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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

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.

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  • $\begingroup$ Pure interfaces reason do not take into consideration possible common ions effects, where common means whenever there is an equilibrium that links precipitation and pH. $\endgroup$
    – Alchimista
    Commented Apr 13, 2019 at 7:46

2 Answers 2

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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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