[Aman24]

A 0.2536 g sample of the Cr-complex was thermally decomposed and the NH3 released was collected in a 250 mL Erlenmyer flask containing 30.00 mL of 0.3065 M HCl solution and about 20 mL of deionized water. The solution was then titrated with standardized NaOH solution having a concentration of 0.2816 M. It required 11.92 mL of NaOH to reach the bromcresol green end point. Determine the %NH3 in the sample.

I'm not even sure how to start this problem. I have learned solution/titration stoichiometry last semester (although i forgot it), i think maybe somehow that can be applied.

Any tips are appreciated

edit: i think i kind of know what to do. It said it took 11.92 of the NaOH and thats with the NH3 added. So if i find the amount NaOH needed for just 0.3065 M HCl then i can subtract those two amounts and ill have the difference of NaOH needed when NH3 is involved. Not sure what i'd do after that though, or if thats even useful...

Edit2: the most confusing part is the 20 mL of water. So if the acid is 30 mL of 0.3065 M HCl, the number of moles is 0.009195. So now add the 20 mL of water, the molarity will be different but the moles remains the same. If i do a neutralization calculation (molarity -> mole -> mole -> mL) then there is no difference between adding the water, and not adding the water. I still get the same answer. The problem with this is that the solution is neutralizing as you add water to it. How am i supposed to factor in the water neutralizing some of the HCl? Or maybe im not supposed to factor that in...

Alright well somehow i figured this out... I got 39.46%. Which makes sense because i had 19.97% Chromium and 40.8% Chloride.

[Borek]

For calculations this additional water doesn't matter at all, it just dilutes the acid.

[Aman24]

Yeah it didn't matter, i wonder why they'd include that.