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chemistry lab (college level)

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So, this is going to be more like 2 or 3 questions. Here goes.
First, what is the net-ionic reaction for Fe (III) + K4Fe(CN)6 be? I've been agonizing over this for HOURS with no answer in sight. (I don't even know what to put down as a product!)

Second, what would the net-ionic reaction be for Cu2+ + NH4OH --> [Cu(NH3)4]?

Third, if trying to find chloride in an unknown, would it be sensible to use AgCl? (I think this is what we actually did use when we ran the lab in class) and how would I show a net-ionic reaction for that? Would it be Ag+ +Cl- --> AgCl (I think I may or may not have been over complicating things)

and lastly, how would I run a test for Nitrate? That isn't the brown ring test?

I'm so sorry this is so very long! I'd really greatly appreciate any and all help given!

  • chemistry lab (college level) - ,

    First: Prussian blue is the compound formed. The ideal formula is Fe4[Fe(CN)6]3 and I usually call this ferric ferrocyanide but the new IUPAC name would be iron(III) hexacyanoferrate(II).

    2. I would write it as
    Cu^+2 + 4NH3(aq) ==> Cu(NH3)4^+2(aq).
    Where did the NH4OH go? Technically,
    NH3 + HOH ==> NH4^+ + OH^- but that is a weak base and not many of the ions are present. Aqueous NH3, which I wrote above as NH3(aq), is NH3 in water solution. Years ago, and I learned it this way too, we assumed NH3 reacted with H2O to form NH4OH and that dissociated into NH4^+ and OH^-. While it is true that the NH4^+ and OH^- ions are formed, it has been concluded that the compound, NH4OH, does not exist. So we write the equilibrium as
    NH3 + HOH ==> NH4^+ + OH^- and we do not show the intermediate of NH4OH.
    In fact, we write the Kb for NH3 as

  • chemistry lab (college level) - ,

    Third question. When trying to determine the amount of chloride in an unknown, yes, one would use the AgNO3 addition to form AgCl. The net ionic equation is as you wrote it. And yes, the brown ring test is the nitrate test.

  • chemistry lab (college level) - ,

    I see. Thanks so much for your help!

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