posted by james on .
Determine the moles of ammonia in the following cobalt ammine complex: A sample of .1500 g of the cobalt ammine complex is places in a flask and 25.00mL of .200M HCL is added. a few drops of bromocresol green is added and the titration with .100M to NaOH required 16.42mL.
I figured out that .00500 mol of hcl added, .001642 M of NaOH in titration, and .003358 mol of HCL is excess.
i don't know where to go from here
James--I'm not completely satisfied with this but it makes sense to me. Check me out--I'll point out my reservations later.
First, your 0.005 moles HCl is correct.
Your other numbers are correct, also, BUT you have them labeled incorrectly. The 0.001642 moles is the excess HCl present and the 0.003358 moles is the amount of the ammine present.
What I have done now is to convert 0.003358 moles NH3(that is what was neutralized with the HCl) to grams.
0.003358 x molar mass NH3 = grams NH3.
Subtract from 0.1500 g to obtain grams Co. Then convert them to percentages OR just convert those grams to moles.
moles NH3 = g/molar mass NH3.
moles Co = g/molar mass Co.
Then take the ratio of the two. That gives me, if I didn't goof, Co(NH3)2. What are my reservations? Mostly, that the ratio I obtained is 1.0 Co to 2.12 NH3 and I rounded that to 1.0 to 2.0; however, with masses and buret readings to 4 significant figures, I think it should come closer to 1.0 to 2.0. I'm an analytical chemist and I think I can do better than that with that kind of precision. Of course this could be a made up problem, anyway, and it was made to come out this way. At any rate, take a look and see what you think but keep in mind that the failure to come closer to whole numbers nags at me. I have spent about the last hour trying to work around it and this is what I always come up with. After you find the "correct" answer, I would appreciate you posting it to my attention. Thanks.
There was a typo on the problem. and cobalt was never meant to be used. It should have read copper which went along with previous questions. Hence, everyone's confusion. Thanks