ajanson
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chloride measurement in bleach
How can I measure the chloride content of diluted bleach?
We used the Hach method 8113 (mercuric thiocyanate addition) to measure chloride in a sodium hypochlorite solution containing 1900 mg/l free chlorine.
The result was 1500 mg/l. Was this chloride actually present as a contaminant in the bleach? Or, was the chloride instead produced by decomposition
of the hypochlorite (meaning it wasn't there to start with)? The test method requires the pH of the sample to be 2 after the reagents are added so
the hypochlorite is no longer there as NaOCl. Is the hypo converted to chlorine gas in the Hach test? If it is, then the 1500 mg/l chloride could
actually have been a contaminant in the diluted bleach solution. Any insight into the following is requested:
a) How can I accurately measure chloride in a bleach solution?
b) Is the chloride I found due to chlorite decomposition or was it present as a contaminant in the bleach
c) In industrial bleach solutions, what kind of contaminant chloride levels can I expect to see? Thanks in advance,
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woelen
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Answering question (a) is not easy for me. You need to find a metal ion, which precipitates with chloride, but not with hydroxide and hypochlorite.
Maybe thallium (I) ions may do the job (TlCl is insoluble, TlOH is very soluble, I don't know, however, whether Tl(I) is stable with ClO(-) ions
around).
Questions (b) and (c): Commercial bleach solutions, including the ones, found at homes, are alkaline solutions, ideally containing Cl(-) and ClO(-) in
a 1 : 1 molar ratio. They are made electrolytically, or by bubbling Cl2 through a solution of NaOH. In both cases, the main reaction is:
Cl2 + 2OH(-) --> Cl(-) + ClO(-) + H2O
So, you always have chloride in such solutions. On aging of the solution, more chloride is formed, but also chlorate:
3ClO(-) ---> 2Cl(-) + ClO3(-)
Besides that, some oxygen loss occurs, but this is only marginal:
2ClO(-) --> 2Cl(-) + O2
The bleach solutions usually are quite alkaline. Excess NaOH is used. High pH makes the bleach more stable. So, freshly prepared bleach contains one
chloride ion for each hypochlorite ion.
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ajanson
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Regarding the metal for precipitation, silver nitrate addition is one method listed in Standard Methods for Water and Wastewater (Argentometric
Method 4500 Cl- (B).
The question then evolves to: When the pH of the sample is reduced to 2 by the Hach reagents, is the hypochlorite converted to Cl2 gas or does it
decompose to chloride ions. In the one case, it will appear as chloride and the other not. Typically acid + hypo => chlorine gas.
Regarding the manufacture, I agree that the chemistry says one mole of NaCl is produced per mole of NaOCl - but that would mean my 12.5% bleach also
contains about 10% NaCl? That is extremely salty too. Some manufacturers claim their process produces "low salt" bleach - but I have not got any
written clarification on that.
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ajanson
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How much sodium chloride can there be in commercial bleach?
The MSDS label indicates it is 12.5% available chlorine (minimum) by weight, 15% by volume and has a density of 1.202. If it is 12.5% OCl, it would
be about 18% NaOCl by weight. If the sodium chloride is in a 1:1 mole ratio, it would mean that there is 14% by weight of NaCl too. Can I have 18%
by weight NaOCl and 14% NaCl and still have a specific gravity of only 1.2?
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woelen
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Available chlorine is not defined as percentage OCl(-) by weight, but it is defined as follows:
Given 100 gram of the hypochlorite-containing material (be it liquid or solid does not matter) and excess hydrochloric acid is added, then X grams of
chlorine are produced. The percentage available chlorine then is equal to X. Certain compounds even can have over 100% 'available chlorine'. An
example is solid pure LiOCl, which according to definition has appr. 124% available chlorine.
So, 100 grams of your solution will produce 12.5 grams of elemental chlorine, Cl2, when excess HCl is added. Your solution then will contain 12.5 *
MW(NaOCl)/MW(Cl2) grams of NaOCl and 12.5*MW(NaCl)/MW(Cl2) grams of NaCl.
MW(NaOCl) = 74.5
MW(Cl2) = 71
MW(NaCl) = 58.5
So, you'll have 13.1 grams of NaOCl in your solution and 10.3 grams of NaCl in your solution.
[Edited on 25-1-06 by woelen]
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ajanson
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Thank you very much for taking the time to reply. It is much appreciated.
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