Influence of pH on precipitate crystallite sizes?

It's normally assumed that a higher pH of solution would slow nucleation rates and hence lead to larger crystal formation. I've found literature that it is due to solid-liquid inter-facial tension. Which is a concept that I do not fully understand.


In my experiments, higher pH solutions led to the formation of smaller crystallites. I can't make sense of why its contradicting the literature. Here are the images from my experiments:


[![Images of crystals at different pH](https://i.stack.imgur.com/TBPmK.jpg)](https://i.stack.imgur.com/TBPmK.jpg)


My samples are precipitates produced from mixing aqueous $\ce{CaCl2}$ +$\ce{MgCl2}$ and $\ce{Na2CO3}$ solutions. The elevated pH images were produced through the addition $\ce{NaOH}$ to the previously mentioned mixture.


If you have any comments or ideas, it would be greatly appreciated. I am a engineering student who is trying his hand at inorganic chemistry. 


Thanks in advance.




At low pHs there is little "free" $\ce{CO3^{2-}}$ to precipitate the $\ce{Mg^{2+}}$ and $\ce{Ca^{2+}}$ cations. Most of the carbonate species are dissolved $\ce{CO2}$, $\ce{H2CO3}$ and $\ce{HCO3-}$. Thus the precipitate forms slowly and you get relatively large crystals. 


At high pHs there is a lot "free" $\ce{CO3^{2-}}$ to precipitate the $\ce{Mg^{2+}}$ and $\ce{Ca^{2+}}$ cations. Thus the precipitate forms relatively rapidly and you get relatively small crystals. 


I have no idea what literature would lead you to believe otherwise. 


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Aside of influence on the crystallization rate, you have also consider the direct chemical influence.


The sodium carbonate hydrolyzes in not enough alkaline solutions.
$$\ce{Na2CO3 + H2O <=> NaHCO3 + NaOH}$$


Precipitation of carbonate is interfered by solubility of Bicarbonates.
$$\ce{CaCO3 v + H2O + HCO3- <=> Ca(HCO3)2 + OH-} $$


Addition of the hydroxide affects both processes. It shifts the equilibrium towards precipitation and also speeds up the precipitation.


The same for magnesium. 


For magnesium, there is possibility of a side reaction of precipitation the insoluble hydroxide. But lower carbonate solubility will probably overrule this.


$$\ce{MgCl2 + 2 NaOH -> Mg(OH)2 v + 2 NaCl}$$


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