[jeepack]

Hi! I understand chemistry but I don't remember how the ion reactions work.

The equation I am working with is Na2CO3 + Ca(OH)2 = 2 NaOH + CaCO3

My objective here is to use exactly the right amount of water in the initial Na2CO3 dissolved sollution in order for the end result to be a predetermined concentration of NaOH liquid with precipitated CaCO3.

My problem is that the solubility of Na2CO3 and Ca(OH)2 are very different. I am suppose to reach about 10% NaOH (in water solution) without having to add water or evaporate it. When I do the math, it says that if I add the correct amount of water in order for the NaOH solution to be a 10% solution, then I will only have enough water in the mix to completly dissolve one of the reactants, the Na2CO3 and not all of the Ca(OH)2. In fact, after complete dissolving of Na2CO3, there is enough space left in the water for about 1% of the total Ca(OH)2 solids to dissolve correctly; anything more than 1% would theoretically saturate the solution.

Can someone please help me?

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My guess: My guess is that all it takes is that 1%, and then the precipitation starts and every time one molecule precipitates, it leaves free water in order to dissolve more of the reactant, hence making the reaction a progressive reaction where the Ca(OH)2 will eventually completly dissolve, as long as the precipitation into CaCO3 is also taking place... This is just a guess, I would like to know if I am correct, and what kind of calculations would demonstrate this. Also, I am wondering if it is better to mix the Ca(OH)2 in a heterogenous water solution when adding it to the dissolved Na2CO3 solution, or if it is ok to drop in the Ca(OH)2 powder directly, as long as done with small quantities...
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Thanks in advance