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I'm currently running an electrolysis experiment and I have a few questions regarding crystallization of zinc sulfate monohydrate $\ce{ZnSO4 * H2O}$.

I have prepared a saturated solution of zinc sulfate, will it crystallize as monohydrate, or only heptahydrate $\ce{ZnSO4 * 7 H2O}$ will?

And if so, what is the reason behind it? Does it has to do with the monohydrate and heptahydrate part?

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    $\begingroup$ You need to crystallize monohydrate at elevated temperature, above 70 °C. Between 40 and 70 °C you obtain hexahydrate, below 40 °C – heptahydrate. $\endgroup$
    – andselisk
    Nov 9, 2017 at 10:05
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    $\begingroup$ There is no such thing as water solution of $\ce{ZnSO4.H2O}$. There is just water solution of $\ce{ZnSO4}$, period. You might have used solid $\ce{ZnSO4.H2O}$ to prepare it, but that's irrelevant. As soon as $\ce{ZnSO4}$ is dissolved, it has no idea that it comes from a monohydrate. $\endgroup$ Nov 9, 2017 at 10:16
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    $\begingroup$ I'd use a warm water bath and a saturated at 75 - 80 °C solution. $\endgroup$
    – andselisk
    Nov 9, 2017 at 10:16
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    $\begingroup$ @IvanNeretin From what I understood, it's about crystallizing the monohydrate, not keeping it in solution. $\endgroup$
    – andselisk
    Nov 9, 2017 at 10:18
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    $\begingroup$ No, you keep it above 70 °C as you grow the crystals. If you just let it cool off, you end up with heptahydrate. At least that's how I interpret that information. $\endgroup$
    – andselisk
    Nov 9, 2017 at 10:34

1 Answer 1

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TL; DR: obtaining a reasonably big single crystal of $\ce{ZnSO4 * H2O}$ seems to be a tricky task, whereas one can easily obtain $\ce{ZnSO4 * H2O}$ as a result of thermal dehydration of heptahydtare above $\pu{70 ^\circ C}$ in a form of melt, and above $\pu{120 ^\circ C}$ – as a polycrystalline solid material.


I finally got across to suitable references to back up my comments. In nature zinc sulfate monohydrate occurs as mineral gunningite, always in a form of efflorescence deposition on the surface of sphalerite and other minerals and doesn't form single crystals [1]:

Surface and ground waters carrying dissolved oxygen attack the sphalerite, yielding soluble sulphates of zinc [...]:

$$\ce{ZnS + 2 O2 -> ZnSO4}$$

[...] In the presence of abundant water all of the soluble salts are removed. If, however, the conditions are such that the supply of water is restricted or evaporation takes place, the metal-bearing solutions become supersaturated with the consequent precipitation of a number of hydrates of zinc, manganese, and iron of which gunningite, $\ce{ZnSO4 * H2O}$, is one.

Thermal dehydration among zinc sulfate hydrates has been studied with DTA in several publications. The most relevant and thorough investigation [2, p. 471] suggests the following scheme:

$$\ce{ZnSO4 * 7 H2O (s) ->[\pu{37 ^\circ C}] ZnSO4 * 6 H2O (s) ->[\pu{70 ^\circ C}] ZnSO4 * H2O (l) ->[\pu{120 ^\circ C}] ZnSO4 * H2O (s) ->[\pu{285 ^\circ C}] ZnSO4 (s)}$$

Original text in German:

Festgestellt werden vier endotherme Effekte: bei etwa $\pu{37 ^\circ C}$ verwandelt sich das Heptahydrat in Hexahydrat; bei etwa $\pu{70 ^\circ C}$ geht das Hexahydrat in Monohydrat über, wobei teilweise Schmelzen im Kristallwasser auftritt. Bei etwa $\pu{120 ^\circ C}$ siedet die Schmelze, Wasser scheidet sich ab, bis das feste Monohydrat zurückbleibt; bei etwa $\pu{285 ^\circ C}$ erfolgt der Übergang des Monohydrates in wasserfreies Salz.

The tricky part is that monohydrate can be isolated in crystal form, but it initially isolates as a melt, and only above $\pu{120 ^\circ C}$ makes a polycrystalline solid phase (occurs as white powder or granules).

Aqueous solubility $\omega_2$ for zinc sulfate $\ce{ZnSO4}$ in mass% of a solute ($\omega_2 = m_2/(m_1 + m_2)$, where $m_2$ is the mass of solute and $m_1$ the mass of water) as a function of temperature ($T$) from CRC Handbook [3, p. 5-172]:

\begin{array}{r|rrrrrrrr} T,\,\pu{^\circ C} & 0 & 10 & 20 & 25 & 30 & 40 & 50 & 60 & 70 & 80 & 90 & 100 \\ \hline \omega_2,~\% & 29.1 & 32.0 & 35.0 & 36.6 & 38.2 & 41.3 & 43.0 & 42.1 & 41.0 & 39.9 & 38.8 & 37.6 \end{array}

With this in mind, what can be suggested is to prepare a saturated at $\pu{70 ^\circ C}$ solution, and then slowly increase the temperature (e.g. using a thermostat), hoping that zinc sulfate monohydrate starts to crystallize (using the decrease in solubility starting from $\pu{60 ^\circ C}$).

According to [4, entry 3575, p. 470], solubility of $\ce{ZnSO4 * H2O}$ also decreases in this interval: $\pu{101 g}/\pu{100 g}~\ce{H2O}$ at $\pu{70 ^\circ C}$, and $\pu{87 g}/\pu{100 g}~\ce{H2O}$ at $\pu{105 ^\circ C}$. The difference is rather small, so I wouldn't expect rapidly-grown large crystals.

References

  1. Jambor, J. L.; Boyle, R. W. The Canadian Mineralogist 1962, 7 (2), 209–218.
  2. Balarew, C.; Trendafelov, D.; Gerganova, M. Monatshefte für Chemie 1971, 102 (2), 465–473. DOI 10.1007/BF00909340 (in German).
  3. Haynes, W. M.; Lide, D. R.; Bruno, T. J. CRC handbook of chemistry and physics: a ready-reference book of chemical and physical data.; 2017; Vol. 97.
  4. Perry, D. L. Handbook of inorganic compounds, 2nd ed.; Taylor & Francis: Boca Raton, 2011. ISBN 978-1-4398-1461-1.
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