Question

0.448 g of an unknown diprotic acid is dissolved in about 60 mL of water in a beaker. The solution is transferred to a 100.00 mL volumetric flask, which is then filled up to the mark. The solution is mixed by inverting multiple times.
25.00 mL of this solution is then transferred to an Erlenmeyer flask for the titration. What mass of the diprotic acid is in the 25.00 mL that is transferred?

Answer 1

According to the Question:

We initially had 0.448 grams of the unknown diprotic acid, which was added to about 60 mL of water and that solution was added to a 100  mL flask and filled to the 100 mL mark

The above is as mentioned in the question, our 0.448 grams of diprotic acid is basically diluted to 100 mL [it's volume has been increased to 100mL]

but the amount of the acid is still the same

Which means that we now have 0.448 grams of the diprotic acid in a 100mL solution

Percent of the diprotic acid that will be present in the Erlenmeyer Flask:

From this 100 mL solution, 25 mL is transferred to the Erlenmeyer Flask

Which means that (25/100) * 100 = 25 % of the 0.448 grams of diprotic acid is present in the 25 mL sample

Mass of di-protic acid in the 25 mL solution:

Since 25% of the initial amount remains in the final solution,

Mass of the diprotic acid in the final solution = 0.448 * 0.25

Mass in the final solution = 0.112 grams

Therefore, 0.112 grams of the di-protic acid will be present in the Erlenmeyer Flask