Jump to content

dchak

Members
  • Posts

    5
  • Joined

  • Last visited

About dchak

  • Birthday 01/25/1987

Retained

  • Lepton

dchak's Achievements

Lepton

Lepton (1/13)

10

Reputation

  1. i think there might be a problem with that!!you see,the slt mixture is in HCl solution,so Copper would probably react with HCl to form CuCl too.If I heat the solution to dryness then i obviosly cant react the CuCL2 with Cu as both will be solids...however here's a thought......CuCl is'nt soluble in water right?..whereas CuCl2 is......so what if i heat the solution mixture to dryness and disolve the residue in water....i can then filter out the CuCl salt right??
  2. i think i know what happened when i heated the solutions. The CuCl reacted with HCl to form CuCl2 as it got the required energy to do so when i heated it.on getting heated to dryness it gave a brownish-yellow residue,which is the colour of anhydrous CuCl2.As this residue cooled,it simultaneously absorbed moisture to form CuCl2 crystals which is bluish green. I guess i'll try finding a characteristic rection for Cu2+ ions.
  3. actually am doing a project.so yes i'm in dire need of the answer i was actually tried evaporating the solution to dryness to find the weight of the residue.what i found was very confusing:the entire residue was first of a brownish yellow colour.when it cooled the entire thing became green,almost like anhydrous CuCl. i then made seperate solutions of CuCl in HCl and CuCl2 in HCl and heated them both to dryness.i got more or less the same result.
  4. here's my problem-i have a mixture of CuCl and CuCl2 in HCl.Could you suggest a reaction which i could use for titration so as to determine how much CuCl and how much CuCl2 is present in that.
  5. hi guys, can you please suggest some reaction to distinguish Cuprous chloride from Cupric chloride.
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.