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# Synthesis of Sodium Hydroxide

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I am "electrolizing?" a Saturated solution of Sodium bicarb using Carbon electrodes just in case. Wouldn't this produce Sodium Hydroxide? If so, how would I know when the reaction is complete. The way I see it, $\cf{CO2}$ Production will taper off and $\cf{O2}$ production will start as the Sodium bicarb is converted. Is there any way to test for sodium Hydroxide? Thanks.

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just do the same with nacl solution

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I was told that Sodium hypochlorite and other complexes were made when using NaCl... Is this not true?... Besides, I just wanted to know if my current setup worked.

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ya i think ull get naoh at the end if u use bicarbonate

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will there be any way of my knowing that all bicarb has been converted to hydroxide? Other than leaving it there for a week and hoping for hte best I guess.

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Take a sample of the solution and add some vinegar to it. If there is bubbling, then you still have some bicarbonate/carbonate in there.

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At the anode you'll get CO2 and O2 simultaneously:

4HCO3(-) --> 4CO2 + O2 + 2H2O + 4e

At the cathode you'll get H2 and OH(-)

2H2O + 2e --> 2OH(-) + H2

So, theoretically you'll end up with NaOH solution only, in practice it will be REALLY hard to get it completely free of carbonate. When you are close to the end, then any CO2 formed at the anode will be absorbed by the hydroxide at once and you get carbonate again:

CO2 + 2OH(-) --> CO3(2-) + H2O

The solution also will pick up CO2 from the air. In practice you'll never succeed in making it free of carbonate, not even with rigorous exclusion of air.

Use of NaCl does not work very well, unless you use some method of separating the anode liquid from the cathode liquid and use some special membrane, which is permeable for sodium ions and chloride ions, but not for hydroxide. When the liquid is allowed to mix, then you get hypochlorites and chlorates in solution as well.

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Nice Woelen, you have helped me a lot, in more than one forum I believe. So thanks a lot. I really do appreciate all the help all of you guys give. I'm trying my best to participate also but my knowledge is somewhat lacking. Anyhow, I threw in some acetic acid and it bubbled... Where would I get one of these semi-permeable membranes? Could I make one with agar or something?

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If your trying to make NaOH out of NaHCO3 then why even electrolyse it? Just heat it in solution to decompose it to sodium hydroxide and CO2. Sure your product wont be that pure but it will be in the same area of quality as your relectrolysed version.

~Scott

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I did some research and multiple source indicate that NaHCO3 thermally decomposes to Na2CO3, not NaOH. Is the same thing happening during electrolysis also?

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I did some research and multiple source indicate that NaHCO3 thermally decomposes to Na2CO3, not NaOH. Is the same thing happening during electrolysis also?

Indeed, you're right, it decomposes to Na2CO3. In electrolysis you'll get the same, but there the process continues and you'll end up with a solution of NaOH with still quite some Na2CO3 in it.

Heating of NaHCO3 will first give you H2O, CO2 and Na2CO3 at fairly low temperature (around 100 C). In order to drive off the other molecule of CO2 requires very strong heating. I'm afraid that even pointing the flame of a propane torch directly on the Na2CO3 still does not drive off the CO2, so that is not a useful way of getting NaOH.

But now for something else, is this just you want to play around and see if you can make NaOH yourself, just out of curiousity, or are you really interested in the NaOH itself?

If you are after NaOH, then I would suggest buying it. It is sold as drain cleaner in many places (hardware stores, drugstores and some super markets) in the form of white pellets. It also is cheap (around EUR 2 per pound, or \$2 per pound).

Another thing you could try is to use a salt bridge instead of a membrane. For this purpose, take two beakers with NaCl solution. Take a rubber tube and fill this with cotton and completely soak this with salt-solution. Now do the electrolysis with the anode in one beaker and the cathode in the other beaker. The salt bridge assures that there is no mixing of liquids, but a disadvantage is the larger resistance of the total circuit, you'll loose more electrical energy as heat.

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Basicly, I just like the Idea of having a chemical in a bottle and saying, I MADE this. I don't have a use for it at the moment. My chemicals are seriously lacking at the moment. During my Copper (II) Chloride experiments last summer gave my grandma the strange impression that I was making Meth. It's so sad, ignorance. I say, if you don't understand it, you can't judge it or make stupid assumptions. Honestly, name me a chemist who can create meth with copper, H2O2 and HCl and I'll give you a very large sum of money. Ignorance shouldn't be an excuse... This goes for multiple things. Anyhow, I really like your Salt bridge Idea. I saw one in this cell that had copper and zinc, in teir respective sulphate solutions with a salt bridge between them. I will try this. I do have yet another question though. Why would'nt the chemicals eventually mix? I mean Liquid obviously permeates cotten. I understand that the Ions would flow through. But I'd think if you left the salt bridge in long enough, liquids would mix, because of... Ok I don't remember the law but it talks about liquids moving from an area of high conc. to low. Anyhow, thanks.

EDIT: I know NaOH is a very common chemical, but what uses does it have for a chemist? I understand that this is a broad question, but in general, what would you use it for.

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I did some research and multiple source indicate that NaHCO3 thermally decomposes to Na2CO3, not NaOH. Is the same thing happening during electrolysis also?
Yeah thats true for when it is by itself but in aqueous solution it behaves differently and you will get NaOH by heating.

Indeed, you're right, it decomposes to Na2CO3. In electrolysis you'll get the same, but there the process continues and you'll end up with a solution of NaOH with still quite some Na2CO3 in it.

Heating of NaHCO3 will first give you H2O, CO2 and Na2CO3 at fairly low temperature (around 100 C). In order to drive off the other molecule of CO2 requires very strong heating. I'm afraid that even pointing the flame of a propane torch directly on the Na2CO3 still does not drive off the CO2, so that is not a useful way of getting NaOH.

I'm sorry why would you do that? You need to heat it in solution not dry, Why do you think the hydrogen would come from if you were just blasting sodium carbonate with heat? Also the product will be very similar in purity to the electrolized one.

~Scott

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naoh is cool for makin soaps and can do cool acid base reactions as well as metals + naoh like aluminum.

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Actually, aluminum is one of the few metals that will react with NaOH.

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I'm sorry why would you do that? You need to heat it in solution not dry, Why do you think the hydrogen would come from if you were just blasting sodium carbonate with heat? Also the product will be very similar in purity to the electrolized one.

I strongly doubt that, actually, I do not believe this. If you heat a solution of NaHCO3, then indeed bubbles of CO2 will be driven out, until carbonate remains:

2HCO3(-) ---> H2O + CO2(driven off) + CO3(2-)

The carbonate will be in equilibrium with a very small amount of hydroxide:

CO3(2-) + H2O <----> HCO3(-) + OH(-)

So, continued boiling may indeed remove a little more CO2 than one would expect for pure carbontate to remain, but this will only be a very weak effect (carbonate is only a weak base and OH(-) is a very potent obserber of CO2).

If you continue heating, then all water will boil away and a dry residue of sodium carbonate remains with at most a few percents of NaOH, but I even doubt that, my bet would be just a few tenths of percents of NaOH.

Have you actually tried this way of making NaOH?

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I strongly doubt that, actually, I do not believe this. If you heat a solution of NaHCO3, then indeed bubbles of CO2 will be driven out, until carbonate remains:

2HCO3(-) ---> H2O + CO2(driven off) + CO3(2-)

The carbonate will be in equilibrium with a very small amount of hydroxide:

CO3(2-) + H2O <----> HCO3(-) + OH(-)

So, continued boiling may indeed remove a little more CO2 than one would expect for pure carbontate to remain, but this will only be a very weak effect (carbonate is only a weak base and OH(-) is a very potent obserber of CO2).

If you continue heating, then all water will boil away and a dry residue of sodium carbonate remains with at most a few percents of NaOH, but I even doubt that, my bet would be just a few tenths of percents of NaOH.

Have you actually tried this way of making NaOH?

Your acid base reaction is incorrect. You say.

2HCO3(-) ---> H2O + CO2(driven off) + CO3(2-)

Here the HCO3- is acting as an acid this is not what occurs it acts as a base in this reaction (it is amphiprotic)

HCO3- + H2O -> H2CO3 + OH-

The carbonic acid quickly decomposes yeilding CO2 and H2O

H2CO3 -> H2O + CO2

The OH- formed in the first reaction will react with the Na+ spectator ions and precipitate out as NaOH as the solution is left to evaporate. This will produce much more then a 'few tenths of a percent'.

~Scott

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pity you didnt have any Mercury, you could make sodium amalgam via electrolysis, and then add water to it to leave NaOH soln and your original Mercury.

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Your acid base reaction is incorrect. You say.

2HCO3(-) ---> H2O + CO2(driven off) + CO3(2-)

Here the HCO3- is acting as an acid this is not what occurs it acts as a base in this reaction (it is amphiprotic)

HCO3- + H2O -> H2CO3 + OH-

The carbonic acid quickly decomposes yeilding CO2 and H2O

H2CO3 -> H2O + CO2

The OH- formed in the first reaction will react with the Na+ spectator ions and precipitate out as NaOH as the solution is left to evaporate. This will produce much more then a 'few tenths of a percent'.

~Scott

I still do not agree, but let's give a turn to this discussion:

A counter question: What would you expect if you boil a solution of Na2CO3 to dryness, instead of NaHCO3? You also expect NaOH in that case or do you expect Na2CO3 as residue?

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I know NaOH is a very common chemical, but what uses does it have for a chemist? I understand that this is a broad question, but in general, what would you use it for.

It is one of the chemicals I use most (together with HCl and H2SO4). Almost every aqueous experiment I do either involves an acid or a base. If I need a base I take NaOH, if I need an acid I take HCl when the complexing of chloride does not disturb the experiment, otherwise I take the more expensive and harder to obtain H2SO4.

Many redox reactions either consume H(+) or OH(-), that's why I need acid or base. Many organic acids also can only be brought in solution as their sodium salt or potassium salt. For this reason I also use NaOH (sometimes KOH, but the latter is MUCH more expensive and harder to get).

NaOH also is very interesting for dissolving elements like sulphur and silicon. These solutions have really interesting properties.

In fact, there are too many uses of NaOH to name them all. Have a look at my site and look at the experiments I do. Then you'll see a lot of experiments with one of the chemicals being NaOH:

http://woelen.scheikunde.net/science/chem/exps/exppatt.cgi?compound=sodiumhydroxide

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I still do not agree, but let's give a turn to this discussion:

A counter question: What would you expect if you boil a solution of Na2CO3 to dryness, instead of NaHCO3? You also expect NaOH in that case or do you expect Na2CO3 as residue?

Interesting question, I would say that it still will occur but only slower probably quite slower. This is because the carbonate ion will has to act as a base twice for the reaction to occur example

CO3(2-) + 2H2O <-> HCO3- + OH- + H2O <-> H2CO3 + 2OH-

The bolded carbonic acid wil then decompose as such:

H2CO3 => H2O + CO2

Now if you didn't let the CO2 escape your solution would just reach equilbrium and not too much NaOH would be formed, However by removing the CO2 the equilbrium is constantly shirted to the right and favours the production of more OH-, Now the only reason this will take longer then the NaHCo3 is because the CO3(2-) must react as a base twice before the CO2 can begin to be removed. When using the Bicarbonate it must only act as an base once (From the italic step in the eqn)therefore the reaction will preceed faster.

~Scott

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Interesting question' date=' I would say that it still will occur but only slower probably quite slower. This is because the carbonate ion will has to act as a base twice for the reaction to occur example

CO3(2-) + 2H2O <-> [i']HCO3-[/i] + OH- + H2O <-> H2CO3 + 2OH-

The bolded carbonic acid wil then decompose as such:

H2CO3 => H2O + CO2

Now if you didn't let the CO2 escape your solution would just reach equilbrium and not too much NaOH would be formed, However by removing the CO2 the equilbrium is constantly shirted to the right and favours the production of more OH-, Now the only reason this will take longer then the NaHCo3 is because the CO3(2-) must react as a base twice before the CO2 can begin to be removed. When using the Bicarbonate it must only act as an base once (From the italic step in the eqn)therefore the reaction will preceed faster.

~Scott

Well, now I hope that you see the gap in your reasoning in case of the solution of the bicarbonate.

Suppose some hydroxide is formed, according to your proposed mechanism, that hydroxide will react with other bicarbonate in solution immediately, forming carbonate:

HCO3(-) + OH(-) <----> H2O + CO3(2-)

This equilibrium is very much to the right, CO3(2-) only is a weak base.

Now another gap in your reasoning:

A solution of a bicarbonate is weakly acidic.

A solution of a carbonate is fairly basic.

A solution of hydroxide is very basic.

You expect the hydroxide to form fast from the somewhat acidic solution of bicarbonate, while it is formed much slower from the more alkaline carbonate. Think of the bicarbonate being converted to hydroxide, then alkalinity of the liquid moves through the medium situation with the carbonate and the process comes to a halt.

You see this reasoning? So, making hydroxide from bicarbonate is harder than making hydroxide from carbonate.

In fact, the carbonate ion is quite stable. Think of a cettle with hard water in it. Hard water contains calcium and magnesium bicarbonate in solution. When you boil the water, then the bicarbonate decomposes to water, carbon dioxide and carbonate ion. When the liquid cools down, then the solubility of the calcium/magnesium carbonate drops to a low point, such that it precipitates out of solution. This is the main problem with hard water. You don't get calcium hydroxide. If the latter would be formed, then we would not have problems with hard water, because calcium hydroxide is sufficiently soluble and does not precipitate.

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Also, don't forget that carbon dioxide is VERY soluble in water. Any that is formed will not be leaving as a gas. If it did, then every time you put NaHCO3 in neutral or basic water, you'd see bubbling almost immediately. In reality, the CO2 remains in the water and never leaves the system, so the equillibrium doesn't shift in any measurable way.

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Well, now I hope that you see the gap in your reasoning in case of the solution of the bicarbonate.

Suppose some hydroxide is formed, according to your proposed mechanism, that hydroxide will react with other bicarbonate in solution immediately, forming carbonate:

HCO3(-) + OH(-) <----> H2O + CO3(2-)

This equilibrium is very much to the right, CO3(2-) only is a weak base.

Yes it will go both ways, as all equilibriums do ,but due to the removal of CO2 does not go that way very much.

Now another gap in your reasoning:

A solution of a bicarbonate is weakly acidic.

A solution of a carbonate is fairly basic.

A solution of hydroxide is very basic.

You expect the hydroxide to form fast from the somewhat acidic solution of bicarbonate, while it is formed much slower from the more alkaline carbonate. Think of the bicarbonate being converted to hydroxide, then alkalinity of the liquid moves through the medium situation with the carbonate and the process comes to a halt.You see this reasoning? So, making hydroxide from bicarbonate is harder than making hydroxide from carbonate.

I'm not realy sure what you mean by medium situation? not familiar with that term. But that stuff you said about a solution... ect does not realy matter as the removal of CO2 will continualy produce more OH- even if bicarbonate is weekly acidic.

In fact, the carbonate ion is quite stable. Think of a cettle with hard water in it. Hard water contains calcium and magnesium bicarbonate in solution. When you boil the water, then the bicarbonate decomposes to water, carbon dioxide and carbonate ion. When the liquid cools down, then the solubility of the calcium/magnesium carbonate drops to a low point, such that it precipitates out of solution. This is the main problem with hard water. You don't get calcium hydroxide. If the latter would be formed, then we would not have problems with hard water, because calcium hydroxide is sufficiently soluble and does not precipitate.

What your saying makes sense in that carbonate ions don't usually produce OH- to any grewat degree this I can agree with but still it makes sense to me that they can. I am still undecided I think we should just throw some sodium biocarbonate in water ans boil it and settle this thing.

Also, don't forget that carbon dioxide is VERY soluble in water. Any that is formed will not be leaving as a gas. If it did, then every time you put NaHCO3 in neutral or basic water, you'd see bubbling almost immediately. In reality, the CO2 remains in the water and never leaves the system, so the equillibrium doesn't shift in any measurable way.

The water in this situation is boiling so alot of that CO2 will be liberated from solution.

Scott

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I am still undecided I think we should just throw some sodium biocarbonate in water ans boil it and settle this thing.

Indeed, this evening I'll put this question in front of our final jury, being Nature itself

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