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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)

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    $\begingroup$ What exactly are troubles in getting the result? The old good mass conservation law and Na2O/NaOH molar mass ratio would do. $\endgroup$
    – Poutnik
    Aug 17, 2021 at 16:26
  • $\begingroup$ @Poutnik Sure, that's probably what i want help with. I'm trying to make a graph where i can plot added NaOH against different weight ratios. I have already done this for 100% NaOH (was hoping to confirm my answer) but not for the 50% liquid caustic which I'm having trouble with. $\endgroup$
    – Hendrix13
    Aug 17, 2021 at 23:13
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    $\begingroup$ It is rather a simple algebra at high school level. Readers may hesitate to help you unless you put your resolution attempts to the question and describe your troubles particularly. Solving everything is not preferred way to help here. $\endgroup$
    – Poutnik
    Aug 18, 2021 at 4:20
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    $\begingroup$ Improving the question by editing is definitely preferred to reposting. // In contrary to some other Q/A or forum sites, answers on CH SE site are figuratively paid by the user's own effort. When you ask, it is expected you have already searched for and thoroughly thought about the topic, providing explicit compact summary of partial answers or at least ideas or thoughts you have got until then. Effort not shown may be considered as effort not done and such a question may be closed. $\endgroup$
    – Poutnik
    Aug 18, 2021 at 9:34
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    $\begingroup$ Focus on the fact it is rather a mathematical then chemical task, with simple math skills more needed than particular chemical knowledge. The only CH knowledge needed is NaOH/Na2O molar mass ratio, what you already have to recalculate ratios toward SiO2. $\endgroup$
    – Poutnik
    Aug 18, 2021 at 9:55

2 Answers 2

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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).
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    $\begingroup$ Brilliant comment that's very easy to follow! Thank you for making the whole thing clearer for myself and any other readers :) $\endgroup$
    – Hendrix13
    Feb 17 at 13:58
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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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  • $\begingroup$ Hi @Maurice, thanks for the comment! I don't think your answer is right but I couldn't fit my working/explanation in this comment box. Instead, I've edited my original post above briefly; refer to the "EDIT FROM MAURICE REPLY 1" above. I'm very curious about what the answer is now because I could be overlooking something...? Not sure if I have done anything wrong. Keen for more of your input :) $\endgroup$
    – Hendrix13
    Feb 17 at 8:21

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