Assuming that the lab procedure was conducted in an identical manner in all other respects, if a large quantity of vinegar had been taken initially for analysis, for example, a 50.0 ml instead of 25.0mL discuss briefly how each of the following would change.
A. the moles of acetic acid in each titrated sample
B. The volume of NaOH required to reach the phenolphthalein endpoint.
C. The molarity of the vinegar.
D. the mass percent of acetic acid in the vinegar
I don't see that this problem differs that much from the previous problem. If you will explain what you don't understand perhaps I can clue you in.
To discuss how each of the following would change in the given scenario, we need to understand the relationship between the variables involved and the principles of titration.
A. Moles of acetic acid in each titrated sample:
The number of moles of acetic acid can be calculated using the equation: moles = molarity x volume. If the volume of vinegar is doubled, from 25.0 mL to 50.0 mL, the moles of acetic acid in each titrated sample would also double. This is because the moles of a substance are directly proportional to the volume.
B. Volume of NaOH required to reach the phenolphthalein endpoint:
The volume of NaOH required is determined by the stoichiometry of the reaction between acetic acid and NaOH. Assuming the reaction is 1:1, the molar ratio is 1 mol acetic acid to 1 mol NaOH. Therefore, the volume of NaOH required will also double when the volume of vinegar is doubled.
C. Molarity of the vinegar:
Molarity is defined as moles of solute per liter of solution. If the amount of acetic acid remains the same, but the volume of vinegar is doubled, the molarity would decrease by half. This is because the moles of solute (acetic acid) remain the same, but the liters of solution increase.
D. Mass percent of acetic acid in the vinegar:
To calculate the mass percent, we need to know the mass of acetic acid and the total mass of vinegar. Assuming the density of vinegar is approximately 1 g/mL, the mass of acetic acid in 25.0 mL of vinegar can be calculated. If the volume of vinegar is doubled to 50.0 mL, the mass of acetic acid would double as well. Therefore, the mass percent of acetic acid in the vinegar would remain the same since the percentage is calculated based on the mass of acetic acid divided by the total mass of vinegar.
In summary, if the volume of vinegar is doubled in the given scenario:
A. The moles of acetic acid in each titrated sample would double.
B. The volume of NaOH required to reach the phenolphthalein endpoint would double.
C. The molarity of the vinegar would decrease by half.
D. The mass percent of acetic acid in the vinegar would remain the same.