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    Can someone please help me,

    A sample of solid chromium(III) hydroxide displays amphoteric character when treated separately with dilute hydrochloric acid and with dilute aqueous sodium hydroxide.Write an ionic equation for each of these reactions. Include the formula of each complex ion formed. Describe the changes that you would observe in each reaction.

    Where chromium(III) hydroxide = Cr(OH)3

    Im stuck

    For HCl I thought it was: Cr(OH)3 + 3H+ --> Cr^3+ + 3H2O which is wrong
    For NaOH I thought it was Cr(OH)3 + OH- --> Cr(OH)4-
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    I think the key here is the word complex. The ion Cr3+ isn't a complex so you need to identify the complex that would be present in solution and the ligands in it need to be present in the reactants to balance it all out.
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    (Original post by Sniperdon227)
    Can someone please help me,

    A sample of solid chromium(III) hydroxide displays amphoteric character when treated separately with dilute hydrochloric acid and with dilute aqueous sodium hydroxide.Write an ionic equation for each of these reactions. Include the formula of each complex ion formed. Describe the changes that you would observe in each reaction.

    Where chromium(III) hydroxide = Cr(OH)3

    Im stuck

    For HCl I thought it was: Cr(OH)3 + 3H+ --> Cr^3+ + 3H2O which is wrong
    For NaOH I thought it was Cr(OH)3 + OH- --> Cr(OH)4-
    I think it's easier to understand if you remember Cr(OH)3 is in effect Cr(H2O)3(OH)3

    Adding H+ ions... they are donating protons=
    Cr(OH)3 + 3H+ + 3H2O --> [Cr(H2O)6]3+

    (basically Cr(H2O)3(OH)3 + 3H+ ---> [Cr(H2O)6]+3 ) this is in the textbook.. I believe the mark scheme would accept this too??

    In the case of adding aqueous NaOH:
    Cr(OH)3 is in effect Cr(H20)3(OH)3... If you add OH- ions to it (base), it takes a proton from a water ligand so you get [Cr(H20)2(OH)4]-

    I'm not confident on this so anyone out there feel free to correct
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    (Original post by Signorina)
    I think it's easier to understand if you remember Cr(OH)3 is in effect Cr(H2O)3(OH)3

    Adding H+ ions... they are donating protons=
    Cr(OH)3 + 3H+ + 3H2O --> [Cr(H2O)6]3+

    (basically Cr(H2O)3(OH)3 + 3H+ ---> [Cr(H2O)6]+3 ) this is in the textbook.. I believe the mark scheme would accept this too??

    In the case of adding aqueous NaOH:
    Cr(OH)3 is in effect Cr(H20)3(OH)3... If you add OH- ions to it (base), it takes a proton from a water ligand so you get [Cr(H20)2(OH)4]-

    I'm not confident on this so anyone out there feel free to correct
    Your answers are bang on! Correct! I was hesitant to write Cr(OH)3 as Cr(H20)3(OH)3 so it was unclear what was going on now I see you can do as suggested in the MS ill do so from now on, thanks. Im now intrigued what Al(OH)3 aka Al(H20)3(OH)3 would do? In HCl and aq NaOH respectively
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    (Original post by Sniperdon227)
    Your answers are bang on! Correct! I was hesitant to write Cr(OH)3 as Cr(H20)3(OH)3 so it was unclear what was going on now I see you can do as suggested in the MS ill do so from now on, thanks. Im now intrigued what Al(OH)3 aka Al(H20)3(OH)3 would do? In HCl and aq NaOH respectively
    Yeah I wouldn't have written it Cr(OH)3 myself. I'm so used to writing it out full and it's much easier to visualise and understand what's happening with the ligands!

    I believe it would be the same
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    (Original post by Signorina)
    Yeah I wouldn't have written it Cr(OH)3 myself. I'm so used to writing it out full and it's much easier to visualise and understand what's happening with the ligands!

    I believe it would be the same
    Same thing really

    Cr(H20)3(OH)3 + OH- ---> [Cr(H20)2(OH)4]- + H2O

    Cr(OH)3 + 2H2O + OH– → [Cr(H2O)2(OH)4]–
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    Al(H2O)3(OH)3 + 3H+ ---> [Al(H2O)6]+3 (White PPT to Colourless Soln)

    Al(H20)3(OH)3 + OH- ---> [Al(H20)2(OH)4]- + H2O or Al(OH)3 + 2H2O + OH– → [Al(H2O)2(OH)4]– (White PPT to Colourless Soln)
 
 
 
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