(Hive Addict)
03-04-03 17:57
No 413717
(Rated as: excellent)
The following procedure might be helpful for some bees... It has been retrieved in Vogel (yes pH, there still are a couple of sections without sticky pages :P ), viz. A text-book of practical organic chemistry including qualitative organic analysis, A Vogel, 3rd Edition
p 809 [...] To determine the exact perbenzoic acid content of the solution, proceed as follows. Dissolve 1.5 g of sodium iodide in 50 mL of water in a 250 mL reagent bottle and add about 5 mL of glacial acetic acid and 5 mL of chloroform. Introduce a known weight or volume of the chloroform solution of perbenzoic acid and shake vigorously. Titrate the liberated iodine with standard 0.1 N sodium thiosulphate solution in the usual manner.
1 mL of 0.1N Na2S2O3 = 0.0069 g of perbenzoic acid. [...]
Although Vogel is a cool book, the procedure doesn't explain in detail how the titration works. So, allow me (and correct me if I'd be wrong):
Peracids can liberate iodine from an acidic sodium iodide solution:
2H+ + RCOOOH + 2I- ---> RCOOH + I2 + 2H2O
However, sodium thiosulfate (Na2S2O3) will react with I2 (redox) and bring iodide back into solution:
2 Na2S2O3 + I2 ---> Na2S4O6 + 2 NaI
Now, using the following numbers:
M(PBA) = 138 g/mol (molecular weight perbenzoic acid)
M(STS) = 152 g/mol (molecular weight Na2S2O3)
n(STS) = 2n(PBA)
m(PBA) = n(PBA)*M(PBA) = 0.5*n(STS)*M(PBA)
So, for 1 mL of 0.1N Na2S2O3 solution, you detect 0.5*(0.1*0.001)*138 g perbenzoic acid. However, most bees use performic and/or peracetic acid. To calculate the concentration of performic/peracetic, simply substitute M(PBA) by the molecular weight of performic acid (62 g/mol) or peracetic acid (76 g/mol).
Here is a practical example I did minutes ago: 1.5 g NaI is dissolved in 50 mL distilled water. To this solution, you add 5 mL glacial acetic acid and 5 mL CHCl3. I added 0.250 mL performic acid solution (made by mixing 5 mL formic acid 98% and 5 mL hydrogen peroxide 30%, allowing to stand for 45 minutes) and the NaI solution became purple-like (typical I2-colour). I titrated until the solution became colourless. For this, I used 7.5 mL Na2S2O3 0.1N solution.
So, let's have a look at our known data:
M(PFA)=62 g/mol (molecular weight performic acid)
V(PFA)=0.250 mL
M(FA)=46 g/mol (molecular weight formic acid)
V(FA)=5 mL
d(FA)=1.22 kg
V(STS)=7.5 mL
V(H2O2)+V(FA)=10 mL=V(tot)
The amount of performic acid, according to the previously mentioned formula, must be:
m(PFA) = V(STS)*0.5*0.1*0.001*M(PFA)/V(PFA)
m(PFA) = 7.5*0.5*0.1*0.001*62/0.25 = 93 mg (per mL peracid solution)
If we compare the number of moles peracid formed per mole aliphatic acid used:
n(PFA) = m(PFA)/M(PFA) = 0.93/62 = 15 mmol
n(FA) = m(FA)/M(FA) = V(FA)*d(FA)/M(FA) = 6.1/46 = 133 mmol
n(PFA)/n(FA) = 15/133 = 0.11
If we take a look at JACS 68 (1946) 907 (hosted somewhere on Rh's site as pdf... The convenient preparation of per-acids, by FP Greenspan), than we see that in their test, n(PFA)/n(FA) = 0.9 to 0.12 from 0.5h to 1h on.
Voila, now you all know how to calculate the peracid concentration of your solution. If someone finds an error in my calculations, pls correct.
Abusus non tollit usum
(Hive Addict)
03-07-03 00:04
No 414412
OK, sorry ppl, but the forementioned post is not completely correct. The maths and theorem are only valid if there is NO H2O2 present in you solution. This means that you can't use the calculations for your in-situ prepared performic or peracitc acid.
Although there must be a way around. Just need to find it I guess. I found a mentioning in CA today where it was said that H2O2 can be titrated by KMnO4 and peracids cannot. Maybe that is something worth reading/testing about. Or is there anyone with factual knowledge on this particular subject?
Abusus non tollit usum