Sciencemadness Discussion Board

Measuring hypochlorite concentration

vmelkon - 1-6-2012 at 18:41

How would one measure the concentration of sodium hypochlorite in water?
I suppose I could add H2O2 and see how much oxygen is evolved or measure how much H2O2 I would have to add so that no more O2 is evolved.

H2O2 + ClO- => H2O + O2 + Cl-

The H2O2 would be 3% (from the pharmacy).

Unfortunately, I don't know the molarity of the H2O2. 3% I think is the v/v.
I guess 1.279 M.

Perhaps, I could add HCl and see chlorine gas evolve and keep on adding until no more comes out.

Any other chemical techniques?

PS: I'm an amateur

elementcollector1 - 1-6-2012 at 18:59

Where are you getting the hypochlorite? You can usually find concentration on the label, or an MSDS if the label doesn't say.
A possible method would be to add a soluble iron salt and measure how many moles of iron (III) hydroxide were precipitated out, then compare that with how many moles of NaOCl one would need to precipitate that amount of Fe(OH)3...

barley81 - 2-6-2012 at 03:33

In "Illustrated Guide to Home Chemistry Experiments" by Robert Bruce Thompson, titration of bleach is described. The sample is mixed with KI and acetic acid, and titrated with sodium thiosulfate with starch indicator.

If you mix an iron salt with bleach, some ferrate ion (FeO<sub>4</sub><sup>2-</sup>;) is formed.

vmelkon - 2-6-2012 at 05:28

I am preparing the bleach myself by electrolysis of NaCl.
Quote:
The sample is mixed with KI and acetic acid, and titrated with sodium thiosulfate with starch indicator.


Does that mean the reaction starts off like this?
2HI + ClO- => Cl- + H2O + I2

AJKOER - 2-6-2012 at 05:34

Here is a link to "Illustrated Guide to Home Chemistry Experiments" by Robert Bruce Thompson on the titration of bleach:

http://books.google.com/books?id=9_2idzksARMC&pg=PA71&am...

The equations are:

OCl- + 2 CH3COOH + 2 I- ----> Cl- + 2 CH3COO- + I2 + H2O

forming brown Iodine water. Upon adding starch, an intense blue color develops. The back titration:

2 S2O3-- + I2 ---> 2 I- + S4O6--

removes Iodine decreasing the solution's coloration which remains visible in even in very dilute amounts.

AJKOER - 2-6-2012 at 06:03

Quote: Originally posted by vmelkon  

Does that mean the reaction starts off like this?
2HI + ClO- => Cl- + H2O + I2


The answer on whether the reaction "starts" this way is, in my opinion, no. However, a possible (I2 yield varying with pH, concentrations, temperature, relative molar amounts and even light), net ionic equation is:

2 H+ + 2 I- + ClO- => Cl- + H2O + I2

Possible intermediates include (but not limited to) HIO3, HOI and HI. For those interested in more details see my recent comments on Bromine & Iodine chemistry and links provided.


[Edited on 2-6-2012 by AJKOER]

blogfast25 - 3-6-2012 at 05:54

Quote: Originally posted by AJKOER  
Quote: Originally posted by vmelkon  

Does that mean the reaction starts off like this?
2HI + ClO- => Cl- + H2O + I2


The answer on whether the reaction "starts" this way is, in my opinion, no. However, a possible (I2 yield varying with pH, concentrations, temperature, relative molar amounts and even light), net ionic equation is:

2 H+ + 2 I- + ClO- => Cl- + H2O + I2

[Edited on 2-6-2012 by AJKOER]


In acidic conditions this reaction proceeds quantitatively from left to right to completion. The ΔG is very considerable, at a guess. Iodide is easily oxidised and hypochlorite ('chlorate-1' some call it now) a very powerful oxidiser.

Intermediates like HIO3 are very unlikely (or at best very short lived): the equilibria are what they are and not affected by feverish imaginations. ;)

[Edited on 3-6-2012 by blogfast25]

unionised - 3-6-2012 at 07:07

Since the chlorine is reduced from a positive, to a negative, oxidation state there must be an intermediate species with chlorine in a zero oxidation state. As the total chlorine concentration is rather low this intermediate can not be Cl2.

It's fair to say there is some very peculiar intermediate.

AJKOER - 3-6-2012 at 07:47

Quote: Originally posted by blogfast25  

Intermediates like HIO3 are very unlikely (or at best very short lived): the equilibria are what they are and not affected by feverish imaginations. ;)


I will avoid commenting on Blogfast25's characterization and instead present what I think is an interesting paper on Iodine chemistry.

Reference: "Rate Constants for Reactions between Iodine- and Chlorine-Containing Species...", J. Am. Chem. Soc., Vol. 118, No. 15, 1996, by Lengyel et al. Link: http://hopf.chem.brandeis.edu/pubs/pub234%20rep.pdf

Now, per equation (M6) Table I, page 3716:

HOCl + I- ==> HOI + Cl-

where in the current context, HOCl can be produced by the action of a weak acid on a hypochlorite.

Further, to quote from page 3710:

"Part (d) is a slow disproportionation of HOI and HIO2 to
molecular iodine and iodate. The stoichiometries of the
disproportionations are

5HOI = 2I2 + IO3- + 2H2O + H+ (6)

5HIO2 = I2 + 3IO3- + H2O + 3H+ (7)"

I previously mentioned some of the factors influencing equation (6).

Now, the Iodate can be removed as follows, to quote "This is the rate determining step. The iodate in excess will oxidize the iodide generated above to form iodine:

IO3− + 5 I− + 6 H+ → 3 I2 + 3 H2O

Source: Wikipedia. Link:
http://unblock.xpulsenetwork.com/en.wikipedia.org/wiki/Iodin...

Hence, my reference to the HIO3 intermediate. To prove my net ionic equation, note the following three equations from above:

5 HOCl + 5 I- ==> 5 HOI + 5 Cl-
5HOI ==> 2I2 + IO3- + 2 H2O + H+
IO3− + 5 I− + 6 H+ ==> 3 I2 + 3 H2O
-----------------------------------------------------------------
Net: 5 HOCl + 10 I- + 5 H+ ==> 5 I2 + 5 Cl- + 5 H20

or: HOCl + 2 I- + H+ ==> I2 + Cl- + H2O

as was to be shown. Here is an independent source that verifies this reaction based on HOCl, Acetic acid (HOAc) and Potassium iodide:

HOCl + 2KI + HOAc → I2 + KCl + KOAc + H2O

Source: Kaohsiung Medical University, "20 Applications of Oxidation/Reduction Titrations", top of page 145. Link:
http://sjlin.dlearn.kmu.edu.tw/20.pdf


[Edited on 3-6-2012 by AJKOER]