How to fix or calculate Na2O (sodium oxide) content in a mixture of sodium silicate and sodium hydroxide with known solid contents? For example; 8%, 10% and 12%. The ratio of sodium silicate and molarity of sodium hydroxide are known.
The use of Na2O is archaic and, in my view, completely useless. There is no Na2O content in an aqueous solution - there are Na+ ions and OH- ions. This material does not exist except in certain special conditions. It hydrolyzes virtually immediately in the presence of water in the atmosphere. It certainly cannot exist in aqueous solution. There are better ways of describing sodium content in a system.
You need to know:
1. Sodium silicate - the weight % of Na2O in the water glass- usually about 10%.
2. How much grams NaOH is used to prepare the NaOH solution. From each gram NaOH you will have 0.774794 grams Na2O ( from 2NaOH->Na2O+H2O).Then convert to %.
The sum of both is total Na2O in w/w %,.
The best is to use table in excel to make the calculations.
It is easier not to make a mathematical equation related to required % (8, 10 or 12), better the table should calculate the % of Na2O and find the needed amount by guessing .
PS: Be aware of w% or mol %. If you divide the number by the mollar mass of Na2O you will get the molar ratios.
1) The % of Na2O in NaOH = 0.775 x 100 % = 77.5 %
2) The % of Na2O in Na2SiO3 = 0.508 x 100 % = 50.8 %
So if you have 10% Na2SiO3 with 90% NaOH, the % of Na2O in this mixture = (10.0 % x 0.508) + (90.0 % x 0.775) = 0.0508 % + 0.698 % = 0.748 x 100 % = 74.8 %
But if you have 10% NaOH and 90% Na2SiO3, the % of Na2O in this mixture = (10.0 % x 0.775 + (90.0 % x 0.508) = 0.0775 %+ 0.457 % = 0.535 x 100 % =53.5 %
Prof. Ibrahim Abdel-Rahman. Please kindly put some light on, how to calculate SiO2 concentration if we are using LSS
In the previous question I forgot to multiply the last answer with 100 %.
I did it again, so the answer of your question will be as follows:
1) The % of SiO2 in Na2SiO3 = 100 % - Na2O % = 100 % - 50.8 % = 49.2 %
2) The % of SiO2 in NaOH = 0.0 %
So if you have 10% Na2SiO3 with 90% NaOH, the % of SiO2 in this mixture = (10.0 % x 0.492) + (90.0 % x 0.0) = 0.0492 % + 0.0 % = 0.0492 x 100 % = 4.92 %
But if you have 10% NaOH and 90% Na2SiO3, the % of SiO2 in this mixture = (10.0 % x 0.0) + (90.0 % x 0.492) = 0.0 % + 0.443 % = 0.443 x 100 % = 44.3%
The use of Na2O is archaic and, in my view, completely useless. There is no Na2O content in an aqueous solution - there are Na+ ions and OH- ions. This material does not exist except in certain special conditions. It hydrolyzes virtually immediately in the presence of water in the atmosphere. It certainly cannot exist in aqueous solution. There are better ways of describing sodium content in a system.
Alan F Rawle
His question in solid solution and not aqueous solution (Solid mixture).
Ibrahim Abdel-Rahman Na2O still does not exist in the presence of water.
Alan F Rawle
Yes, sure there is no Na2O in the presence of water. This is only calculations to find how much the percentage of Na2O in the fromula NaOH. The question is: when you mix the NaOH with sodium silicate and heat them to get dry solid mixture (No water), then, what is the percentage of Na2O in this mixture?
Hi Professor Ibrahim Abdel-Rahman , I have a doubt, that calculus is appropriate in the case of the one-part geopolymers, but in the case of the conventional way for geopolymer preparation (mixing of alkali activator solution with aluminosilicates precursors), is still useful to consider an equivalent Na2O content by the addition of NaOH? even when by this procedure there is not heating for obtaining of the sodium oxide. Thank you
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