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boomstickbob

making naoh?

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Hey can any one give me some good information on making sodium hydroxide? ive spent the last 2 days stalking the web to only find sodium hypochlorite production, if any one has any info please feel free to share, thank you.

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Take 59-60 grams of salt NaCl, 18 grams of water, mix,

plug some electrodes that don't react with neither HCl nor NaOH, and pass direct current through it,

you will smell Chlorine gas (it always reminds me primary school swimming pool) on positive electrode,

and there will be Hydrogen gas on negative electrode.

 

With steady current I=1 A you will need 26 hours 45 minutes to change all above to NaOH (in theory).

 

Really easier is to buy it (but less fun).

Edited by Sensei

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Ho Hum.

For a start, the salt won't dissolve in that little water.

Then there's the problem of oxygen- rather than chlorine being produced at the anode.

Add to that the fact that, once there's any NaOH present, it will tend to react with chlorine to produce things like NaClO and NaClO3.

 

Oh, while I'm at it, a copper wire won't react with NaOH or with HCl, but it won't serve as an anode in that cell because it would be oxidised by chlorine.

So, in summary, mainly wrong.

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Did you performed it ever in your life?

 

I was doing it dozen of times (last time week ago).

Usually with current I between 2A to 5 A. And high voltage.

 

It is normal procedure of production of NaOH and HCl (indirectly - from gaseous Cl)

 

No oxygen is produced this way- because it cannot be produced - because there is so little water.

You can't breath after 5-10 seconds since start up with I=5 A, because of Chlorine gas.

 

If water would be in significant amount, right, oxygen would be produced, but it would be no longer Hydrogen+Chlorine production but regular electrolysis with Hydrogen+Oxygen produced.

 

ps. I didn't say a word about copper. It would be indeed bad idea to use it. I have tried it with bad results.

Cu immediately reacts with -OH and we will have at least bluish Cu(OH)2.

(Regular electrolysis with a lot of water with copper electrodes is instantly producing Cu(OH)2 . I had it plenty in containers. After week or two it all disintegrated to CuO and H2O. Easily visible because it changed color from bluish to dark)

 

Add to that the fact that, once there's any NaOH present, it will tend to react with chlorine to produce things like NaClO and NaClO3.

 

 

Right. But whole point of this setup is to produce Chlorine gas.

We want to Cl2 bubbles to escape it as soon as they're formed on positive electrode.

And then (usually) collect it in another container with fresh water, so HCl and HOCl will be produced in it.

 

I wouldn't care much about NaClO. If gaseous Cl2 dissolved in solution (which is undesired), there will be also HCl present. Which will react with NaClO.

NaClO + 2 HCl → Cl2 + H2O + NaCl

Edited by Sensei

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Among a whole lot of wrongness

 

Usually with current I between 2A to 5 A. And high voltage.

 

It is normal procedure of production of NaOH and HCl (indirectly - from gaseous Cl)

 

No oxygen is produced this way- because it cannot be produced - because there is so little water.

You can't breath after 5-10 seconds since start up with I=5 A, because of Chlorine gas.

 

If water would be in significant amount, right, oxygen would be produced, but it would be no longer Hydrogen+Chlorine production but regular electrolysis with Hydrogen+Oxygen produced.

 

ps. I didn't say a word about copper. It would be indeed bad idea to use it. I have tried it with bad results.

Cu immediately reacts with -OH and we will have at least bluish Cu(OH)2.

(Regular electrolysis with a lot of water with copper electrodes is instantly producing Cu(OH)2 . I had it plenty in containers. After week or two it all disintegrated to CuO and H2O. Easily visible because it changed color from bluish to dark)

 

 

Right. But whole point of this setup is to produce Chlorine gas.

We want to Cl2 bubbles to escape it as soon as they're formed on positive electrode.

And then (usually) collect it in another container with fresh water, so HCl and HOCl will be produced in it.

 

I wouldn't care much about NaClO. If gaseous Cl2 dissolved in solution (which is undesired), there will be also HCl present. Which will react with NaClO.

NaClO + 2 HCl → Cl2 + H2O + NaCl

If you are using a high voltage, you are just wasting power.

The excess heat will contribute to shifting the product from hypochlorite to chlorate

and so on, but I think it's fair to say that the killer brain fart is where you say

 

"Right. But whole point of this setup is to produce Chlorine gas."

 

Do you want someone to read the title of the thread for you?

 

Another remarkable bit of evidence of a lack of clear thinking is where you say "If gaseous Cl2 dissolved in solution (which is undesired), there will be also HCl present."

Well, if the system generates any NaOH (and I accept it will produce a bit of the stuff with a lot of impurities) then the concentration of HCl is going to damn all isn't it?

I realise you didn't mention copper.

What you said was "some electrodes that don't react with neither HCl nor NaOH"

Need I remind you that copper doesn't react with HCl or NaOH? So, while you didn't mention it by name, it should work according to your woefully inept procedure.

Most amateur scientists probably did the "two copper wires a battery and a jar of salt water" experiment as a kid.

Most of us remember that the wires corroded.

 

 

Would you like to try again?

Edited by John Cuthber

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Do you want someone to read the title of the thread for you?

 

At least I tried to answer "how to make NaOH at home".

Your answer was "go to shop and buy"..

 

In post #3 I just described "home version" of Chloralkali process in easy to understand words

http://en.wikipedia.org/wiki/Chloralkali_process

 

You can read in the article that positive electrode should be made of titanium.

 

Another remarkable bit of evidence of a lack of clear thinking is where you say "If gaseous Cl2 dissolved in solution (which is undesired), there will be also HCl present."

Well, if the system generates any NaOH (and I accept it will produce a bit of the stuff with a lot of impurities) then the concentration of HCl is going to damn all isn't it?

 

Rethink once again what was initial quantity of salt and water.

59-60 g of NaCl and

18 g of H2O

That's exactly the same number of molecules = 6.022141*10^23 of both. Each NaCl has corresponding it H2O. They are in pairs.

 

Now, negative electrode is producing H2 which is instantly escaping our system and never comes back.

There is 2x -OH and 2NaCl left.

-OH joins with Na+ and we have NaOH + Cl-

2 Cl- on positive electrode are joining together and neutral gaseous Cl2 is produced (are you now inspired how important is shape of titanium positive electrode in this process to get rid of Chlorine in gaseous state?)

 

If I see 1 Liter volume of Hydrogen produced in one container, and 1 Liter of Chlorine volume in another container, then I am pretty sure that quantity of impurities is pretty low.

 

If some Cl will not escape it, but joins with water, equivalent it molecule of NaCl won't be able to change (there is equal number of pairs of NaCl and H2O after all).

So we will have: NaCl, NaCl, HCl, HClO.

 

That's why so important to produce gaseous Cl2 ,which I said:

"But whole point of this setup is to produce Chlorine gas.

We want to Cl2 bubbles to escape it as soon as they're formed on positive electrode."

 

If both gases escaped setup at the same rate, the same volume, remain should be NaOH mostly.

 

Impurities produced by this method are quite meaningless (especially if it's just boomstickbob's home experiment) in comparison to business production using Chloralkali process, which can even leave traces of mercury in NaOH.

 

Quote from the nearly beginning of article:

"Additionally, the chlorine and sodium hydroxide produced via the mercury-cell chloralkali process are themselves contaminated with trace amounts of mercury. The membrane and diaphragm method use no mercury, but the sodium hydroxide contains chlorine, which must be removed."

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"At least I tried to answer "how to make NaOH at home".

Your answer was "go to shop and buy".."

And so did you- it was one of the few things you said that was correct.

"Really easier is to buy it"

 

"In post #3 I just described "home version" of Chloralkali process in easy to understand words"

A version that wouldn't work.

 

"Rethink once again what was initial quantity of salt and water.

59-60 g of NaCl and
18 g of H2O"

It doesn't matter who thinks about it, nor how often. The stuff still won't dissolve.

 

"2 Cl- on positive electrode are joining together and neutral gaseous Cl2"

Still ignoring the oxygen then?

Do you realise that, as the reaction goes on there's less Cl- so there's less Cl2 production?

Eventually, you end up making oxygen.

 

"So we will have: NaCl, NaCl, HCl, HClO."

Will you please stop pretending that you will end up with acids like HCl in the presence of NaOH?

 

And the problem of reaction of Cl2 and NaOH still remains whatever shape the electrodes are- because they still have to be in the solution.

You need something to keep them separate.

That kind of seems to be totally and utterly missing from your so called ""home version" of Chloralkali process"

 

So, when it comes down to it you gave a method that (without a whole lot of details, which you missed out) won't actually work.

 

Anyway, buying the stuff is the simple method and adding slaked lime to washing soda solution may well be the most practical home chemistry version.

Edited by John Cuthber

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Sensei, JC is completely right here. Your chemistry is a bit off on this.

 

To the solubility issue. I'm not sure if you mistyped or misremembered the amounts, but there is no way you are dissolving 60 g of NaCl in 18 mL of water. The solubility of NaCl is 359 g/L, which means that in 18 mL of water you'd be looking at a maximum of about 6.5 g dissolved in solution. Furthermore, and I may be wrong in this not having done any electrolysis beyond learning the theory, but would 18 mL of water not boil off fairly quickly with that much current? If not quickly, then certainly before the 26 hours is up.

 

You will not get HCl forming in the presence of strongly basic conditions. That should be pretty obvious.

 

The wiki article you linked uses a membrane, diaphragm or mercury cell to separate the two products (or constantly remove in the case of the latter) and as far as I can see, you did not describe any such thing in your process, which leads me to believe that you are not making what you think you are making. As John Cuthber noted, if you are making anything (ignoring your solubility issues) it's quite likely the -OCl or -OCl3 salts and even they would be contaminated with some percentage of NaCl.

 

Also worth noting, especially for the benefit of the OP, is that you should not be smelling chlorine or allowing yourself to come into contact with it. Not because it's not there to smell, but because chlorine gas is quite toxic.

 

To the OP: NaOH is trivially easy to buy and I would not bother trying to make it from scratch. Was there any particular reason you were trying to do this?

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Sensei missed the point that when overpotential applied, oxygen, the thermodynamic product, would dominate.

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For a damnedly slow, but easily done way of making pure sodium hydroxide,place two electrodes (anode carbon, cathode whatever you feel like) in two beakers full of saturated salt water. Soak a paper towel in salt solution, and place both ends into the solutions. Pass current through, and the Cl- ions will migrate to the anode, the Na+ to the cathode, and concentrations of NaOH and HCl/HOCl (or chlorine in water, whatever you prefer to write it as) will slowly increase over time.

Back when I had to do this (and didn't just buy it), I used tap water in place of salt water at one end, which would slowly become basic as NaOH was formed at the cathode.

Did I mention this is really slow? Not only due to the initial near-lack of conductivity of the tap water I used, because I did use saltwater in both beakers in one try.

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Did I mention this is really slow? Not only due to the initial near-lack of conductivity of the tap water I used, because I did use saltwater in both beakers in one try.

What voltage you used?

Try high voltage at beginning, and then lower it with time.

This can be done automatically by stabilizer.

 

If you ever produced f.e. iron oxides, aluminum oxides, copper oxides (or hydroxides) in higher amounts using electrolysis, you probably noticed how current increases over time (and very fast). (couple times I had 15-17 A while iron oxides production after >20-30 minutes)

Stabilizer is the only feasible way to have good rate of production, and you won't have to control it all the time..

Edited by Sensei

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What voltage you used?

Try high voltage at beginning, and then lower it with time.

This can be done automatically by stabilizer.

 

If you ever produced f.e. iron oxides, aluminum oxides, copper oxides (or hydroxides) in higher amounts using electrolysis, you probably noticed how current increases over time (and very fast). (couple times I had 15-17 A while iron oxides production after >20-30 minutes)

Stabilizer is the only feasible way to have good rate of production, and you won't have to control it all the time..

High voltage would mean oxygen not chlorine at anode.

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What voltage you used?

Try high voltage at beginning, and then lower it with time.

This can be done automatically by stabilizer.

 

If you ever produced f.e. iron oxides, aluminum oxides, copper oxides (or hydroxides) in higher amounts using electrolysis, you probably noticed how current increases over time (and very fast). (couple times I had 15-17 A while iron oxides production after >20-30 minutes)

Stabilizer is the only feasible way to have good rate of production, and you won't have to control it all the time..

High voltage means the strip of wet towel gets hot.

You can certainly char the paper that way.

So you waste a lot more energy and risk starting a fire.

 

What you actually need isn't a voltage stabiliser, but a current stabiliser.

 

Still mainly wrong.

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