A stock solution was prepared by dissolving exactly 0.4000 g of pure ASA (180.16 g/mol)

Question

in 10.00 mL of NaOH and heating the solution to a gentle boil. After cooling to room
temperature, the solution was poured into a 250-mL volumetric flask and diluted to the mark
with DI water. Before coming to lab, calculate the concentration of ASA in the flask.

I got my molarity as 0.00888M:
0.400g/180.16=0.00222mols/0.250= 0.00888M
however am I suppose to take into account the 10.0ml of NaOH and if so how would I do that?

Answer 1

Your calculated molarity is correct. No, you don't need to "correct" for the 10 mL NaOH because that 10 mL is part of the solvent used when made to the 250 mL mark. That is, the 10 mL is part of the 250 mL.

Answer 2

Okay thank you that makes a lot more sense!