dchak Posted June 17, 2008 Share Posted June 17, 2008 hi guys, can you please suggest some reaction to distinguish Cuprous chloride from Cupric chloride. Link to comment Share on other sites More sharing options...
hermanntrude Posted June 17, 2008 Share Posted June 17, 2008 you can tell by looking at the colours. Copper (I) and copper (II) look very different Link to comment Share on other sites More sharing options...
frosch45 Posted June 18, 2008 Share Posted June 18, 2008 Copper(I) chloride (cuprous chloride) is a greenish color Copper(II) chloride (Cupric chloride) is tealish Link to comment Share on other sites More sharing options...
dchak Posted June 18, 2008 Author Share Posted June 18, 2008 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. Link to comment Share on other sites More sharing options...
hermanntrude Posted June 18, 2008 Share Posted June 18, 2008 that's a toughie. You can't titrate for the Chloride ions because of the HCl. I guess you'll have to find something which a) reacts only with cupric or cuprous ions b) either changes pH upon reaction or changes colour good luck! Link to comment Share on other sites More sharing options...
DeanK2 Posted June 18, 2008 Share Posted June 18, 2008 Are you in dire need of an answer, or simply curious? Hopefully it is the latter as the neccesary substances with the properties that hermanntrude has mentioned will be almost impossible to find. Link to comment Share on other sites More sharing options...
dchak Posted June 19, 2008 Author Share Posted June 19, 2008 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. Link to comment Share on other sites More sharing options...
YT2095 Posted June 19, 2008 Share Posted June 19, 2008 try these, for basic data on the differences/similarities: http://en.wikipedia.org/wiki/Copper(II)_chloride http://en.wikipedia.org/wiki/Copper%28I%29_chloride http://www.webelements.com/compounds/copper/copper_chloride.html http://www.webelements.com/compounds/copper/copper_dichloride.html Link to comment Share on other sites More sharing options...
dchak Posted June 19, 2008 Author Share Posted June 19, 2008 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. Link to comment Share on other sites More sharing options...
YT2095 Posted June 19, 2008 Share Posted June 19, 2008 a simple way would be to add a chunk of Copper metal to the mixture of a known weight, allow it to finish reacting by: CuCl2 + Cu -> 2 CuCl and then weight the metal afterwards to find out how much was used. the rest is simple stoichometry. Link to comment Share on other sites More sharing options...
hermanntrude Posted June 19, 2008 Share Posted June 19, 2008 a simple way would be to add a chunk of Copper metal to the mixture of a known weight, allow it to finish reacting by: CuCl2 + Cu -> 2 CuCland then weight the metal afterwards to find out how much was used. the rest is simple stoichometry. excellent idea YT, assuming the reaction goes to completion, it'd give you a simple method, without even titrating. Gravimetric rather than volumetric. I was just about to suggest it myself Link to comment Share on other sites More sharing options...
dchak Posted June 20, 2008 Author Share Posted June 20, 2008 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?? Link to comment Share on other sites More sharing options...
hermanntrude Posted June 20, 2008 Share Posted June 20, 2008 bummer. I forgot about the HCl CuCl is insoluble? really? I googled it and you're right... one of the exceptions to the solubility rules LINK OK so how about this: neutralise the solution with NaOH, recording the amount you use, then filter it and weigh the dreid solid obtained. Then evaporate the neutral filtrate to dryness and calculate how much of it is copper (II) chloride either by subtraction, using the amount of NaCl you know you formed, or by titration for Cu(II), if there is such a thing, or by both methods to prove some level of accuracy. Link to comment Share on other sites More sharing options...
Tartaglia Posted June 20, 2008 Share Posted June 20, 2008 I would probably precipitate out all the copper as CuI which releases I2 and titrate the iodine with thiosulphate using a starch indicator Cu2+(aq) + 2I-(aq) -> CuI(s) + 1/2 I2(aq) I2 + 2S2O32- -> S4O62- + 2I- Link to comment Share on other sites More sharing options...
hermanntrude Posted June 20, 2008 Share Posted June 20, 2008 I would probably precipitate out all the copper as CuI which releases I2 and titrate the iodine with thiosulphate using a starch indicator Cu2+(aq) + 2I-(aq) -> CuI(s) + 1/2 I2(aq) I2 + 2S2O32- -> S4O62- + 2I- but then you'd only know the total amount of copper in the solution, not the amount of copper (I) and copper (II) Link to comment Share on other sites More sharing options...
Tartaglia Posted June 20, 2008 Share Posted June 20, 2008 Its pretty simple really, you divide it in two. With one lot you determine the Cu(II) in the solution. With the other you oxidise it with a suitable weak oxidant so that all is converted into Cu(II). Precipitate and titrate again The Cu(I) can be determined by difference This is a typical A level question Link to comment Share on other sites More sharing options...
hermanntrude Posted June 20, 2008 Share Posted June 20, 2008 the trouble is that things are a bit extra-confused by the presence of HCl. Adding oxidisers may make more than one reaction occur Link to comment Share on other sites More sharing options...
Tartaglia Posted June 20, 2008 Share Posted June 20, 2008 Get your standard emf book out. I haven't got one infront of me but I suspect a little H2O2 would do and the excess would be destroyed by the Cu2+ ion. Stir overnight There are plenty of other possibilities Link to comment Share on other sites More sharing options...
hermanntrude Posted June 20, 2008 Share Posted June 20, 2008 not everything can be entirely predicted by the book. You don't know the concentration of the Chloride ions or the concentration of the cupric/cuprous ions, so you can't predict the likelihood of a reaction. the "emf" values you refer to are only valid at standard temperature and pressure and at one molar concentration or, in the case of gases, one bar of pressure. Link to comment Share on other sites More sharing options...
Tartaglia Posted June 20, 2008 Share Posted June 20, 2008 More commonly known as the nernst equation Are you truly telling me you would have difficulty finding something that oxidises Cu+ to Cu2+, without oxidising Cl-? Link to comment Share on other sites More sharing options...
hermanntrude Posted June 20, 2008 Share Posted June 20, 2008 probably not, actually. what bothers me is that it's not you that's doing it. I know the nernst equation, but it won't help you if you don't know the concentration of the HCl or the two Cu ions. Link to comment Share on other sites More sharing options...
Tartaglia Posted June 20, 2008 Share Posted June 20, 2008 Here is an easier approach Divide sample in two Add I- to the first solution and titrate released I2 with S2O32- --> Cu2+ Stir 2nd sample with nitric acid. This should convert all copper ions to Cu2+. Raise PH with NaOH and decant supernatent leaving insoluble precipitate of Cu(OH)2 free from Cl-. Dissolve Cu(OH)2 in dilute H2SO4 add excess NaI and titrate produced iodine with S2O32- --> Cu+ and Cu2+ Link to comment Share on other sites More sharing options...
John Cuthber Posted June 21, 2008 Share Posted June 21, 2008 It might be easier to do the oxidation with H2O2 then boil the solution to remove the excess. Link to comment Share on other sites More sharing options...
Tartaglia Posted June 21, 2008 Share Posted June 21, 2008 Funnily enough I very nearly put H2O2 as I had suggested it earlier. There was just enough doubt in my mind as to whether the Cu2+ might catalytically decompose the H2O2 before the insoluble CuCl was oxidised There are ofcourse many other alternatives Link to comment Share on other sites More sharing options...
vishal8415 Posted February 2, 2009 Share Posted February 2, 2009 Copper (I) Chloride can not be greenish, if it is greenish than its purity have gone down and it has turned into oxychloride form. Copper (I) Chloride is a Greyish to Off-White or Slightly browinsh Color product. Copper (II) Chloride is a greenish product. To make Copper (II) Chloride or Cupric Chloride you need HCl and Nitric Acid and Chlorine. Vishal Link to comment Share on other sites More sharing options...
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