I'm not a chemist and don't remember much chemistry from school, so please forgive a naive question (I've read the Wikipedia article on sodium silicate though).

I'm researching this topic to make a positive photo resist developer for making PCBs.

  1. Will dissolving $\ce{Na2SiO3}$ (the anhydrous form) in water produce essentially the same solution as dissolving $\ce{Na2SiO3*5H2O}$ but in different concentration or are these intrinsically different?

  2. Will $\ce{Na2SiO3}$, if left in contact with open air, eventually hydrate to $\ce{Na2SiO3*5H2O}$?

  3. Also, the Wikipedia article talks about the $\ce{SiO2:Na2O}$ ratio affecting the pH of the water solution. Is the information about this ratio already present in the chemical formula so if I buy $\ce{Na2SiO3*5H2O}$ it is always the same ratio or is this a different parameter that is specified in addition to the chemical formula? (I understand that different concentrations of the solution will produce different pH, but consider equal concentrations)

I believe I need the developer to be an alkaline solution so I need to understand what will affect the pH when dissolving sodium metasilicate in water (e.g. if I need $\ce{Na2SiO3}$ or $\ce{Na2SiO3*5H2O}$, if I need to consider other specifications besides the chemical formula, if I need to consider the shelf life of the chemical if it changes with time e.g. gets hydrated from $\ce{Na2SiO3*5H2O}$ into $\ce{Na2SiO3*9H2O}$). Searching electronics forums on the internet I haven't found a clear answer. For some people using this sodium metasilicate as a developer works for some it doesn't develop at all, some add $\ce{NaOH}$ to it but there are places that advertise their sodium metasilicate developer as sodium hydroxide free.

  • $\begingroup$ 1)In the short run they may be different. They key point is that anhydrous sodium metasilicate contains chains, while hydrated sodium metasilicate is actually dihydro-orthosilicate. I guess, the first one will transform into the second one over time in solution, but it will take some time. $\endgroup$
    – permeakra
    Aug 9, 2015 at 13:30

1 Answer 1

  1. As explained in the comment by permeakra, due to the structural differences between anhydrous and hydrated Na2SiO3, their solutions will also be intrinsically different, at least for a while.

  2. Whether anhydrous salt will turn to hydrate or vice versa depends on your ambient humidity and temperature. Anyway, I'd not recommend leaving either of them in contact with open air, for fear they might react with CO2. Silicic acid is so weak that it can be displaced even by carbonic acid.

  3. Well, theoretically, the information about the SiO2:Na2O ratio is already present in the chemical formula Na2SiO3 (with or without water). However, the industrial-grade product may not quite match that formula. In fact, it may be a wild mixture of different silicates, where each individual compound cannot and need not be characterized. For many large-scale applications it suffices to know just the said ratio. Not sure about your application, though.


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