Sodium Hypochlorite Properties

As far as I know, sodium hypochlorite is common bleach. If you are not getting satisfactory answers from one supplier, I would try another. There are many out there and I am sure they will answer you emails and give you correct TDS to get your business.

If you know more about their chemical then they do, should you be dealing with them?

I have had similar problems when purchasing sodium hydroxide solution and phosphoric acid solution, when my questions were not satisfactorily answered, I found someone else that could answer them.

Thanks for comments. You're right, it is common bleach. Only thing is for household use it's usually sold at ~ 6% available chlorine, but for industrial use 14/15%

I've tried to find other suppliers and only come up with 3. On Google there are vast nos of entries for safety data sheets, MSDS, COSHH sheets etc, but very sparse on detailed info about the product. Case of the health & safety obsession wagging the dog.

Incidentally one of the others had what I thought was a mistake in figure for dilution to give 250 ppm recommended for disinfection. I told them and they agreed! But their DS didn't give a comparison for my other queries.

Your first concern is the definition of available chlorine, correct?

1. Your definition that it is the elemental chlorine converted to hypochlorous acid,
2. Their definition that it is the molecular chlorine consumed in the entire process needed to form said hypochlorous acid.

I would say that both are correct, depending on the application. Your version is accurate in a teaching situation where you are focusing on that portion of the overall reaction or when you are trying to determine the elemental chlorine content of the hypochlorous acid/hypochloride equilibrium product, and their version is industry specific to how much molecular chlorine is needed to make it in the first place. As this process requires a dispoportionation of the chlorine to form the chloride/hypochloride ions, therefore requiring molecular chlorine, it can be argued either way. Can you say more about why you're concerned about it?

Regarding the density- your description seems a better way of putting it, but you should consider that it may contain additional NaOH, Na Cl, Na2CO3 etc. that can also increase the density.

I don't think there's a problem with the definition of available chlorine. All the chlorine in NaOCl is available. Chlorine in chloride ions is not available chlorine.

No, I can't agree that 2. is correct. There are 2 chlorine atoms in the reaction, but only one ends up in the NaOCl as available chlorine. That's what I'm trying to tell them.

Ref impurities, the supplier in question doesn't mention, but another one gives

Analysis

Sodium Hydroxide (as % NaOH w/w) < 1.0

Available chlorine (as % Cl2 14.5 ± 0.5 (at despatch)

Max impurities

Sodium Chlorate (as % NaOH w/w) < 1.0

Sodium Carbonate (as % NaOH w/w) < 1.0

Iron (ppm) < 2

Total quite low compared with NaOCl content (specially on my figures). It doesn't mention NaCl, so I assume the NaCl formed in the reaction is removed somehow.

Ref concentration basis - it's pretty fundamental that % w/w is X kg chemical in 100 kg of product as supplied (not X kg in (X + 100) kg), but they don't seem to understand!

As stated in your chemical equation, 71 grams of Cl2 will in fact produce 74 grams of NaClO.

As far as hypochlorite in solution: ClO- ion is a stong conjugate base of the weak acid HOCl. If one dissolves the pure substance in water, there is a noticeably alkaline shift in pH. Depending on the buffer strength of the water involved, acidity must now be added to convert some of the hypochlorite to the active acid form to increase the biotoxicity of the treatment (if you are interested in killing organisms).

As far as the density of solution, usually this would be measured in their quality control lab, and the available chlorine is back-calculated from the result of titration of the hypochlorite ion.

You say "As stated in your chemical equation, 71 grams of Cl2 will in fact produce 74 grams of NaClO." Yes, that's correct, but as I've said, only 35.5 gram (1 gm-atom) ends up in the NaClO and so contributes to available chlorine. I appreciate it releases chlorine if pH falls, that's why it's kept alkaline in solution to improve shelf life, that doesn't affect the discussion.

I assume the solution density is a measured value, and I'm not arguing with the figure, 1.243 kg/litre at 15% av. chlorine. What I'm saying is - if solution at 15% av. chlorine has 15.75 % NaOCl as is claimed, the density cannot be as high as that.

The equation Cl₂ + 2 NaOH = NaClO + NaCl + H₂O is a stoichiometric correct entry. That is 2 atoms of chlorine (1 mole of chlorine gas) react with 2 moles of NaOH to produce the reactants listed. Or 71 grams of chlorine gas react with 78 grams of NaOH to produce the reactants. You seem to be forgetting the other reacting agent 2(NaOH) in your assumptions. You cannot use the CL2 = NaClO without embracing all products and reactants. That appears to be the bases of your assumption. There is only 1 atom of chlorine in the wanted reaction product NaClO and the other atom of Cl is used to form NaCl.

I would refer you to Clifford White's book; The Handbook Of Chlorination as a reference. If you are in the business, it is an essential book but you can also go to the library.

I know 2 atoms of chlorine (1 mole of chlorine gas) react with 2 moles of NaOH to produce the reactants listed. But as I've said, only 35.5 gram (1 gm-atom) ends up in the NaClO and so contributes to available chlorine. So the mass of NaOCl/mass of Cl in the product solution is 74.5/35.5, not 74.5/71. The other chlorine atom forms NaCl, as in your equation, the chloride ion in which is not available chlorine. You confirm this by saying "There is only 1 atom of chlorine in the wanted reaction product NaClO and the other atom of Cl is used to form NaCl." Part of your post agrees with me, another part doesn't.

You cannot drive the equation without using Cl2 with an atomic weight of 71. That is essential to the process. Think about it, you need 71 grams of Cl2 to get 74.5 grams of NaOCl. You are over reading the per mass as the product. It is referring to the reactants.

I know, I've said several times there are 71 kg chlorine in the reaction to get 74.5 kg NaOCl. But as I've also said, only one Cl atom goes into the NaOCl, the other into NaCl, and does not constitute available chlorine. How many times do I have to repeat it? The reaction details have no direct bearing on the characteristics of the product NaOCl. Cl₂ has molecular weight 71, not atomic weight. Is this what's confusing you?

If you agree with the supplier that 15% w/w available chlorine contains 15.75% w/w NaOCl, (see definition of w/w in my posts) how do account for solution density 1.243 kg/l?

I am not sure if you aware also, at least in the United States, most of the sodium hypochlorite is produced as an offshoot of the chlor-alkali business. In that business, the plant produces chlorine and caustic soda solution from electrolysis of salt brine. One then need only combine the gaseous chlorine with the sodium hydroxide solution (caustic soda) to arrive at sodium hypochlorite (properly diluted to 10% for industrial bleach, or ~5% for household).

Due to environmental considerations where the EPA does not chlorine use in potable water (and also in wastewater treatment), where the chlorine can react with organics in solution to produce halogenated organics that seem to have a great deal to do with cancer (particularly in the south portion of the Mississippi River), the industry has been scaling back.

I owe an apology. I was wrong

I started looking at oxidation reactions of Cl2 and NaOCl, hoping to confirm my earlier point, but it turned out to show the opposite. Oxidising SO₂ to H₂SO₄

Cl₂ + SO₂ + 2H₂O = 2HCl + H₂SO₄

NaOCl + SO₂ + H₂O = NaCl + H₂SO₄

so 1 molecule NaOCl does the same as 1 molecule Cl₂, even though it only has 1 atom of chlorine, and supplier's formula is correct, % w/w NaOCl = % w/w av. chlorine*74.5/71.

To account for the solution density, I assume the NaCl formed in reaction

Cl₂ + 2NaOH = NaOCl + NaCl + H₂O is not removed. (I don't know how you'd remove it anyway, but I'm sure the chemical companies could if they wanted to).

That means for say 15% w/w av. chlorine, there is 15*74.5/71 NaOCl + 15*58.5/71 NaCl = 15.75 + 12.36 = 28.1% w/w total dissolved solids, and density 1.243 kg/l is quite reasonable.

I still can't say I'm impressed by the response from supplier's technical people. If they understand how their formula is derived, they didn't explain it, and they don't seem to know how w/w solutions work, which is basic in their line of work.

Thanks for replies, sorry to waste folks's time, but I've learnt something, maybe otheres have too.

No Problem!

Keep in mind re. the NaCl/ density I don't think they normally bother removing it as bleach is such a cheap commodity, and also due to the continual breakdown NaOCl (UV etc.)→ NaCl+ O·. Since half lives indicate the largest mass will be formed right after removing it anyways, they probably don't worry too much about it.

P.S.- the lousy customer service is a prevalent issue regardless.

I hate to do this to you but:

Cl₂ + H₂O => HOCl + HCl (first thing that happens to gaseous chlorine upon dissolving in water)

then the equivalence with SO₂ is the same 1:1 ratio for a 2 electron reaction. The technical support at the manufacturers had it right to start with.

Not sure what you're saying. I acknowledged in #12 that the manufacturer's data is right.

From equation Cl₂ + H₂O => HOCl + HCl it doesn't jump out at me that available chlorine is 2 x the actual chlorine in NaOCl (though I now realise it's the case). From the technical department's response I'm not convinced they understand it too well, or they don't explain it well.

As far as solution density is concerned, they don't seem to realise there's anything that needs explanation. I still believe if 15.75 gm NaOCl is dissolved in 84.25 litre water (15.75% w/w NaOCl, 15% w/w av. chlorine) the density is < 1.243 kg/l, so it's the NaCl content that increases it.

The whole entire point:

Cl2 has molecular chlorine gas in the "0" oxidation state, such that a reagent like SO2 that requires two electrons to reach the "+6" oxidation state (SO4), will use up one molecule of Cl2 per molecule of SO2. NaOCl has chlorine in the "+1" oxidation state, and is therefore a two electron reagent (in being coverted to chloride). Thus we have the 74.45/70.9 mass ratio of equivalence "as available chlorine".

The equation I put up to demonstrate chlorine gas dissolving in water, was to also demonstrate that hypochlorous acid (HOCl) is a two electron reagent.

Yes I agree with that, now I've thought about it, as I said in #12.

What do you think about the density question? The supplier I referred to in #4 says 1.245 kg/l (14.5% av. chlorine), but doesn't mention NaCl as an impurity.

Sodium hypochlorite is always going to at least have as many equivalents of salt as sodium hypochlorite, due to the manufacture of it, no matter what the process, one ends up neutralizing the HCl part with NaOH (at the same time the HOCl is neutralized), or the gas is dissolved directly into NaOH solution with the same result.

As the solution ages, there will be more NaCl produced.