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    Can anyone explain how transition metal substitution works? How many OH^- ions replace water, why and how? My teacher says it's similar to acid/base reactions? Thanks.
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    (Original post by SpohieAlice)
    Can anyone explain how transition metal substitution works? How many OH^- ions replace water, why and how? My teacher says it's similar to acid/base reactions? Thanks.
    ligand substitution?

    one out one in ...
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    Imagine a copper ion with six water ligands attached.

    Along comes an OH- ion and steals an H+ off one of the waters.

    What's left?
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    (Original post by Pigster)
    Imagine a copper ion with six water ligands attached.

    Along comes an OH- ion and steals an H+ off one of the waters.

    What's left?
    Ok, so it's [Cu(H20)5(OH-)]?
    How many times does this happen, or does it depend?
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    (Original post by SpohieAlice)
    Ok, so it's [Cu(H20)5(OH-)]?
    How many times does this happen, or does it depend?
    [Cu(H2O)5(OH)]+ fixed - you don't show the charge on the OH ion ligand, you should it on the complex AND you forgot about the charge on the central ion.

    If you pull off another H+, you'd end up with [Cu(H2O)4(OH)2], which is insoluble. Do you know what happens if you try to pull any more off? i.e. does it redissolve on excess alkali?
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    (Original post by Pigster)
    [Cu(H2O)5(OH)]+ fixed - you don't show the charge on the OH ion ligand, you should it on the complex AND you forgot about the charge on the central ion.

    If you pull off another H+, you'd end up with [Cu(H2O)4(OH)2], which is insoluble. Do you know what happens if you try to pull any more off? i.e. does it redissolve on excess alkali?
    It gains a charge so it redissolves in water with excess: [Cu(H2O)3(OH)3]^-

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    (Original post by SpohieAlice)
    It gains a charge so it redissolves in water with excess: [Cu(H2O)3(OH)3]^-
    Ask your teacher if you can give it a go, to confirm your guess.
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    Ok, thanks for your help. 😀

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