Atwood's was out of 10% chlorine so I had to pick up a few cases of their ordinary bleach (concentration unknown), and I need to know the concentration in order to use Pool Math. I found some great info in older threads for determining bleach strength by creating a 1:10,000 dilution and then performing a standard FC test. However, the dilution part seemed a bit tricky and a source of potential error, so I thought the following might work with some items that are at hand:
1) Carefully pour a little bleach to be tested into a clean, empty Taylor reagent bottle.
2) Fill a 2 liter bottle halfway with chlorine-free water.
3) Use the reagent bottle to add 5 drops of the bleach to be tested into the 2 liter bottle.
4) Completely fill the 2 L bottle, cap it, and shake a bit to mix thoroughly.
5) Test the solution in 2 L the bottle using standard methods.
6) The resulting Cl concentration from the FC test should equal the sodium hypochlorite concentration (%) of the bleach.
I'm relying on the following info for this to work. Previous posts have indicated that Taylor reagent bottles are designed so that 24 drops equal one mL. So, 5 drops would equal 5 * (1/24) => 0.208 mL, which is roughly 1/10,000 of a 2 L bottle. I write "roughly" because it is actually a 1:9,615 dilution, so, we need to multiple by 1.04 to correct for this (1.04 = 10000 / 9615). However, even with out the correction the estimate should be accurate enough for Pool Math purposes. Math and chem geeks, please check my methods and let me know if I've made errors or bad assumptions.
1) Carefully pour a little bleach to be tested into a clean, empty Taylor reagent bottle.
2) Fill a 2 liter bottle halfway with chlorine-free water.
3) Use the reagent bottle to add 5 drops of the bleach to be tested into the 2 liter bottle.
4) Completely fill the 2 L bottle, cap it, and shake a bit to mix thoroughly.
5) Test the solution in 2 L the bottle using standard methods.
6) The resulting Cl concentration from the FC test should equal the sodium hypochlorite concentration (%) of the bleach.
I'm relying on the following info for this to work. Previous posts have indicated that Taylor reagent bottles are designed so that 24 drops equal one mL. So, 5 drops would equal 5 * (1/24) => 0.208 mL, which is roughly 1/10,000 of a 2 L bottle. I write "roughly" because it is actually a 1:9,615 dilution, so, we need to multiple by 1.04 to correct for this (1.04 = 10000 / 9615). However, even with out the correction the estimate should be accurate enough for Pool Math purposes. Math and chem geeks, please check my methods and let me know if I've made errors or bad assumptions.
.) I just now tested my 5 drop method above with Atwood's bleach and got 3.0 ppm Cl with the FAS-DPD method, so I assume it is 3% bleach, which for $1.19 isn't a terrible deal, but not near as good as their 10% liquid chlorine for $2 a gallon.
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