018 \text{ moles} $$
- Determine the moles of NaOH required
Since the reaction between Palmitic acid and NaOH is 1:1 stoichiometry, the moles of NaOH needed are equal to the moles of Palmitic acid: $$ \text{Moles of NaOH required} = 0.001018 \text{ moles} $$
- Calculate the volume of NaOH solution required
Now, we can find the volume of the 0.1N NaOH solution needed to provide the required moles.
Using the formula: $$ \text{Volume (L)} = \frac{\text{Moles}}{\text{Concentration (mol/L)}} $$ Given the concentration of NaOH = 0.1N = 0.1 mol/L: $$ \text{Volume of NaOH} = \frac{0.001018}{0.1} = 0.01018 \text{ L} = 10.18 \text{ mL} $$
- Adjust for the blank titration volume
Finally, we have to subtract the volume of the blank NaOH titration from the total volume: $$ \text{Final volume of NaOH} = \text{Volume of NaOH} - \text{Volume of blank} $$ Substituting the values: $$ \text{Final volume of NaOH} = 10.18 \text{ mL} - 0.2 \text{ mL} = 9.98 \text{ mL} \approx 9.97 \text{ mL} $$
The volume of 0.1N NaOH required to titrate the oil sample is approximately $9.97 \text{ mL}$.
More Information
This calculation illustrates the process of titration in a laboratory context, specifically focusing on the volume of titrant required based on the concentration of the solution and the mass of the sample being tested. It highlights the importance of stoichiometry in determining reactant requirements.
Tips
- Forgetting to convert the final volume from liters to milliliters.
- Miscalculating the percentage of fatty acids present in the sample.
- Overlooking the volume of the blank in the final calculations.
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