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    (Original post by samsimmons)
    I was wondering if anyone could explain it to me? I'm getting a vague idea from textbooks and my teacher conviently forgot to teach it to us
    Sodium thiosulphate is used in the determination of iodine and (indirectly) chlorine and bromine. It can also be used to find concentrations of copper(II) salts by reacting the copper(II) with potassium iodide and then titrating the iodine produced against sodium thiosulphate.

    Sodium thiosulphate is a colourless reducing agent that gets oxidised to the tetrathionate ion:

    2S2O32- --> S4O62- + 2e

    It reacts with iodine in the following way:

    2S2O32- + I2 --> S4O62- + 2I-

    The indicator used to detect the presence of iodine is starch, which turns a deep blue/black colour in the presence of iodine. The freshly prepared starch solution is not usually added until near the end-point to prevent formation of a permanent blue-black colour
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    (Original post by charco)
    Sodium thiosulphate is used in the determination of iodine and (indirectly) chlorine and bromine. It can also be used to find concentrations of copper(II) salts by reacting the copper(II) with potassium iodide and then titrating the iodine produced against sodium thiosulphate.

    Sodium thiosulphate is a colourless reducing agent that gets oxidised to the tetrathionate ion:

    2S2O32- --> S4O62- + 2e

    It reacts with iodine in the following way:

    2S2O32- + I2 --> S4O62- + 2I-

    The indicator used to detect the presence of iodine is starch, which turns a deep blue/black colour in the presence of iodine. The freshly prepared starch solution is not usually added until near the end-point to prevent formation of a permanent blue-black colour
    Very helpful... thank you!
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    Does the starch indicator turn from blue-black to colorless?
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    (Original post by WhyStudy)
    Does the starch indicator turn from blue-black to colorless?
    2S2O32- + I2 --> S4O62- + 2I-
    For this equation, yes
 
 
 
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