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
), viz.
A text-book of practical organic chemistry including qualitative organic analysis, A Vogel, 3rd Editionp 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.