Posted by Chris on Tuesday, February 23, 2010 at 9:48pm.
First I want to point out a serious error although you may have copied the problem exactly as you received it. But just in case you rephrased it, this will critique that. You do not know the concn of the dilute NaOH BECAUSE you must assume the volumes are additive (the problem states you took 50 mL of the stock solution and diluted it WITH 250 mL of water. This is not the way dilutions are performed. You take 50 mL and add ENOUGH water to make the final volume 250 mL. In the latter case, the concn in the diluted solution is 3 M x (50/250) = 0.60 M. In the former case, you can't calculate the concn unless you know the density of the solution and the final volume.
For the second part, write and balance the equation. I will call oxalic acid, to make things simple, H2Ox
H2Ox + 2NaOH ==> Na2Ox + HOH (water)
How many moles do we have in 35 mL of the dilute NaOH? That is M x L = 0.60 x 0.035 = ??
Notice from the equation that it takes only 1/2 the moles of HOx to neutralize the NaOH; therefore, take the ?? moles NaOH and divide by 2. Thaat will give you the number of moles of HOx.
Then grams H2Ox = moles x molar mass.
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