How much Na2O (from NaOH) do I add to sodium silicate with a weight ratio of SiO2/Na2O = 2.34 to get the solution to a weight ratio SiO2/Na2O = 1.9?

There is some confusion in industry as sodium silicate is made with NaOH but given a ratio with respect to Na2O (i.e. SiO2/Na2O not SiO2/NaOH). I presume it follows from this equation:


2NaOH -> Na2O + H2O


Looking at a product information sheet of sodium silicate with a weight ratio displayed as "SiO2/Na2O" the actual ingredients in it are:


* SiO2 wt% = 30.3%
* NaOH wt% = 16.7%
* H2O wt% = 100% - 30.3% - 16.7% = 53%
* SiO2/NaOH = 1.81


This corresponds to SiO2/Na2O = 2.34


So if I'm given a 100g solution of sodium silicate that has these specifications and wanted to change the solution by adding an amount of NaOH so that the final wt% of NaOH would correspond to a final solution weight ratio of SiO2/Na2O = 1.9, what would be the best approach assuming I can add a mass [g] of a liquid specification 50% NaOH and 50% H2O?


**EDIT 1:** Here is my working so far.


First off, to convert the equivalent NaOH to Na2O I do the following
Assuming an initial 100 g of sodium silicate and using the above reaction for molar equivalents then we have:
16.7g NaOH \* (1 mol / 39.99 g NaOH) \* (1 mol Na2O / 2 mols NaOH) \* (61.98 g Na2O / mol) = 12.94 g Na2O


The total water in the system is thus 100 g - SiO2 g - Na2O g = 100 - 30.3 - 12.94 = 56.76%.


Therefore, the sodium silicate with respect to Na2O and not NaOH is:


* 30.3 g SiO2
* 12.94 g Na2O
* 56.76 g H2O


**##**


If I add 7.7 g of 50% NaOH I now have this mass added to my system.


* 2.98 g Na2O (using same logic above)
* 4.72 g H2O


Allowing time for stirring, I could then expect my new desired SiO2/Na2O weight ratio to be (30.3 g) SiO2 /(12.94 + 2.98)g Na2O = 1.90


Does this seem correct? Should I take into account any equilibria considerations from the main reaction because H2O is the solvent?


**EDIT FROM MAURICE REPLY 1:**


Let me be more specific of the small calculation step I missed in between going from the **##** symbol above.


To get my desired SiO2/Na2O ratio of 1.9 I rearrange to get the total Na2O required in the system which is Na2O = 30.3 g SiO2/1.9 = 15.947 g Na2O.


Considering the 12.94 g Na2O in the original solution, I need to add (15.947 - 12.94)g Na2O = 3 g Na2O into the final system to achieve my desired ratio of 1.9.


In 100 g of a 50% NaOH solution (with respect to NaOH) the quantities are:


* 50 g NaOH
* 50 g H2O


With respect to Na2O, these quantities are:


* 38.7 g Na2O
* (100 - 38.7) = 61.25 g H2O


Thus if I want to add 3 g Na2O from the 50% solution then that corresponds to:
3 g Na2O x (1 g NaOH Solution/0.387 g Na2O) = 7.75 g NaOH solution (slight rounding errors in answers before this edit)




Let focus on a generally applicable solution and then on your particular case.





| Quantity | Equation | Value | Legend |
| --- | --- | --- | --- |
| $M\_\ce{NaOH}$ |  | $\pu{39.9971 g mol−1}$ | the molar mass of $\ce{NaOH}$ |
| $M\_\ce{Na2O}$ |  | $\pu{61.979 g mol−1}$ | the molar mass of $\ce{Na2O}$ |
| $w\_\ce{SiO2}$ | The given value | 30.3 % | the reported (equivalent) weight percentage of SiO2 in the stock sodium silicate solution. |
| $w\_\ce{NaOH}$ | The given value | 16.7 % | the reported (equivalent) weight percentage of NaOH in the above solution. |
| $m\_\mathrm{sol}$ | The given value | $\pu{100 g}$ | the total mass of the solution to be adjusted. |
| $r\_{\ce{SiO2}/\ce{Na2O}}$ | The given value | 1.9 | the intended final $\ce{SiO2}$/$\ce{Na2O}$ ratio in the adjusted solution. |
| $r\_{\ce{SiO2}/\ce{NaOH}}$ | $r\_{\ce{SiO2}/\ce{NaOH}}=r\_{\ce{SiO2}/\ce{Na2O}}\frac{M\_\ce{Na2O}}{2M\_\ce{NaOH}}$ | $\approx 1.472$ | the equivalent final $\ce{SiO2}$/$\ce{NaOH}$ ratio in the adjusted solution. |
| $m\_\ce{SiO2}$ | $m\_\ce{SiO2}=m\_\mathrm{sol} \frac{w\_\ce{SiO2}}{100}$ | $\pu{30.3 g}$ | the (equivalent) mass of $\ce{SiO2}$ in the used stock silicate solution |
| $m\_\ce{NaOH}$ | $m\_\ce{NaOH}=m\_\mathrm{sol} \frac{w\_\ce{NaOH}}{100}$ | $\pu{16.7 g}$ | the (equivalent) mass of $\ce{NaOH}$ in the used stock silicate solution |
| $m\_{\ce{NaOH}\mathrm{,add}}$ | $m\_{\ce{NaOH}\mathrm{,add}}=\frac{m\_\ce{SiO2}}{r\_{\ce{SiO2}/\ce{NaOH}}} - m\_\ce{NaOH}$ | $\approx \pu{3.88 g}$ (resp. $\pu{7.76 g}$ of 50 % $\ce{NaOH}$) | the mass of $\ce{NaOH}$ to be added to the stock solution of the mass $m\_\mathrm{sol}$ to obtain the desired $r\_{\ce{SiO2}/\ce{Na2O}}$ ratio. (For obvious reasons, the value must not be negative, otherwise we are out of luck, as there is already too much NaOH). |


  1

I have not understood your calculations. Here are mine.


Let me first rewrite your data in the following way. All your solutions have a mass ratio defined by : $\ce{$r$ = \frac{$m$(SiO2)}{$m$(Na2O)}}$ The initial solution is defined by is $\ce{$r\_o$ = 2.34}$. And you wanted to obtain a final ratio $\ce{$r\_f$ = 1.9}$, by adding $\ce{NaOH}$ and maybe some water, but without changing the amount of $\ce{SiO2}$, i.e. $\ce{30.3 g SiO2}$.


In other words, you wanted a final solution containing $\ce{30.3 g/1.9 = 15.95 g Na2O}$.


In moles, this amount is equal to : $\pu{\frac{15.95g}{62 ~g/mol} = 0.2572}$ mol $\ce{Na2O}$, which is also equivalent to $0.5144$ mol $\ce{NaOH}$.


But $100$ g of the original solution contains already $\pu{16.7 g NaOH}$, or : $\ce{ \frac{16.7 g}{40 g/mol} = 0.4175 mol NaOH}$ .


As a consequence, you have to add $\pu{~ 0.5144 mol - 0.4175 mol = 0.0969 mol NaOH}$, which weighs : $\pu{0.0969 \times ~ 40 g = 3.88 g}$ $\ce{NaOH}$.


Sorry. In my original answer, I made a wrong calculation. I have edited it recently for correcting this mistake.


This corresponds to adding $7.76$ g of a $50$% $\ce{NaOH}$ solution, to $\pu{100 g}$ of the initial solution. This is the same result as Hendricks13 in his reply 1.


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