Thursday
April 24, 2014

Homework Help: CheMistry

Posted by ChEm on Sunday, April 18, 2010 at 11:05pm.

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

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