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 50.0mL instead of 25.0 mL 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 vinegar.

Helppppp

You shouldn't find this that hard. Look at your original data. How did you determine mols initially.

Wouldn't you think that if you doubled the vinegar at the start you would have double the mols?
b. Wouldn't you think it you took x mL to tirate the 25 that it would take 2x mL to titrate 50?
c. Wouldn't you think the M would be the same. M = mol/L. You double mols and double L as in 2*mol/2*L; M stays the same.
etc.

Sure, I'd be happy to help!

a. The moles of acetic acid in each titrated sample: If a larger quantity of vinegar (50.0 mL) is taken instead of the original 25.0 mL, the moles of acetic acid in each titrated sample would be doubled. This is because the amount of acetic acid present in the vinegar is directly proportional to the volume of vinegar used. Therefore, if the volume is increased, the amount of acetic acid will also increase.

To calculate the moles of acetic acid in each titrated sample, you would need to know the concentration of acetic acid in the vinegar, which is typically given as the molarity of the solution. The formula to calculate the moles of acetic acid is:

moles = molarity x volume

b. The volume of NaOH required to reach the phenolphthalein endpoint: The phenolphthalein endpoint is the point at which the color of the solution changes, indicating that the reaction between acetic acid and NaOH is complete. If a larger quantity of vinegar is used (50.0 mL instead of 25.0 mL), the volume of NaOH required to reach the phenolphthalein endpoint would also increase. This is because there is a direct stoichiometric relationship between the acetic acid and NaOH, meaning that a larger amount of acetic acid will require a larger amount of NaOH for complete reaction.

To determine the volume of NaOH required, you would perform a titration by slowly adding NaOH to the vinegar solution until the phenolphthalein indicator changes color. The volume of NaOH added at this point represents the amount needed to neutralize the acetic acid.

c. The molarity of the vinegar: The molarity of the vinegar is a measure of the concentration of acetic acid in the solution. If a larger quantity of vinegar is taken initially (50.0 mL instead of 25.0 mL), the molarity of the vinegar would remain the same. This is because molarity is defined as the number of moles of solute (acetic acid) per liter of solution. As long as the amount of acetic acid in the vinegar does not change, the molarity will remain constant.

To calculate the molarity of the vinegar, you would divide the moles of acetic acid (calculated in part a) by the volume of vinegar used (50.0 mL).

d. The mass percent of acetic acid in vinegar: The mass percent of acetic acid in vinegar is a measure of the mass of acetic acid present in a given mass of vinegar. If a larger quantity of vinegar is used (50.0 mL instead of 25.0 mL), the mass percent of acetic acid in vinegar would not change. This is because the mass percent is calculated by dividing the mass of acetic acid by the mass of the entire vinegar sample, regardless of the volume used.

To calculate the mass percent of acetic acid in vinegar, you would need to know the mass of the acetic acid present (which can be determined by multiplying the moles of acetic acid by its molar mass) and divide it by the mass of the vinegar sample, typically given in grams.