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I need to make a saturated solution of $\ce{Na2CO3}$ and according to the available data I should add $22\ \text{g}$ in $100\ \text{ml}$ of water at $25\ ^\circ\text{C}$.

I wanted to make $300\ \text{ml}$ solution, which would require $66\ \text{g}$. But, even after adding $110\ \text{g}$ of the salt, no turbidity appears which might indicate saturation. So I believe that the solution is still not saturated.

However, when I kept the flask overnight, almost the entire salt precipitated out in large crystalline chunks. I am a bit confused as to what the problem might be. Has anyone faced this kind of trouble?

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  • $\begingroup$ Can you site your source for the solubility of $Na_2CO_3$ as 22 g/100 mL? I wonder if there is a better estimate. $\endgroup$ – bobthechemist Jun 30 '13 at 2:32
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    $\begingroup$ @bobthechemist Ullmann's Encyclopedia of Industrial Chemistry (2007) gives a solubility of 31.3 g / 100 g solution at 32 °C. Etacude, an online resource, provides a solubility curve an a table for the temperature depencency. The solubility at 20 °C is given as 21.8 g / 100 g water. $\endgroup$ – Klaus-Dieter Warzecha Jan 23 '14 at 0:01
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$\ce{Na_2CO_3}$ forms several hydrates with decomposition point close to room temperature (30-40 Celsium). Moreover, the solubility of decahydrate decreases with temperature decrease, and solubility of anhydrous salt almost does not depend on temperature and is much higher than solubility of decahydrate. The relevant pictures are googleable by words 'sodium carbonate solubility curve', see 'images' google service, not general search.

So, I believe you have prepared a solution of the salt at the higher temperature at day, but the salt mostly precipitated in form of decahydrate overnight, because the solution cooled significantly at night. Bonus points for this version if you have significant changes in temperature at your work during the nychthemeron.

Second version is that you indeed prepared seemingly oversaturated solution because it is not saturated for anhydrous $\ce{Na_2CO_3}$, but overnight it crystallized, because it is saturated for decahydrate. I'm a bit sceptical about such possibility, though, because significantly oversaturated solution crystallize even on slightest imperfections, and the still not-dissolved salt is a great collection of such imperfections.

In addition, if the solution stayed overnight open, it indeed may consumed some amount of atmospheric $\ce{CO_2}$ and partially transform into sodium hydrocarbonate with much lesser solubility.

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One possibility that occurs to me is that some proportion of the carbonate anions underwent an acid-base reaction with the water to form bicarbonate, which is substantially less soluble (96g/L for $\ce{NaHCO3}$ vs. 215g/L for $\ce{Na2CO3}$ at 20°C according to the respective Wikipedia pages, meaning the carbonate salt is nearly twice as soluble in terms of molar ratios), and therefore would precipitate out. The acid-base equilibrium would be driven forward by the formation of said precipitate over time.

As for why no precipitate formed initially, even when the solution was supersaturated, I'm not sure. Normally, preparation of a supersaturated solution would require heating to increase solubility. Then, upon cooling, crystals will precipitate out if given nucleation sites around which to form and/or the solution is physically disturbed. The solvation of sodium carbonate is an exothermic process, so the process would raise the temperature of the solution, but I doubt that's sufficient to explain the unexpectedly great solubility that you observed.

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The reason it crystallized was probably because the temperature of the solvent decreased, lowering the solubility level. Also solubility of many solutes are not linearly proportional. Lots of these are solubility curves, so adding 66g to 300ml would not be the same as 33g to 100 ml. When displayed in a graph, the amount of a solute dissolved in a given amount of solvent at the give amount of temperatures, is shown as a quadratic equation. It appears as half a parabola.

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  • $\begingroup$ I think you meant "22g in 100ml" instead of "33g in 100ml". $\endgroup$ – Satwik Pasani Jan 23 '14 at 3:35

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