katy - 20-8-2017 at 13:13
Hi all,
Trying to determine the usefulness of my sad looking old sodium borohydride. What used to be a nice, free flowing, granular solid is now an expanded,
hard, solid mess.
Great thanks to woelen for posting a procedure in another thread...
Posted by woelen:
Why not titrate the material against iodine.
Weigh a certain amount of your NaBH4 (e.g. 50.0 mg). Dissolve in water, to which some NaOH is added. In such solutions, the NaBH4 hardly looses any H2
if the solution is kept cool and only kept around for a short time. If you cannot weigh out 50.0 mg, then weigh out 1.00 gram, dissolve this in 200 ml
of water and take 10 ml of this solution.
Determine how much I2 is needed to oxidize all of the NaBH4 to borate(III). One mole of NaBH4 requires 4 moles of I2. Keep in mind that only a very
small amount of NaBH4 requires a large amount of I2 (one gram of NaBH4 requires almost 27 grams of I2). In the above situation (where you take 0.0500
grams of NaBH4) you need to weigh precisely approximately 1.5 grams grams of iodine and 1 gram of NaOH (which is a good excess amount).
Dissolve the NaOH in some water, allow to cool well, and then dissolve the I2 in this solution. Quickly add this solution to the solution of NaBH4 and
stir well. After a few minutes, add a slight excess amount of acid to this solution, such that free iodine is formed again (brown color).
Titrate the amount of iodine with a solution of Na2S2O3.5H2O. If this titration tells you there was an excess amount of appr. 0.0053 mole of iodine,
then you have nearly pure NaBH4. If much more iodine was left over, then your NaBH4 has less reducing power and is of inferior quality.
This method is much more precise than the production of H2 and is also easier to perform (provided you have a decent scale with which you can weigh
small quantities of chemicals). The reduction is performed in alkaline solution totally and hence no H2 is formed. All of the BH4(-) ions is
tranformed to some borate(III)-species in the alkaline solutions. Iodine is present as hypoiodite and maybe some iodate. Any excess hypoiodite and
iodate quickly are transformed to elemental I2 if the solution is acidified again and the iodine remains in solution, because of all the iodide ions
formed on reduction of NaBH4. With starch as indicator you can do a precise titration
...Thanks woelen, but being a beginner I feel quite lost.
I am hoping someone can please elaborate (or go all the way and provide a nearly idiot proof step by step)?
Allow me to expose my ignorance. Am I to understand that I:
1. Create a basic (pH) solution of .05 g NaBH4 and small amount of NaOH.
2. Create a solution of 1.5g I2, 1g NaOH and then combine with the NaBH4 solution.
2. Add an acid to the solution until it just reaches a brown end point (I assume sulfuric acid is appropriate?). Add starch (to get a nice blue
color).
3. Create a solution of Na2S2O5 of known molarity and add to a burette.
4. Drip Na2S2O5 into borohydride/iodine solution until brown/blue color is extinguished.
5. The number of moles of Na2S2O5 used is related to the number of moles that the the old NaBH4 is deficient/degraded/not active (???).
My understanding is severely lacking..
Any help is appreciated, be it activity determination, purification and/or acquisition of NaBH4.
Appreciate your time!