challenging chemistry problem- are you a chem whiz?

Question

i have a test tomorrow and this is a type of problem that i'm having trouble with:

A student added 50.0 ml of an NaOH solution to 100.0 ml of 0.400M HCl. The solution was then treated with an excess of aqueous chromium(III) nitrate, resulting in formation of 2.06 g of precipitate. Determine the concentration of the NaOH solution.

if you can figure out the answer and explain it a bit that would be great and i will offer you my respect and gratitude.

the problem my professor gave me didn't specify the precipitate. I guess you are supposed to figure it out using solubility rules and stuff. i thought of trying Cr(OH)3 as the precipitate and that seemed to make sense, but the person trying to explain it to me thought it was CrCl3 and i am now confused

i used papastoltes' method and got 1.932M NaOH, but the person trying to teach me did it a different way and got a different answer. if someone could get an answer that i could compare mine to that would be helpful. I need to know which way is correct

Answer 1

You have to figure out the reaction which is leading to a precipitate and what the precipitate is. My guess is that Cr(OH)3 is being ppt out.
So you would take the mole wt of this and divide 2.06 g of ppt by this mole wt to get moles of Cr(OH)3. Then you can find the g-ions of [OH-] = 3*moles of Cr(OH)3. Since 100 ml of 0.4 M HCl is enough acid to neutralize 0.04 g-ions of {OH-} from the NaOH, the initial amount of [OH-] would have been the 3*moles Cr(OH)3 + 0.04. From this and your initial NaOH solution volume, you can figure concentration.

Answer 2

Without knowing what the precipitate was, we're stuck. A plausible guess would be Cr(OH)3, and you can get the number of moles of that; that will tell you the number of moles of un-neutralized OH that were in the acid-base mixture. You can then allow for the 40 mM of H+ from the acid, and determine how much OH- there was.

Answer 3

You can find out the number of moles of NaOH that reacted.
Then you can figure out how many moles of Cr(OH)3 formed.
Then you can figure out how many moles of NaOH were used in the reaction with the Cr(III)
Then you can add those two mole quantities of NaOH together, and divide by the volume of NaOH (in L)
Try it. Let me know how you made out (add a note to your question).