There seems to be only one main iodine tincture available local to me, a very expensive 25ml with 25g/L each of KI and I2 in alcohol, so i thought i might sit down (while buzzing on some presently but probably not much longer to be legal piperazine stimulant
) and work out the precise stochiometry of the reaction so that it can be done more successfully
H2O2 34.0146g/mol
KI 165.9945g/mol
I2 253.809g/mol
HCl 36.5609g/mol
25g/L KI = 0.150607 mol/L
25g/L I2 = 0.098499 mol/L
25g/L=0.025g/ml
in 25ml, by weight:
0.625g (0.003765 mol) KI
0.625g (0.002463 mol) I2
symbolically:
2(KI) + H2O2 + 2(HCl)
------>
2(H2O) + 2(KCl) + I2
molarity:
2x(0.003765) + 0.003765 + 2(0.003765)
the weights of each would be as follows (per 25ml bottle):
KI: 0.625g
H2O2: 0.128g
HCl: 0.137g
the H2O2 needed to provide the
correct amount would be the following,
depending on the concentration of the
H2O2:
1% (0.01g/ml) 12.8ml
3% (0.03g/ml) 4.26ml
6% (0.06g/ml) 2.13ml
10% (0.10g/ml) 1.28ml
45% (0.45g/ml) 0.284ml
the amount of HCl solution at varying
(common) concentrations:
28% (0.28g/ml) 0.4892ml
30% (0.3g/ml) 0.4566ml
31% (0.31g/ml) 0.4419ml
32% (0.32g/ml) 0.4281ml
33% (0.33g/ml) 0.4151ml
Once the proper proportions of reagents are mixed together, the potassium iodide will be turned into potassium chloride and the iodide ion turns into iodine. The usual thing to do from here is to add lots of water to dilute the alcohol and then the iodine will precipitate from the solution. However, this is inefficient in two ways - one, the alcohol, no matter how dilute, is going to cause some amount of iodine to remain in the solution. And every ml of water, which reduces the amount held by the alcohol dissolves iodine to the rate of 0.03 g/100 mL at 20 C, that's 300mg dissolving into a litre of water.
I think the best way to proceed would be to first convert the iodine into salts, then boil it down and all the pesky alcohol goes away. The precise molar amount of base to use on the 25ml tincture bottle to convert the iodine is 0.004926 moles, if NaOH (molar weight 39.99707g/mol) were used, this would mean 0.197g of NaOH per bottle of tincture.
The total number of moles of iodide ions in the solution would thus be
moles KI + 2x moles I2 = moles I+
0.625g (0.003765 mol) KI
0.625g (0.002463 mol) I2
0.003765 + 2x(0.002463) = 0.008691 moles of iodide ions
Okay, so now we have a mixture of sodium iodide and potassium iodide, we can safely boil it down to dryness without fear of losing I2 or stinking and staining up the kitchen
both of them melt at 600+ degrees C, so there's no concern of decomposition or anything, one could bake it in the oven at 250°C until it absolutely has no moisture left in it.
Thus we have defeated the first source of losses.
then, we have 0.008691 moles of iodide ions which we can turn into half as many moles of I2, so we can then turn it back into iodide ions...
back to the formula:
2(KI) + H2O2 + 2(HCl) ------> 2(H2O) + 2(KCl) + I2
okay so that's
iodide ions 0.008691 moles
H2O2 0.0043455 moles (0.1478g)
HCl 0.008691 moles (0.3177g)
ppl can convert their for their own acid and peroxide, but mine's 280g/L HCl and 3% peroxide, so for me that means:
HCl (280g/L) 1.1348ml
H2O2 (30g/L, 3%) 4.9266ml
per 25ml tincture bottle you start with.
Now with this home made Na/K/iodide salt, which is nice and dry, you can add those two reagents, and since you hardly added any water, it's not hard to dry it again.
Because the whole thing is done with proper molar ratios, one should get around 95% yield. For each little 25ml bottle of tincture, theoretical yield of total I2 is 0.0043455 moles or 1.1029g , so it should be possible to get 1.047g out of each little bottle.