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    I would like to clarify the following few facts:

    Halide salts are colourless in solution. Their precipitates are white (with a few well known exceptions: PbI2 and the silver halides).

    F2 is colourless in water, having reacted to form HF; Cl2 (aq) is pale green in water; Br2 (aq) is yellow when dilute and red-brown when concentrated (but is generally said to be "orange"); I2 (aq) is pale brown. Generally, F2 (g) is yellow; Cl2 (g) is green; Br2 (g) is orange and Br2 (l), standard state, is red; I2 (g) is purple and I2 (s), standard state, is grey-black.

    Halogens dissolve in hexane much more than in water, and with the same colours, except that I2 dissolves with purple colour.

    I2 dissolves, more than in water, in KI solution, with a pink colour. The I3- ion, which is occasionally created, is red-brown in solution. Thus, the I2 solution in KI will finally be red again. (Yellow if dilute.)

    Is this right? Please note that my specification prefers using words with qualifiers like "pale", and only writing 1 colour, so several times I have avoided using longer descriptors in favour of a single word. With that in mind, are there any corrections you would make?
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    (Original post by Big-Daddy)
    I would like to clarify the following few facts:

    Halide salts are colourless in solution. Their precipitates are white (with a few well known exceptions: PbI2 and the silver halides).

    F2 is colourless in water, having reacted to form HF; Cl2 (aq) is pale green in water; Br2 (aq) is yellow when dilute and red-brown when concentrated (but is generally said to be "orange"); I2 (aq) is pale brown. Generally, F2 (g) is yellow; Cl2 (g) is green; Br2 (g) is orange and Br2 (l), standard state, is red; I2 (g) is purple and I2 (s), standard state, is grey-black.

    Halogens dissolve in hexane much more than in water, and with the same colours, except that I2 dissolves with purple colour.

    I2 dissolves, more than in water, in KI solution, with a pink colour. The I3- ion, which is occasionally created, is red-brown in solution. Thus, the I2 solution in KI will finally be red again. (Yellow if dilute.)

    Is this right? Please note that my specification prefers using words with qualifiers like "pale", and only writing 1 colour, so several times I have avoided using longer descriptors in favour of a single word. With that in mind, are there any corrections you would make?
    The triiodide ion solution is red, rather than pink. It isn't "occasionally" created, it is formed by dissolving iodine in iodide solutions. The triiodide ion is in equilibrium with iodine and iodide and acts as a quantitative source of dissolved iodine (as iodine reacts the equilibrium shifts to form more etc etc)
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    (Original post by charco)
    The triiodide ion solution is red, rather than pink. It isn't "occasionally" created, it is formed by dissolving iodine in iodide solutions. The triiodide ion is in equilibrium with iodine and iodide and acts as a quantitative source of dissolved iodine (as iodine reacts the equilibrium shifts to form more etc etc)
    I'm guessing this equilibrium is along the lines of:

    I- + I2 ⇌ I3-

    So the more I2 you add, the more I3- you get. So if we want (let's take a fixed concentration of KI in the solution) we can write this as a ratio constant with changing concentrations: Kc*CKI=I3-/I2. Left hand side is constant for a certain concentration of KI.

    Adding I2 will not therefore change the ratio of I3- to I2 at equilibrium, so it shouldn't make any difference how much I2 we add, except that at very low concentrations of I2 we will also have very low concentrations of I3- so we may not see much colour.

    That leaves me with two main questions, if the above is correct:

    What colour is I2, specifically, in KI solution? Pale brown, as in aqueous solution?

    What colour is the I3-, specifically, in KI solution?

    Since the ratio of these species must be the same, I guess the effect the concentration has on the overall colour the solution appears must come from whether low concentrations of KI were present (note that the ratio is directly proportional to KI concentration) and whether low concentrations of I2 were added. But neglecting all this , what should I write in an exam, or expect to see, for the colour of this solution in general? And what is the broad colour range, e.g. yellow to the concentrated colour you just mentioned, possibly?
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    (Original post by Big-Daddy)
    I'm guessing this equilibrium is along the lines of:

    I- + I2 ⇌ I3-

    So the more I2 you add, the more I3- you get. So if we want (let's take a fixed concentration of KI in the solution) we can write this as a ratio constant with changing concentrations: Kc*CKI=I3-/I2. Left hand side is constant for a certain concentration of KI.

    Adding I2 will not therefore change the ratio of I3- to I2 at equilibrium, so it shouldn't make any difference how much I2 we add, except that at very low concentrations of I2 we will also have very low concentrations of I3- so we may not see much colour.

    That leaves me with two main questions, if the above is correct:

    What colour is I2, specifically, in KI solution? Pale brown, as in aqueous solution?

    What colour is the I3-, specifically, in KI solution?

    Since the ratio of these species must be the same, I guess the effect the concentration has on the overall colour the solution appears must come from whether low concentrations of KI were present (note that the ratio is directly proportional to KI concentration) and whether low concentrations of I2 were added. But neglecting all this , what should I write in an exam, or expect to see, for the colour of this solution in general? And what is the broad colour range, e.g. yellow to the concentrated colour you just mentioned, possibly?
    I believe that the triiiodide ion is formed in nearly stoichiometric proportions, i.e. the equilibrium constant is very much larger than 1.

    The colour in solution is red, a kind of attractive dark wine red.

    As it is equilibrium, the solution behaves as an iodine reservoir, which can be formed as it is used up until there is no triiodide left.
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    (Original post by charco)
    I believe that the triiiodide ion is formed in nearly stoichiometric proportions, i.e. the equilibrium constant is very much larger than 1.

    The colour in solution is red, a kind of attractive dark wine red.

    As it is equilibrium, the solution behaves as an iodine reservoir, which can be formed as it is used up until there is no triiodide left.
    OK, thanks for clarifying.

    Are the rest of my colours and notes correct?
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    (Original post by Big-Daddy)
    OK, thanks for clarifying.

    Are the rest of my colours and notes correct?
    yes
 
 
 
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