I need to stain biological samples (blowfly pupae) with 0.5M iodine for Micro CT Scanning to analyse internal morphology.
Use potassium iodide (KI).
KI = 40 + 127 = 167 grams/mole
A 0.5 M solution of KI has 0.5 moles in 1000 ml = 0.5 mole x167 grams/mole = 84.5 grams of KI. So dissolve 84.5 g KI in 1000 ml water.
I hope this helps!
Andrew :-)
if you put this into a volumetric flask and filled it to exactly 1 litre you would have a 1 molar solution So to get a 0.5M solution simply put 63.45g iodine into a volumetric flask and fill it up to 1 litre exactly You now have a 0.5M Iodine solution
Thank you so much for answering. i need to prepare 500 ml iodine solution. somewhere i read to make 0.5M iodine solution firstly 10g of KI should be dissolved in 100 ml distilled water secondly 63.45g of iodine should be added in aquoeous KI solution. Finally add distilled water in order to make the volume 500ml. Is this the correct way to prepare it?
MW of KI = 166g in this case you needed 0.5M or 1N with 500 ml?
Molarity (M) and normality (N) of KI are the same values as 166g/L. So if you need 0.5M with 0.5L (500ml), you have to weigh 166 x 0.5 x 0.5 = 41.5g KI and dissolve in 500 ml of DI, for this you can dissolve all KI with 100 ml, then adjust volume to 500 ml in volumetric flask. If you need 1N with 500ml, you have to weigh 166x0.5 = 83g KI and dissolve it with the same technique.
there is a different between iodine solution and iodide solution. iodide is prepared from KI. but iodine is prepared from both KI and I2. for iodide solution 0.5M=0.5N and for iodine solution 0.5M=1N.
how to prepare 0.5M/1N iodine solution?
Dissolved 200g of KI in about 350 mL of ultra-pure water. Add 127g of iodine(I2) and mix until all the iodine crystals have dissolved. Transfer the solution into a 1000 mL volumetric flask and fill to the mark with ultra-pure water. standardize this solution using Sodium thiosulphate as a primary standard.
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