Primarygun Posted November 22, 2005 Share Posted November 22, 2005 How do you get Iron(III) chloride from a solution of acidified Iron(II) sulphate. Please ensure there's no sulphate ion left. What does "Oxidizing flame" mean? Link to comment Share on other sites More sharing options...
woelen Posted November 22, 2005 Share Posted November 22, 2005 Probably the best way is to precipitate the iron (II) sulfate in water and precipitate this with sodium hydroxide. You also add some H2O2 to oxidize all to Fe2O3. Next, filter the precipitate, rinse with some water and then dissolve it in HCl. Do not dry the precipitate. Next, precipitate another time with a solution of NaOH. Again, filter and rinse with water. Then dissolve again in HCl. That should give you pure FeCl3, dissolved in excess HCl. The reason for the extra dissolve/precipitate step is that the initial precipitate still contains quite some sulfate, due to co-precipitation. Making dry solid FeCl3 is not possible from aqueous solution. When the solution is boiled to dryness, then first the hydrate FeCl3.6H2O is formed. On further heating, this compound gives off water and HCl and a basic chloride remains. "Oxidizing flame" denotes a flame, in which the gas-mix is oxidizing (e.g. excess oxygen). Link to comment Share on other sites More sharing options...
Primarygun Posted November 24, 2005 Author Share Posted November 24, 2005 I thought of getting out the sulphate by adding excess lead(II) nitrate. Then, add excess con. HCl to remove the lead(II) ion. And then, for the iron(II) ion, bubble chlorine gas into it. Lastly, to remove the nitrate ion, adding excess hydroxide into it. Due to the redox reaction, nitrate is reduced to nitrogen dioxide. But I am not sure whether it is correct. Probably not as chlorine forms hypochlorite ion and not all nitrate could be removed completely. Link to comment Share on other sites More sharing options...
woelen Posted November 24, 2005 Share Posted November 24, 2005 I thought of getting out the sulphate by adding excess lead(II) nitrate.Then' date=' add excess con. HCl to remove the lead(II) ion. And then, for the iron(II) ion, bubble chlorine gas into it. Lastly, to remove the nitrate ion, adding excess hydroxide into it. Due to the redox reaction, nitrate is reduced to nitrogen dioxide. But I am not sure whether it is correct. Probably not as chlorine forms hypochlorite ion and not all nitrate could be removed completely.[/quote'] Why use such difficult to handle, expensive and toxic reagents if you also can accomplish it with simple HCl and NaOH? You also add lots of other impurities, such as lead and nitrate, which also are hard to remove completely. Lead chloride is sparingly soluble, not insoluble. Link to comment Share on other sites More sharing options...
Primarygun Posted November 25, 2005 Author Share Posted November 25, 2005 Why use such difficult to handle, expensive and toxic reagents if you also can accomplish it with simple HCl and NaOH? You also add lots of other impurities, such as lead and nitrate, which also are hard to remove completely. Lead chloride is sparingly soluble, not insoluble. Mine is a poor idea. Thanks Link to comment Share on other sites More sharing options...
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now