Sciencemadness Discussion Board - Better/Advanced Chlorate Formation

Better/Advanced Chlorate Formation

AJKOER - 30-10-2013 at 04:54

I appear to have relatively easily (I just decanted, not even filtered) in a closed system (to reduce Chlorine exposure) created clear KClO3 crystals that appear like cactus spikes. I did, interestingly, employ some recent published work (discussed later with references) to expedite the formation.

The logic is based on applying Hypochlorous acid to Mg(ClO)2 and Mg(OH)2 (to control pH so as to be not too acidic thereby reducing the presence of HOCl and OCl- in favor of gaseous Cl2). Any Mg(ClO)2 formed is highly unstable disproportionating readily into Mg(ClO3)2 and MgCl2 with very little heat or even treatment with air/O2. A reference for background, please see "The Manufacture of Sulphuric Acid and Alkali: Ammonia-soda, various ..." by Georg Lunge, an online googlebook, link: http://books.google.com/books?id=FnrTAAAAMAAJ&pg=PA669&a... . To quote from page 669:

"Now chlorine was passed into a milk of- magnesium hydroxide and water, at temperatures between o° and 100°. Even at 0°, together with magnesium hypochlorite, much chlorate was formed, more than corresponding to half of the chlorine entering into the reaction. At 15° a little more chlorate was formed, together with much hypochlorate, some of which was changed into chloride, with evolution of oxygen. In both solutions the hypochlorite is easily converted into chlorate, not merely by heating to 50°, but even by prolonged agitation by a current of air at ordinary temperatures. At 70° C, from the first mostly chlorate was formed, with a little chloride, produced by loss of oxygen. Hence magnesium hypochlorite in statu noscendi does not possess much stability and is easily transformed into chlorate."

The intended path (I got something a bit different as I will explain) was as follows. First, prepare a concentrated slurry of Dibasic Magnesium hypochlorite (see my thread at http://www.sciencemadness.org/talk/viewthread.php?tid=26741 ) as follows:

9 MgSO4 + [ 6NaClO + 12NaOH ] --> 9 Na2SO4 + 3 Mg(ClO)2 .2Mg(OH)2 (s)

Cool to collapse the colloidal suspension of DBMH and decant and proceed to acidify the extracted DBMH as follows:

2 Mg(ClO)2 .2Mg(OH)2 + 12 NaHSO4 --> 6 Na2SO4 + 6 MgSO4 + 4 HOCl + 8 H2O
2 Mg(ClO)2 .2 Mg(OH)2 + 4 HOCl --> 2/3 Mg(ClO3)2 + 4/3 MgCl2 + 4HCl + 4 Mg(OH)2
4 HCl + 2 Mg(OH)2 --> 2 MgCl2 + 4 H2O

For a net reaction:

4 Mg(ClO)2 .2Mg(OH)2 + 12 NaHSO4 --> 6 Na2SO4 + 6 MgSO4 + 2 Mg(OH)2 + 2/3 Mg(ClO3)2 + 10/3 MgCl2 + 12 H2O

I actually based my stoichiometric calculations on the above equation.

However, I used a new extra strength 8.25% bleach (Great Value brand from Walmart ) that actually has added Na2CO3 (imitating Chlorox bleach). As a result, I immediately witness a creation of some Basic magnesium carbonate hydrate that nearly consumed all of my solution. Likely reaction:

6 Na2CO3 + 4 MgSO4 + 10 H2O --> 2 MgCO3.Mg(OH)2 .3H2O + 4 NaHCO3 + 4 Na2SO4

I resolved this issue by adding 5% vinegar listening for a cessation in any bubbling (around 20 cc). If I have not done so, the acidifying would have proceeded as follows:

4 NaHSO4 + 2 MgCO3.Mg(OH)2 .3H2O --> 2 Na2SO4 + 2 MgSO4 + 8 H2O + 2 Mg(OH)2 + 2 CO2

After acidifying with vinegar/NaHSO4, I also added NaCl. I then heated the solution mildly with two hot water baths of recently boiled water. I also subject the mixture to sunlight for 4 hours with an occasional shaking to increase contact between any uv treated Chlorine gas and the solution (no longer being heated). I then cooled the mixture and decanted any residue (lost around 10% of the solution). I then added this mix to a jar containing around 20 cc of K2SO4 (better would be KCl) and cooled again. With 30 minutes, the clear cactus like crystals of KClO3 appeared. I just poured out the solution leaving the crystals and took the picture below [EDIT] Actually that picture was too large, attached is a small alternate picture.
--------------------------------------------------------------------

Now, I did apparently employed some recent chemistry on the suggested preparation path to a chlorate. In particular, the use of Acetic acid/acetate, NaCl and sunlight. Some sources: to quote a 2000 paper (link: http://pubs.acs.org/doi/abs/10.1021/ic991486r ):

"Acetic acid has a second catalytic role through the formation of acetyl hypochlorite, which is much more reactive than HOCl in the transfer of Cl+ to ClO2- to form ClOClO."

Another article, "Effect of Chloride Ion on the Kinetics and Mechanism of the Reaction between Chlorite Ion and Hypochlorous Acid" link:
http://www.researchgate.net/publication/23141635_Effect_of_c... , to quote:

"Moreover, they found that acetate ion accelerates the formation of ·ClO2 enormously." and also "It was interpreted by a steady-state formation and further reactions of acetyl hypochlorite. "

With respect to the role of chlorides in promoting chlorate formation, the authors states in the abstract, to quote:

"It is found that the presence of the chloride ion significantly increases the initial rate of ·ClO2 formation."

Cited reactions involving active chlorine species to accelerate the formation of a chlorate include:

Cl2O2+ H2O --> ClO3- + Cl- + 2 H+ (page 2 eq 5)
Cl2O2 + HOCl --> ClO3- + Cl2 + H+ (page 2 eq 8)

where Acetyl hypochlorite is expected in having a catalytic role.

Photolysis most likely proceeds along the following paths involving the species ·Cl, ·OH and ·ClO:

Cl2 + hv --> 2 ·Cl
2 ·Cl + 2 HOCl --> 2 ·OH + 2 Cl2 (g)
2 ·OH + 2 HOCl --> 2 H2O + 2 ·ClO (forming some Cl2O2)
Cl2O2 + HOCl --> HClO3 + Cl2(g)

where the presence of sunlight, free Chlorine and Hypochlorous acid are required in this particular chain.

Source: See for example "Photolysis of free chlorine species (HOCl and OCl- ) with 254 nm ultraviolet light" page 281 attached, and also Table 2, page 797 at http://www.geosci-model-dev-discuss.net/3/769/2010/gmdd-3-76... .

Attachment: 278FengSmithBolton2007.pdf (171kB)
This file has been downloaded 682 times

[Edited on 30-10-2013 by AJKOER]

blogfast25 - 30-10-2013 at 09:30

This could be a lot more interesting to quite a few people, if:

* you could write up a recipe for a practical method, based on the asserted chemistry
* you determine Actual Yield % of KClO3 for a typical run
* you determine approx. cost per kg of the product

elementcollector1 - 30-10-2013 at 09:34

Given that the only reagents were Na₂CO₃, MgSO₄, K₂SO₄, acetic acid and bleach, I'm very interested to see a comparison of this to electrolytic KClO₃ yield.

AJKOER - 30-10-2013 at 10:10

Quote: Originally posted by blogfast25

This could be a lot more interesting to quite a few people, if:

* you could write up a recipe for a practical method, based on the asserted chemistry
* you determine Actual Yield % of KClO3 for a typical run
* you determine approx. cost per kg of the product

Good points, and here is my immediate responses:

1. Per my discussion of exactly what did occur (the unexpected Na2CO3 surprise in the Bleach), my quoted net reaction actually did not take place. As such, I am hesitant to suggest quantities for a reaction (via Dibasic magnesium hypochlorite) that I did not precisely perform and may differ significantly in outcome due to an important pH difference (the not precisely known quantity of Na2CO3 in the Bleach). Also, my correction method of listening for bubbles to add sufficient dilute vinegar is not precise either. Bottom line, more work has to be done including repeating the experiment with a different (non-Na2CO3) bleach.

2. I have only a rough estimate of yield, which appears to be equal or perhaps exceeded by my postulated net reaction. Again, more analysis required.

3. With respect to cost, I would state that my chlorate preparation is more likely expensive as compared to electrolysis as MgSO4 is more costly than KCl (the latter being available in bulk for water purification). I am not considering the cost of replacing electrodes, transformers, .etc. I am also assuming that one can ignore the cost of time (around 6 hours to extraction of the KClO3 for my preparation versus I would guess at least a day for electrolysis) and the vary cost of electricity. Another important cost related variable is batch size limitation. Other than the employment of photolysis for potential yield improvement, my preparation is easily scalable. I will let others comment on the electrolysis method for KClO3 formation on this last point.

One could argue that my preparation is more efficient as Cl2 is not allowed to escape and is used to foster the formation of HOCl and chlorate. Also, one may be able to recover/recycle the MgSO4.

[Edited on 30-10-2013 by AJKOER]

bbartlog - 1-11-2013 at 08:13

If the decomposition of magnesium hypochlorite to chloride and chlorate is rapid, wouldn't it make sense to just add *some* magnesium and acetic acid to your commercial bleach, rather than try for stoichiometric amounts? In that case it would basically be functioning as a catalyst. Maybe add 1/20 of an equivalent of MgSO4 to your commercial 8.25% NaOCl solution, then slowly add acetic acid just until all of the precipitate dissolved (or all the turbidity cleared). Just seems like it would be more economical than using all of that NaHSO4 and magnesium sulfate.
Also, I notice in your equations that you've assumed that the molar concentration of NaOH in bleach is twice the concentration of the actual NaOCl. What is the basis for that? Just from memory I seem to recall that they only use about 1% NaOH to stabilize the bleach against decomposition. It's potentially a dangerous assumption to make, because if someone works from that basis and tries to neutralize such a large quantity of NaOH in their bleach with something acidic (like NaHSO4

they will end up gassing off quite a lot of chlorine.

AJKOER - 1-11-2013 at 09:01

I agree with bbartlog comment that the proposed preparation perhaps needless consumes too much NaHSO4.

I am working on a process with the goal of producing HOCl/Cl2 in one reaction (which employs NaHSO4) and then reacting the products with a mix of Mg(ClO)2/Mg(OH)2/MgCO3. My listed reaction citing NaOH is not from existing Sodium hydroxide in bleach (I concur there is a little NaOH added to bleach for stability), but I am actually adding another ingredient (NaOH) to replicate the formation of Dibasic magnesium hypochlorite. However, I am having second thoughts on this NaOH addition as it does eventually requires more NaHSO4, per your point. Further, until I replicate a success in this potentially more efficient path, I am not disclosing the reaction chain and amounts employed.

It has been overcast where I am for the last two days, but I am expecting possibly an adequate supply of sunlight tomorrow. The photolysis of HOCl to chlorate is a known problem for the maintenance of HOCl level in swimming pools on sunny days. To me, the interesting part of this observation is that photolysis proceeds in even dilute solutions (like the swimming pool) as a function of uv levels. As such, I will experiment with removing the photolysis part of the preparation when I have replicated KClO3 formation.

I also find my preparation a bit different from the usual chlorination paths (for example, the treatment of a thick Mg(OH)2 slurry with chlorine) as my solutions are not slurries or especially concentrated. However, I am employing highly ionic solutions rich in MgCl2, a salt which is known to increase the activity level of even dilute acids. Interestingly, NaCl has a similar effect on activity level, but to a lesser extent.

[Edited on 1-11-2013 by AJKOER]

MrHomeScientist - 1-11-2013 at 11:41

Quote: Originally posted by AJKOER

Further, until I replicate a success in this potentially more efficient path, I am not disclosing the reaction chain and amounts employed.

Why not? That seems rather against the spirit of scientific discourse.
Lots of people are surely interested and would like to try replicating your work to confirm its "better"-ness (poor attempt to use a word from the thread title). Are you seeking a patent or something?

woelen - 1-11-2013 at 12:48

You really need to give a procedure, like blogfast25 suggested. Experimental setup, precise quantities used, etc. If this really works, even in fairly low yield, then it is very interesting, because it opens up a method of obtaining chlorates for many people in small quantities, without the need to construct a cell with suitable anode and power supply, and without the need of using hard to obtain or restricted chemicals.

If you do not want to disclose amounts used and other relevant details, then you should not have posted this at all in the first place. An experimenter's site like sciencemadness is for discussing things, such that other people can comment on your procedures and may improve on it. Non-disclosure is OK to me, but then keep the whole thing for yourself and stay away from here.

[Edited on 1-11-13 by woelen]

bbartlog - 1-11-2013 at 20:06

But for those who just want chlorate in small yield and don't care about efficiency, it is easier and simpler to just simmer commercial sodium hypochlorite bleach overnight (or better 24 hours), add a saturated solution of KCl, and stick the container in the refrigerator/freezer to precipitate potassium chlorate. I thought I had posted something about the time I did this, but apparently not (at least I can't find it in my posts).
The method proposed by AJKOER may be faster or have somewhat better yields, but in my opinion it's not enough of an improvement to justify the use of the additional reagents and the greater complexity.

AJKOER - 2-11-2013 at 13:15

For those who want to know what I am trying to do now, here is a summary:

Preparation of the 1st Solution:

4 NaClO + 2 MgSO4 --> 2 Mg(ClO)2 + 2 Na2SO4
(2-x) Mg(ClO)2 + 2(2-x) NaHSO4 --> 2(2-x) HOCl +...
x Mg(ClO)2 + (2x) HAc --> 2x HOCl + ...

where I am acidifying with both Acetic acid and NaHSO4 to form Hypochlorous acid in the 1st solution. However, for simplicity, assign a value of zero to 'x' (moles of Acetic acid) and use the full amount of NaHSO4 and an arbitrary non-zero amount of vinegar.

Net reaction:
4 NaClO + 2 MgSO4 + 2(2-x) NaHSO4 + (2x) HAc --> 4 HOCl + (4-x) Na2SO4 + x Mg(Ac)2 + (2-x) MgSO4

Second mixture to be combined with the 1st is prepared by combining:

2 NaClO + MgSO4 --> Mg(ClO)2 + Na2SO4 with
3 Na2CO3 + 2 MgSO4 + 5 H2O --> MgCO3.Mg(OH)2 .3H2O + 2 NaHCO3 + 2 Na2SO4

and also add NaCl, for a net composition of the 2nd solution of:

2 NaClO + 3 Na2CO3 + 3 MgSO4 + 5 H2O + y NaCl--> Mg(ClO)2 + MgCO3.Mg(OH)2 .3H2O + 2 NaHCO3 + 3 Na2SO4 + y NaCl

Upon combining Solution 1 and 2, the hypochlorite with mild heating (hot water bath) and sunlight hopefully forms chlorate:

Mg(ClO)2 + 2 HOCl --> 2 HCl + Mg(ClO2)2
Mg(ClO2)2 + 2 HOCl --> 2 HCl + Mg(ClO3)2

forming chlorate as the HCl is removed (via MgCO3, NaHCO3 and Mg(OH)2) thereby fostering the creation of HOCl (to create Mg(ClO3)2 by acting on Mg(ClO)2).

For the record, the actual reaction paths, I suspect, are more complex, but the above net reaction equations may provide good yields.
----------------------------------------------------

Some numbers as requested:

So, for a target on Mg(ClO3)2 (1 mole = 191.2080 g per mole is 10 cc. 10*2/191.208 implies a target of .105 moles.

Required total NaOCl.46H2O with MgSO4 is 901.8181 * 6* .105/1.1 density = 516.5 cc with 2/3 in 1st solution and 1/3 in the latter. Half reaction amounts: 172.1 cc and 86.1

Required MgSO4: 5*.105*246.47/1.68 = 46.2/3*5 cc = 77 cc where 2/5 for 1st and 3/5 for 2nd. Half reactions amount: 15.4 cc and 23.1 cc.

Required NaHSO4.H2O = 4* .105* (120.06+18)/2.742 density = 21.14 cc all in the 1st solution. Half: 10.6 cc.

Required Na2CO3 is 106* 3* .105/2.54 density = 13.1 cc all in 2nd solution. Half 6.6 cc.

Add NaCl equal to moles of HOCl: 4*.105*58.44/2.165 = 11.3 cc to the 2nd solution.
==============================================================================

I am currently trying to execute this preparation. I probably will have to repeat as I have some accidentally mixing of the Na2CO3 and NaHSO4, slow to add Acetic acid and NaCl.

No assertion of the success of this modified procedure (hence my inclination to wait).

[Edited on 3-11-2013 by AJKOER]

AJKOER - 2-11-2013 at 14:26

Quote: Originally posted by bbartlog

But for those who just want chlorate in small yield and don't care about efficiency, it is easier and simpler to just simmer commercial sodium hypochlorite bleach overnight (or better 24 hours), add a saturated solution of KCl, and stick the container in the refrigerator/freezer to precipitate potassium chlorate. I thought I had posted something about the time I did this, but apparently not (at least I can't find it in my posts).
The method proposed by AJKOER may be faster or have somewhat better yields, but in my opinion it's not enough of an improvement to justify the use of the additional reagents and the greater complexity.

Actually, I recall reading that the best temperature for the NaClO3 formation is around 70 C, or may be a bit higher, but not boiling unless the intent is also to concentrate the solution. However, cooling the solution and extracting a hypochlorite hydrate may be a better path for concentrating as some aqueous chlorates will undergo some decomposition (releasing oxygen) on heating/boiling.

Now, with respect to the cost of the additional reagents, I am 'borrowing' NaHCO3 (and heating to form Na2CO3), vinegar and even the Epsom salts from the household supplies (an economist could call these sunk costs), so one could argue, for small scale preparations, to omit the cost factor of these extra ingredient.

Now on the complexity argument, I agree to some extent true. But, if one is following a recipe, there is only marginally more complexity by adding to the ingredient list. Understanding what one is doing, now that is another matter.

[Edited on 2-11-2013 by AJKOER]

bbartlog - 2-11-2013 at 20:31

Decomposition of hypochlorite to chlorate speeds up by a factor of 3.5x for every 10 degrees celsius increase in temperature. There is no optimum at 70C. Although the decomposition to chlorate (rather than oxygen) predominates over the decomposition to NaCl + O2 at every temperature and pH in the absence of catalytic metal ions, it's possible to favor it even more by dropping the pH to 9 or so.

blogfast25 - 3-11-2013 at 04:30

Quote: Originally posted by AJKOER

----------------------------------------------------

Some numbers as requested:

So, for a target on Mg(ClO3)2 (1 mole = 191.2080 g per mole is 10 cc. 10*2/191.208 implies a target of .105 moles.

Required total NaOCl.46H2O with MgSO4 is 901.8181 * 6* .105/1.1 density = 516.5 cc with 2/3 in 1st solution and 1/3 in the latter. Half reaction amounts: 172.1 cc and 86.1

Required MgSO4: 5*.105*246.47/1.68 = 46.2/3*5 cc = 77 cc where 2/5 for 1st and 3/5 for 2nd. Half reactions amount: 15.4 cc and 23.1 cc.

Required NaHSO4.H2O = 4* .105* (120.06+18)/2.742 density = 21.14 cc all in the 1st solution. Half: 10.6 cc.

Required Na2CO3 is 106* 3* .105/2.54 density = 13.1 cc all in 2nd solution. Half 6.6 cc.

Add NaCl equal to moles of HOCl: 4*.105*58.44/2.165 = 11.3 cc to the 2nd solution.

I doubt if anyone here could possibly replicate your experiments based on that information. What do you mean by 'cc'? Cubic centimeter?

And "NaOCl.46H2O": is that a typo?

"106* 3* .105/2.54 density = 13.1 cc", so I assume this actually means:

(106 g/mole x 3 x 0.105 mole) / 2.54 g/cm³ = 13.1 cm³. But why express the quantities of reagents used in volume units when they are solids??? If you don't have any scales, you could use volumes instead but would have to use the apparent density of the product, not the actual density of solid anhydrous Na2CO3. Apparent densities are always lower. And were you actually using anhydrous Na2CO3? The household version of Na2CO3 is a mixture of hydrates, mainly 10 and 7 I think.

[Edited on 3-11-2013 by blogfast25]

AJKOER - 3-11-2013 at 05:47

Sorry for the format issues, the calculations where largely copied and pasted from my work file.

OK, NaOCl has a molar mass of 74.442 and H2O has a molar mass of 18.015. So the percent NaOCl in a hypothetical hydrate (I do work with actual hypochlorite hydrates, but this is more shorthand) is 100*(74.442/(74.442 + 46*18.015)) = 8.24% hypochlorite by weight. The extra strength bleach labels claims to be 8.25% NaOCl by weight. More correctly, and I should adjust for (but have not so far, hence the approximate volumetric route) the fact that the bleach is at most 8.25% (probably less due to decomposition) with varying amounts of added NaOH and/or Na2CO3 between brands.

The Na2CO3 was prepared by heating (with stirring) NaHCO3.

Now, I argue for a quick synthesis (and the ease of quickly replicating) when mixing two solids, I use volumes corrected by the actual density for both solids. In other words, if the apparent/actual density is the same ratio for two compounds, the stoichiometric equation remains accurate. The error arises from the absolute difference in the respective factors that corrects the actual density to the apparent density for each solid. Compressing solids with largely different lattice structures and intermolecular attractions may reduce this error. The latter is technically termed tapped density (which I have been employing) and for a free flowing salt, the term is apparent or bulk density (as it is also called, see, for example, discussion at http://www.pharmacopeia.cn/v29240/usp29nf24s0_c616.html ).Clearly not precise, but neither is precisely weighing a salt that absorbs moisture and ignoring the water weight.

[Edited on 3-11-2013 by AJKOER]

blogfast25 - 3-11-2013 at 09:50

It amuses me no end how you calculate things using 5 significant digits data and then end up measuring quantities of solids... in volumes (cc), using erm... 'tapped densities'. There's nothing to be gained from that and nobody does it like that: gravimetry is both faster AND more accurate than volumetry, at least in most cases.

Anhydrous Na2CO3 can easily be weighed very accurately. The very pure form is used as a primary standard in acidometry.

AJKOER - 3-11-2013 at 20:16

Got some results from my latest run. Again more cactus like crystals grew (picture inserted below). I modified the procedure from what I discussed previously and was not able to get any benefit for this run of any photolysis (clouds).

Revised Solution 1 Preparation: Add Epsom salt to a 8.25% Bleach with added Sodium hydroxide. This order of reactant additions does not actually form Dibasic magnesium hypochlorite, but a Magnesium hypochlorite in a slurry of white Magnesium hydroxide, which is fine for the current target preparation of Mg(ClO3)2:

3 MgSO4.7H2O + [2 NaOCl + 4 NaOH] --> Mg(ClO)2 + 2Mg(OH)2 (c) + 3 Na2SO4

Second Solution (intended solely for HOCl/Cl2 formation):

4 NaClO + 2 MgSO4 --> 2 Mg(ClO)2 + 2 Na2SO4
(2-x) Mg(ClO)2 + 2(2-x) NaHSO4 --> 2(2-x) HOCl +...
x Mg(ClO)2 + (2x) HAc --> 2x HOCl + ...

Net reaction plus additional NaCl to promote the formation of chlorate:

4 NaClO + 2 MgSO4 + 2(2-x) NaHSO4 + (2x) HAc + 4 NaCl --> 4 HOCl + (4-x) Na2SO4 + x Mg(Ac)2 + (2-x) MgSO4 + 4 NaCl

Upon combining Solution 1 and Solution 2:

Mg(ClO)2 + 2 HOCl --> 2 HCl + Mg(ClO2)2
Mg(ClO2)2 + 2 HOCl --> 2 HCl + Mg(ClO3)2

There is a very obvious significant formation of Chlorine (and I suspect little HOCl) until the addition of another dose of 4 NaOH to remove the HCl and drive the formation of chlorate (I did cooled the reaction vessel in a freezer before attempting this addition).

The next step is to cool the solution and filter out (or even decant) the Na2SO4 clumps. To the clear filtered solution, I then added K2SO4 (not KCl) to eventually form the chlorate crystals. I no longer recommend KCl since if the final solution is acidic, the KCl can reduce any chlorate to a mix Cl2 and ClO2, or hypochlorite to Cl2, both evident by the solution turning green.
--------------------------------------------------------------------------------------------------

Approximate calculations for a target of Mg(ClO3)2 of .105 moles (1 mole = 191.2080 g per mole):

Required 8.25% hypochlorite (NaOCl + 46H2O) is 901.8181 * 5* .105/1.1 density = 430.4 cc with 3/5 in 1st solution (258 cc) and 2/5 (172 cc) in the 2nd Solution. Half reaction: 129 cc and 86 cc.

Required MgSO4: 5*.105*246.47/1.68 = 77 cc with 3/5th for 1st Solution ( 46.2 cc )and 2/5 (30.8 cc) for 2nd Solution. Half reactions: 23.1 and 15.4.

Required NaHSO4.H2O = 4* .105* (120.06+18)/2.742 density = 21.1 cc all in 2nd solution. Half: 10.6 cc.

Required NaOH is 40* 4* .105/2.13 density = 7.9 cc all in 1st solution. Followed by a follow-up addition upon significant Cl2 generation of another 7.9 cc. Half 3.9 cc.

Add NaCl equal to moles of HOCl: 4*.105*58.44/2.165 = 11.3 cc.

If we did not use any NaHSO4 to form the HOCl we would need dilute 5% Acetic acid equal to: 4*.105*(60.05+18*63.386)= 504 cc. Targeting the amount of the vinegar at say 5% of this amount, or 25.22 cc, then we need 95% of the NaHSO4 input, per above, or .95*21.1 = 20 cc NaHSO4.
====================================================================

[EDIT] My assessment of the yield without a photolysis phase in my suggested preparation of chlorate is low.

I may repeat with UV dosing or test out a whole new electrochemical approach based on a previous thread analysis by myself of the so called 'bleach battery' (actually operates on HOCl in the presence of Al and Cu).

[Edited on 4-11-2013 by AJKOER]

PHILOU Zrealone - 5-11-2013 at 07:38

What about Cl2O formation? HOCl is relatively unstable and turns into Cl2O(g).

Isn't the chloride anion conterproductive with the chlorate formation in acidic media?
This is one of the reason why HClO4 can't be produced from NaClO3 and HCl...it turns into Cl2...the conversion of chloric acid to perchloric acid succeeds with NaClO3 and diluted H2SO4...

[Edited on 5-11-2013 by PHILOU Zrealone]

AJKOER - 7-11-2013 at 17:55

Yes, I agree Philou Zrealone, chloride anion is counterproductive with chlorate in an acidic media. In fact, I recall that one can generate ClO2 (along with some Cl2, but not Cl2O) with a chlorate in the presence of plain salt (NaCl acting as a reducing agent) in the presence of a strong acid. Here is a source: "Industrial Inorganic Chemistry" by Karl Heinz, page 173, link: http://books.google.com/books?id=pGDP98XYnA4C&pg=PA173&a...

However, in less acidic conditions where HOCl and OCl- survive, the action of the chloride anion (per the references I supplied in the opening thread) is claimed to be actually catalytic for chlorate formation.

[Edited on 8-11-2013 by AJKOER]

Finnnicus - 7-11-2013 at 23:03

What about the use of Ca(ClO)₂ as the hypochlorite source?
The CaSO₄ should precipitate quite nicely.
Just my 2¢.

PHILOU Zrealone - 9-11-2013 at 11:02

Quote: Originally posted by AJKOER

Yes, I agree Philou Zrealone, chloride anion is counterproductive with chlorate in an acidic media. In fact, I recall that one can generate ClO2 (along with some Cl2, but not Cl2O) with a chlorate in the presence of plain salt (NaCl acting as a reducing agent) in the presence of a strong acid. Here is a source: "Industrial Inorganic Chemistry" by Karl Heinz, page 173, link: http://books.google.com/books?id=pGDP98XYnA4C&pg=PA173&a...

However, in less acidic conditions where HOCl and OCl- survive, the action of the chloride anion (per the references I supplied in the opening thread) is claimed to be actually catalytic for chlorate formation.

[Edited on 8-11-2013 by AJKOER]

I was speaking of Cl₂O from the starting HOCl.
In concentrated environment:
2 HOCl (l)<--==> Cl₂O (g) + H₂O (l)

For the rest:
2 NaOClO₂ + H₂SO₄ --> Na₂SO₄ + 2 HOClO₂
4 HOClO₂ --> 3 HOClO₃ + HCl
but
HOClO₂ + HCl --> ClO₂ + Cl₂ + H₂O
And this explains why the process with H₂SO₄ has a yield of hydrogen perchlorate vs chlorate of 50% instead of 75% and why the same proces with HCl doesn't work at all

[Edited on 9-11-2013 by PHILOU Zrealone]

AJKOER - 28-12-2013 at 20:15

SUCCESS!

I appear to have relatively easily created KClO3 in good yield without nearly any sophisticated equipment employing some recent published work to expedite chlorate formation. The method, however, is based on photolysis and takes days in good uv conditions (like snow, for example).

The process is based on simply acidifying a 8.25% Chlorine bleach with Acetic acid (actually used vinegar), adding NaCl and placing the solution in a glass vase sealed with a thin layer of clear plastic foil (to allow uv rays to react with any escaping Chlorine from the solution that would otherwise be blocked by the thick glass). The volume ratio employed for the 8.25% bleach was 2.07 parts to one part vinegar. Speculation on the reaction chain:

7 NaOCl + 7 HAc --> 7 NaAc + 7 HOCl
2 HOCl --uv--> 2 HCl + O2
HCl + HOCl = Cl2 + H2O
HCl + NaOCl --> NaCl + HOCl

Net:
8 NaOCl + 7 HAc --> 7 NaAc + NaCl + H2O + O2 + Cl2 + 5 HOCl

And, upon adding NaCl also to expedite the chlorate formation:

8 NaOCl + 7 HAc + 5 NaCl --> 7 NaAc + 6 NaCl + H2O + O2 + Cl2 + 5 HOCl

Further photolysis most likely proceeds along the following paths involving the species ·Cl, ·OH and ·ClO:

Cl2 + hv --> 2 ·Cl
2 ·Cl + 2 HOCl --> 2 ·OH + 2 Cl2 (g)
2 ·OH + 2 HOCl --> 2 H2O + 2 ·ClO (forming, at most, one Cl2O2 as there are poisoning reaction paths)
Cl2O2 + HOCl --> HClO3 + Cl2(g)

forming at most one mole of chlorate for each eight moles of NaOCl given adequate sunlight, pH, ionic strength and solution concentrations.

Source: See for example "Photolysis of free chlorine species (HOCl and OCl- ) with 254 nm ultraviolet light" page 281 attached, and also Table 2, page 797 at http://www.geosci-model-dev-discuss.net/3/769/2010/gmdd-3-76... .
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Now, some references on recent chemistry on paths to chlorate. In particular, the use of Acetic acid/acetate, NaCl and sunlight. Some sources: to quote a 2000 paper (link: http://pubs.acs.org/doi/abs/10.1021/ic991486r ):

"Acetic acid has a second catalytic role through the formation of acetyl hypochlorite, which is much more reactive than HOCl in the transfer of Cl+ to ClO2- to form ClOClO."

Another article, "Effect of Chloride Ion on the Kinetics and Mechanism of the Reaction between Chlorite Ion and Hypochlorous Acid" link:
http://www.researchgate.net/publication/23141635_Effect_of_c... , to quote:

"Moreover, they found that acetate ion accelerates the formation of ·ClO2 enormously." and also "It was interpreted by a steady-state formation and further reactions of acetyl hypochlorite. "

With respect to the role of chlorides in promoting chlorate formation, the authors states in the abstract, to quote:

"It is found that the presence of the chloride ion significantly increases the initial rate of ·ClO2 formation."

Cited reactions involving active chlorine species to accelerate the formation of a chlorate include:

Cl2O2+ H2O --> ClO3- + Cl- + 2 H+ (page 2 eq 5)
Cl2O2 + HOCl --> ClO3- + Cl2 + H+ (page 2 eq 8)

where Acetyl hypochlorite is expected in having a catalytic role.
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Some calculations:

For a 5% vinegar solution by weight, 60.05/(60.05+18x)=.05, so 1 + x18/60.05 = 20 or x=19*60.05/18=63.386, implying a molecular weight for dilute vinegar of 1,201 g/mol. Relative to dilute 8.25% Bleach (NaOCl+46H2O), 8/7*901.8181*1.01/(1.1*1201)=0.79 at least. I assumed 2.07 parts in volume of NaOCl to each part of Vinegar for the current run. Then, 3.07y = 1575 cc, which is the volume of the vase, implies y = 513 cc of Vinegar and 1,062 cc NaOCl. Moles of NaOCl= 1,062cc*1.1/901.8181 = 1.295 moles.

So expected chlorate is at most: 1.295/8 = .162 moles (or, KClO3 = .162*122.5495/2.34 = 8.5 cc).

Required NaCl = 5/8*1.295*58.44/2.165 = 21.8 cc.
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Below, please find a picture on the yield which is difficult to measure as I have apparently exceeded my expectations based on perhaps added NaCl falling out of solution (I used lite salt KCl/NaCl as my source of KCl in this run). I measured some 30cc of the moist precipitate as seen in the upside down picture.

During the experiment when in sunlight, I witness evident O2 evolution which suggested using Mg(ClO)2 in place of NaOCl in the future given its increase sensitivity to the action of oxygen. To quote a reference see "The Manufacture of Sulphuric Acid and Alkali: Ammonia-soda, various ..." by Georg Lunge, an online googlebook, link: http://books.google.com/books?id=FnrTAAAAMAAJ&pg=PA669&a... on page 669:

"Now chlorine was passed into a milk of- magnesium hydroxide and water, at temperatures between o° and 100°. Even at 0°, together with magnesium hypochlorite, much chlorate was formed, more than corresponding to half of the chlorine entering into the reaction. At 15° a little more chlorate was formed, together with much hypochlorate, some of which was changed into chloride, with evolution of oxygen. In both solutions the hypochlorite is easily converted into chlorate, not merely by heating to 50°, but even by prolonged agitation by a current of air at ordinary temperatures. At 70° C, from the first mostly chlorate was formed, with a little chloride, produced by loss of oxygen. Hence magnesium hypochlorite in statu noscendi does not possess much stability and is easily transformed into chlorate."

[EDIT] I tested the KClO3 by wrapping a small amount in Al foil and heating. The foil was entirely consumed. I also mixed equal amounts of brown sugar and the KClO3. On heating, a very impressive self-sustaining reaction.
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Interestingly, I also came across a 2012 paper suggesting that the process does not stop at chlorate formation solely. Per "Perchlorate production by photodecomposition of aqueous chlorine solutions" (link: http://www.ncbi.nlm.nih.gov/pubmed/22962844 ), to quote:

"The amount of ClO(4)(-) produced depends on the starting concentrations of Cl(2,T) and ClO(3)(-), UV source wavelength, and solution pH, but it is independent of light intensity. We hypothesize a mechanistic pathway derived from known reactions of Cl(2,T) photodecomposition that involves the reaction of Cl radicals.."

[Edited on 29-12-2013 by AJKOER]

blogfast25 - 29-12-2013 at 09:58

You don't seem to mention introducing the KCl anywhere.

What was the run time for this run?

Now if you transcribe that post into a simple recipe with gram for solids and ml for liquids, a sober description of methodology ('do this, then do that'), I think you may have some takers wanting to try that. I for one might give it a shot.

An estimated overall net reaction equation would also be useful.

I'm sceptical about the amount of UV you've actually introduced into the system but I'm no expert on UV in sunlight/daylight. If UV really does play such an important part then input could be maximised by operating in shallow white trays, covered with PE cling and 'aimed' at the sun.

Purity of the KClO3 could be measured relatively easily by capturing the amount (volume) of O2 that is released per gram of KClO3, by prolonged modest heating. Even weight loss would work as it contains about 39.2 % oxygen, if pure.

[Edited on 29-12-2013 by blogfast25]

AJKOER - 29-12-2013 at 12:08

Quote: Originally posted by blogfast25

You don't seem to mention introducing the KCl anywhere.

What was the run time for this run?

Now if you transcribe that post into a simple recipe with gram for solids and ml for liquids, a sober description of methodology ('do this, then do that'), I think you may have some takers wanting to try that. I for one might give it a shot.

An estimated overall net reaction equation would also be useful.

I'm sceptical about the amount of UV you've actually introduced into the system but I'm no expert on UV in sunlight/daylight. If UV really does play such an important part then input could be maximised by operating in shallow white trays, covered with PE cling and 'aimed' at the sun.....

On the net reaction, the uncertainty in the photolysis paths to the formation of HClO3 (and chlorate) is such that a precise moles of HOCl to KClO3 depiction is probably not appropriate in my opinion. However, based on a paper studying the UV effects on swimming pool chlorination may provide some guideline. To quote from "Effects of UV-dechloramination of swimming pool water on the formation of disinfection by-products: A lab-scale study" by Nicolas Cimetiereab and Joseph De Laatc available at http://hal.archives-ouvertes.fr/docs/00/91/65/00/PDF/Effects...

"The data showed that UV irradiation led to a 90 % photodecay of free chlorine for UV doses ranging from 13 to 20 kJ m-2, to the formation of chlorate ion (0.05-0.11 mole of chlorate/mole of free chlorine decomposed) and to a significant increase in the chlorine demand of pool water during the post-chlorination step."

Now, the run time also is a bit unclear, as the snow did not persist (usual warm spell rapidly removed it) and the skies where not always clear (although, even cloudy days allow UV radiation exposure). The experiment, nevertheless, lasted 6 days in total.

Sunlight, itself, as I am speculating in my future experiment, may not be the only 'exposure' factor, perhaps the direct oxidation of the hypochlorite (and the chloride?) with oxygen per the reaction:

HOCl ---uv--> HCl + O

Also, the very shape of reaction vessel (a long vase) permitting the bubbling of a column of freshly released oxygen through the solution, may be significant. Also, the very thin plastic covering on the large mouth of the vessel may promote the formation of Chlorine radicals contributing to the chlorate formation.

Yes, I should have been a little clearer on the use of Lite salt (KCl, NaCl,...) to extract the chlorate. The precipitate may consist of KCl.NaCl.2H2O in addition to KClO3.

In the future Mg(ClO)2 run (awaiting the snow), I may test the use of acetone to capture the Mg(ClO3)2 as was suggested in a patent.
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[EDIT] Based on the lab study referenced above, for those insisting on best reaction sequence:

19 NaOCl + 19 HAc --> 19 NaAc + 19 HOCl
4 HOCl --uv--> 4 HCl + 2 O2
3 HCl + 3 HOCl = 3 Cl2 + 3 H2O
HCl + NaOCl --> NaCl + HOCl

The total reaction so far:
20 NaOCl + 19 HAc --uv--> 19 NaAc + NaCl + 3 H2O + 2 O2 + 3 Cl2 + 13 HOCl

And, upon adding NaCl also to expedite the chlorate formation:

20 NaOCl + 19 HAc + 13 NaCl --> 19 NaAc + 14 NaCl + 3 H2O + 2 O2 + 3 Cl2 + 13 HOCl

Further photolysis most likely proceeds along the following paths involving the species ·Cl, ·OH and ·ClO:

3 Cl2 + hv --> 6 ·Cl
6 ·Cl + 6 HOCl --> 6 ·OH + 6 Cl2 (g)
6 ·OH + 6 HOCl --> 6 H2O + 6 ·ClO (forming, per assumption, one Cl2O2 as there are several poisoning reaction paths)
Cl2O2 + HOCl --> HClO3 + Cl2(g)

So, the estimated total net reaction would be:

20 NaOCl + 19 HAc + 13 NaCl --uv--> 18 NaAc + HAc + 14 NaCl + 9 H2O + 4 O2 (g) + 9 Cl2 (g) + NaClO3

This equation implies that 20 moles of NaOCl is required for each mole of chlorate given adequate sunlight, pH, ionic strength and solution concentrations. This is in line with the lower yield figure suggested by the lab study of around 22 moles as estimated from 1/(90%*.05).

[Edited on 29-12-2013 by AJKOER]

AJKOER - 1-3-2014 at 09:55

An important theoretical point for aqueous chlorate production!

This source, "Inorganic Reactions in Water", by Ronald Rich, page 453 (link: http://books.google.com/books?id=Dv_F03cdKPUC&pg=PA453&a... ), and I recall having seem it in old texts as well (see, for example, http://books.google.com/books?id=jvjmAAAAMAAJ&pg=RA1-PA1... ) which notes the instability of HOCl in presence of NaCl), cites the following interesting reaction:

6 HOCl + Cl- (but only neutral chlorides) --> ClO3- + 3 Cl2(g) + 3 H2O

The Ronald Rich source also notes the reaction:

ClO3- + 5 Cl- + 6 H3O+ ---> 3 Cl2 (g) + 9 H2O

which is a reminder that attempting to harvest a chlorate in the presence of an acid chloride, is potentially problematic. Note, in the presence of an acid, the Hypochlorous acid needed to form the chlorate, is removed via the fast reaction:

HOCl + Cl- + H+ --> Cl2 (g) + H2O

Also, there is also the route to ClO2 (see Wikipedia http://en.wikipedia.org/wiki/Chlorine_dioxide ), to quote:

"HClO3 + HCl → HClO2 + HOCl
HClO3 + HClO2 → 2 ClO2 + Cl2 + 2 H2O"

where for the HCl, one can substitute a chloride salt and a strong acid.

In the present context of chlorate production, for example, if MgCl2 is formed, not being a neutral salt (like KCl and NaCl), be mindful that it doesn't lower pH sufficiently so as to hinder chlorate production.
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[EDIT] Note, the following reaction is usual presented (see, for example, https://www.kbcc.cuny.edu/academicDepartments/PHYSCI/PL/chm1... ) as follows (which is the reverse of the above):

NaOCl(aq)+ HCl(aq) --> HOCl (aq) + NaCl (aq)

Now, adding HOCl to both sides:

NaOCl (aq)+ HCl (aq) + HOCl (aq) --> 2 HOCl (aq) + NaCl (aq)

But, HCl + HOCl --> Cl2(g) + H2O, especially with hot/concentrated conditions, so we have:

NaOCl (aq)+ Cl2 (g) + H2O = 2 HOCl (aq) + NaCl (aq)

and, I am inclined to think that the above reaction may be able to proceed from right to left:

2 HOCl + NaCl --> NaOCl + Cl2(g) + H2O

and with more HOCl:

NaOCl + HOCl --> NaClO2 + HCl

where the HCl can be removed with more HOCl and the NaClO2 can be also oxidized to NaClO3, producing chlorate as required.

Bottom line, I can see how boiling NaCl in a large excess of aqueous HOCl (being of sufficient concentration), could form NaClO3 (and also, necessarily, a lot of chlorine gas which is allowed to vent off). So, the old garage chemist preparation for chlorate production is modified to boiling acidified chlorine bleach (by adding vinegar) together with NaCl. Note, one can also view this reaction in the role of modern theory presented in the opening of this thread. Namely, that a chlorate formation path can be accelerated in the presence of chloride and acetic acid.
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Here is another find, source: “Outlines of industrial chemistry: a text-book for students”, by Frank Hall Thorp and Charles D. Demond, page 120, link: http://books.google.com/books?pg=PA120&lpg=PA120&dq=Higgins+(J.+Soc.+Chem.+Ind.+1887,+248&sig=wZ7GHjb_mRMUerDgSFShQUPde2Q&ei=KHcSU6zbD 8uukAeA1IHIAw&id=svkrAAAAIAAJ&ots=xqHrxEkU3b#v=onepage&q=Higgins%20(J.%20Soc.%20Chem.%20Ind.%201887%2C%20248&f=false . Apparently the extraction of KClO3 by the action of KCl on Mg(ClO3)2 is problematic. The solution (being the product of chlorinating a fine suspension of MgO) should be concentrated down first to crystallize out most of the MgCl2. Else, a crystalline precipitate of a double salt, MgCl2 • KCl • 6 H20, so called artificial carnallite, is formed along with the KClO3. To quote:

“Magnesia* is sometimes substituted for lime, in order to increase the yield of potassium chlorate, since the latter is much less soluble in a magnesium chloride solution than in one of calcium chloride. Thus a yield of 90 per cent can be obtained, owing to more complete separation when crystallizing. In this process chlorine is passed into a "milk" of powdered magnesia in water, forming a solution of magnesium chloride and chlorate, which is then concentrated until crystals of magnesium chloride (MgCl2 • 6 H,0) separate on cooling. These are removed, and the proportion of chlorate to chloride in the mother-liquor is about 1 Mg(ClO3)2 to 2.8 MgCl2. The theoretical quantity of potassium chloride is then added, and potassium chlorate separates, leaving the magnesium chloride in solution. Any excess of potassium chloride must be avoided, since it would combine with the magnesium chloride to form a crystalline precipitate of a double salt (MgCl2 • KCl • 6 H20 — artificial carnallite), which would contaminate the product. The mother-liquors might be worked for chlorine, according to the Weldon-Pechiney process (p. 106)."

[Edited on 2-3-2014 by AJKOER]