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I'm an IB Chemistry student and I'm doing the Winkler's titration for my Internally Assesed Investigation. However, my school doesn't have alkaline iodide azide, so I need to make that myself first. How do you do that?

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    $\begingroup$ See: "A Laboratory Manual for Environmental Chemistry" and" Introduction to Environmental Sciences" They are both in google books. As well as their procedure. Make sure that you are doing this under supervision. A quick look to the MSDS of sodium azide shows that it can explode. $\endgroup$ Commented Dec 4, 2016 at 15:53
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    $\begingroup$ You might need a blast shield to do this. It's not recommended at the IB level... $\endgroup$
    – Zhe
    Commented Dec 4, 2016 at 16:05

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I'm quite sure you meant alkaline iodide sodium azide solution. The preaparation of this reagent is given here:

To prepare this reagent, take 700 g of potassium hydroxide/500 g of sodium hydroxide and add 150 g of potassium iodide/ 135 g of sodium iodide and dissolve in freshly boiled and cooled water and make up tp 1000 ml.

Dissolve 10 g of sodium azide in 40 ml distilled water and add this solution with continuous stirring to the cool alkaline iodide prepared previously.

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    $\begingroup$ The link you sent in your answer sometimes uses alkaline iodide azide and sometimes alkaline iodide sodium azide. I can't tell if they're using them interchangeably or if they are referring to two different things. $\endgroup$
    – ILoveJesus
    Commented Dec 4, 2016 at 16:10
  • $\begingroup$ @ILoveJesus There is nothing called alkaline iodide azide. It is alkaline iodide sodium azide. $\endgroup$ Commented Dec 4, 2016 at 16:12
  • $\begingroup$ the cool alkaline(NaOH/KOH) iodide(NaI/KI) solution they are referring is the 1 litre solution prepared previously. $\endgroup$ Commented Dec 4, 2016 at 16:14
  • $\begingroup$ @NilayGosh I calculated that the molar ratio between the Iodide and the Hydroxide is 1:1. And then I calculated that the ratio of the masses between NaI and NaOH should be 3.75:1. However, in the thing you sent me and in other labs, they always take more NaOH than NaI, and it's actually with the same ratio, just inverted. Could you explain this to me please? $\endgroup$
    – ILoveJesus
    Commented Dec 12, 2016 at 17:44

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