[Tittywahah]

Waiting for HCl at the moment so when that is ready I shall get to work on the above out of interest.  Though I will also use Phenolpthalein just to see what happens don't see any reason why it won't work being that it will turn from pink to colourless at a ph 8 then red below 7 theoretically. I've used it before many times in acid base titrations so do not see what the problem here would be. Can only experiment and if wrong will get back here and confess.

Edit:  This is from wiki answers, this guys explanation is unclear.  I wonder if you could explain it better.

"""""firstly, take 1ml of sodium silicate and add 99ml of water. titrate it with 0.5N HCl and use 1 drop phenophthalein as indicator. the amount of HCl used is converted to gram by using the density and molecular weight calculations.
1.177gm of HCl used for 1 gm of sodium oxide, calculate by multiplying your HCl used for your sodium oxide.

now take 10ml of silicate and weigh it in grams and heat it at 180 degrees celsius for 2 hours to get solid content (water evaporated)
deduct the value of sodium oxide from total solid content obtained.
this is the value of silica in your sodium silicate.""""""


1.calculate by multiplying your HCl used for your sodium oxide - I take this to mean that I should multiply the amount of titrant used by the density of the HCl to determine the grams used?  Thats fine, but what the hell does he mean by this - "used for your sodium oxide. "?
2.now take 10ml of silicate and weigh it in grams - I am dissolving the solid grams in water I then weigh the whole solution of water? The solid that is left is is then deducted from the original weight of the solution - this seems absurd to be honest since and completely incomprehensible.  Not to mention that the solid left over is still oxide and silica?

Can someone please make sense of this procedure. If I don't understand then it is because there is lack of clarification in this explanation.

[Borek]

I will also use Phenolpthalein just to see what happens don't see any reason why it won't work being that it will turn from pink to colourless at a ph 8 then red below 7 theoretically. I've used it before many times in acid base titrations so do not see what the problem here would be.

http://www.titrations.info/titration-end-point-detection

http://www.titrations.info/acid-base-titration-end-point-detection

and curve for silicate titrated with HCl against phenolphthalein:

[Tittywahah]

Hi Borek, does this horizontal axis indicate that between P 11. and 5 one is adding twice as much HCl again before the end point is visible?  If this is so - I have a special range of litmus paper that measures between 8 and 6.5 - I could also use this.  I suppose a bit of experimenting then I can learn for myself.  thankyou for that link by the way, useful reading as well.  I really don't need to do any of this, but I just have to, strange obsession this need to analyse and learn and know.

[Borek]

Hi Borek, does this horizontal axis indicate that between P 11. and 5 one is adding twice as much HCl again before the end point is visible?

Perhaps this is some limitation of my English, but I read what you wrote as "volume of the titrant added between 100% (pH 11.46) and 200% (pH 5.16) is two times larger than the volume added between 0% and 100%". No, these volumes are identical.

I have a special range of litmus paper that measures between 8 and 6.5 - I could also use this.

1. Titrating against pH strips , while possible in theory, in practice doesn't work that good.

2. Your strips won't work, you need an indicator that changes pH on the steep part of the curve, somewhere between 7 and 4. Otherwise your titration results will be quite erratic because you will have serious problems correctly - and reproducibly - detecting end point.

[Intanjir]

Beautiful chart Borek =)
And yes from it I can see that methyl orange would definitely be the appropriate indicator to use thank you.

I agree that adding a drop of pure ethanol wouldn't hurt if the sodium silicate solution is sufficiently dilute. I think that adding to 1% would work, but at 10% I think it might make a nodule. I'll try it out later today out of curiosity.

All of the recipes I have seen for the bouncy ball have used the already dissolved sodium silicate.
I do not think the silicate powder will do much of anything at all if added to concentrated ethanol. At least not the stuff I have, it is hard enough just to get it to dissolve in water.

Since my product is well labeled I intend to cheat by just adding the amount of HCl I calculate will account for the Na, and using the pH meter to record the pH. I'll add some methyl orange too though.

[Intanjir]

So the wiki answers answer appears to presume that you are starting with sodium silicate solution. So this explains second part where he has you dry it. He seems to think that 180 C for 2 hours is enough to get the anhydrous solid, and perhaps he is right. So you might try taking 3g or so of the powder and roasting it at 180C and see if you lose weight. If everything is in order then you should lose around 18% of the weight as you lose your waters of hydration. Anyways I read it here as him having you subtract the weight of sodium oxide content that you think there should be from the weight of the dried powder to get the weight of the silica which is of course fine.

The first part of the answer is interesting. He has you make a fairly dilute solution. I do not think any silica would precipitate at this dilution in the time frame of the titration. So you still have the silicate ions floating around lowering the pH a bit. However from looking at Borek's very helpful chart this is all perfectly fine. All we are after for the titration is finding the steep part of the pH curve, and this corresponds to having just enough HCl to account for each Na. I confess that I was making it all harder than it really had to be, since I had forgotten the importance of the knee in the titration pH curve. Borek's chart also suggests that pheno may just barely work as an indicator here, if you have a good enough eye to discern very pale magenta from perfectly clear. I am skeptical though.

By 'used for sodium oxide' he just means the conversion ratio for the weights assuming that 2 moles of HCl at the steep part of the titration curve corresponds to 1 mole of Na2O.

[Tittywahah]

I was searching for the maths involving how to interpret the data from the titrations that we have been talking about.  From this source: The OxyChem Sodium Silicates Handbook, I found the following:

%Na2O = mLs HCl used x Molarity x 3.1/ sample weight

The only puzzling thing was the 3.1.  So I discovered this earlier on in the document where Baume is recorded with regards to temperature, there is a 3.0 and 3.2.  Now I have absolutely no idea what any of this means, and have probably assumed the connection between the 3.1 and this Baume to be the answer, but I think not.  Do any of you have a clue why the 3.1 is here?

[Intanjir]

3.1 = the molecular weight of Na2O(61.98) divided by 20.

You divide by 2 because it takes two HCl per Na2O, and you further divide by 10 because of the conversions from mL to L and to percentages(ie divide by 1000, and then multiply by 100).

[Tittywahah]

So:   (mLs HCl/2) x M x (61.98/10)  Although why divide by 10 in the first place?  This is the same as the number of moles in 100 mLs, which makes no sense at all. Why do this?  The 61.98 is molecular weight per litre of one mole.  Why divide it by 10?


Edit:  Always do a balanced equation before titration:
Na2SiO3+2HCl=H2O+2NaCl+SiO2
Na2SiO3+2HCl=H2SiO3+2NaCl
the first one we have silicon dioxide precipitate (the Silica gel) and the second one we have the silicic acid but I am thinking that actually the the first reaction is the final one immediately after the second one has taken place because silicic acid loses water very quickly and becomes the gel of the first reaction equation.  Just wanting to be certain about my understanding.

[Borek]

Although why divide by 10 in the first place?

Have you read Intanjir post? He explained it quite precisely:

you further divide by 10 because of the conversions from mL to L and to percentages(ie divide by 1000, and then multiply by 100).

[Tittywahah]

I know that I dumb on the uptake here, ignorant of maths and can't get it because I am lacking in understanding where others take for granted - I am sorry. But I do not understand the principle here at work, why am I converting to Litres and then to percentages?  I see his answer but I can not understand the mathematical reasoning.  I know the 62g/mol is per litre and so this needs to be divided by 1000 to arrive at mol per mL as in other stoichemetry calculations, but to further multiply by 100 I do not understand. Sorry.

[billnotgatez]

@Tittywahah
Forum rules suggest you use appropriate editing to post things like formulas.
You can use the function bars above where you enter the post as well as other features.
You can select the Preview button before posting to make sure you got the editing correct.

For instance you could have posted your formulas this way
Na₂SiO₃+2HCl  H₂O+2NaCl+SiO₂
Na₂SiO₃+2HCl  H₂SiO₃+2NaCl

[Tittywahah]

Ok I stand corrected. Sorry.

[Intanjir]

The multiplication by 100 is simply to convert a fractional proportion into a percent, ie 0.75 is 75%.

[Tittywahah]

Thankyou.