posted by Shelby .
You have an unlimited amount of 12M HCl stock solution. You need to neutralize 2.00 grams of NaOH by slowly adding 0.100M HCl solution. How will you prepare the proper amount of hydrochloric acid solution to add to the beaker of sodium hydroxide using a 500 ml volumetric flask.
2.00 g = 2.00/molar mass NaOH = about 0.05 moles NaOH and that will take 0.05 moles HCl.
mL12M x M12M = mL0.1M x M0.1M
mL x 12 = 500 x 0.1
mL of 12M acid = (500 x 0.1/12) = 4.167.
I would use a 5.000 mL buret, add exactly 4.167 mL of the 12M stock solution of HCl to the 500 mL volumetric flask, then add DI or distilled water to the mark of the flask and mix thoroughly.
(As extra information, I provide the follwing. For whatever it's worth, I would do it a different way and use a 4 mL pipet, transfer exactly 4.00 mL with a pipet to the flask and proceed. That will NOT make 0.1M; the concn will be 0.096 but that's much simpler. Four mL pipets are easily available (one could use 5 mL pipet for a final concn of 0.12M too) and the cocn is very close to what we wanted. I suppose, however, that the intent of the problem is to make you think how to make an exact 0.1M solution. Another point to remember is that if you intend to titrate the 2.00 g NaOH will a buret, using 0.1M HCl will require L = moles/M = 0.05/0.1 = 0.5 L or 500 mL. Requiring such a large volume using a 50 mL buret means refilling the buret about 5 times and that increases the reading error immensely. We don't like to do that.)