Calcium chloride hydrolysis

Question

I was wondering about $ \ce{CaCl2} $ hydrolysis.

The most primitive answer would have been:

$ \ce{Ca(OH)2} $ does not dilute that great, therefore it is not strong enough to have the $ \ce{Ca^2+} $s not hydrolise. Thus:

$$ \ce{Ca^2+ + H2O <<=> CaOH^+ + H+} $$

Yet, as we can see in this answer - $ \mathrm{pK_B^{II}} $ for $ \ce{Ca(OH)2} $ is $ 1.4 $, which is not bad. Therefore, some reckon the diluted part of $ \ce{Ca(OH)2} $ to be a pretty strong base, making hydrolysis impossible. To make things worse, the cation hydrolysis is known to be harder to trace rather than anion hydrolysis.

Still, I have come across some articles on $ \ce{CaCl2} $ hydrolysis (like this one - a top Google search line; yet, this one first shows that $ \mathrm{\Delta G_{900^oC}} $ is $ \mathrm{90.951 > 0} $).

So, is it safe to say that at $ \mathrm{\approx 25^oC} $ $\ce{CaCl2}$ does not undergo hydrolysis?

Answer 1

Just calculating from the head by the simplifying formula $\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log{(\frac {[\ce{A-}]}{[\ce{HA}]})}$, $\mathrm{pH}$ of $\pu{1 M } \ce{CaCl2}$ is approximately 1/2( 12.6 - log 1)=6.3. So concentration $\ce{[CaOH+]}$ is about million times smaller then $\ce{[Ca^2+]}$. Hydrolysis happens, but is fully negligible. Note that the real $\mathrm{pH}$ depends on impurities and dissolved $\ce{CO2}$.

About the calculation of respective fractions, we can use the formula for gradual dissociation of multiprotic acid:

If we formally suppose $$\ce{H2A <=> HA- + H+ <=> A^2- + 2 H+}$$ then respective molar fractions for particular acid forms are

$$x_{\ce{H2A}} = x_{\ce{Ca^2+}} =\frac {[\ce{H+}]^2}{[\ce{H+}]^2 + K_\mathrm{a1} \cdot [\ce{H+}] + K_\mathrm{a1} \cdot K_\mathrm{a2}}$$

$$x_{\ce{HA-}} = x_{\ce{Ca(OH)+}} =\frac {K_\mathrm{a1} \cdot [\ce{H+}]}{[\ce{H+}]^2 + K_\mathrm{a1} \cdot [\ce{H+}] + K_\mathrm{a1} \cdot K_\mathrm{a2}}$$

$$x_{\ce{A^2-}} = x_{\ce{Ca(OH)2}} =\frac {K_\mathrm{a1} \cdot K_\mathrm{a2}}{[\ce{H+}]^2 + K_\mathrm{a1} \cdot [\ce{H+}] + K_\mathrm{a1} \cdot K_\mathrm{a2}}$$

where $K_\mathrm{a1} = K_\mathrm{w} / K_{\mathrm{b1},\ce{Ca(OH)2}}$ and $K_\mathrm{a2} = K_\mathrm{w} / K_{\mathrm{b2},\ce{Ca(OH)2}}$

Answer 2

There is actually a somewhat unique and important aspect of anhydrous Calcium chloride hydrolysis, not yet addressed.

Namely, a very exothermic dissolution reaction that occurs with solid $\ce{CaCl2}$ from replicating the reverse reaction of its formation from Hydrogen chloride acting on Calcium oxide, at least, to a limited extent:

$$\ce{ CaO (s) + 2 HCl(g) <=> CaCl2 (s) + H2O (l) + Energy}$$

So, where does the energy term arise to move the cited equilibrium reaction partially back to the left? The answer for solid calcium chloride apparently includes a lattice dissociation enthalpy of +2,258 kJ/mol.

Here is an example of a supporting report (also found in Concise Encyclopedia of Chemistry by DeGruyter):

Thus, when you add calcium chloride to water, the solution heats. When adding calcium chloride to water, hydrochloric acid and calcium oxide form. You must be careful when mixing the substances due to the heat of the reaction and the acid produced.

The chemical implications of a presence of some insoluble $\ce{CaO}$ also includes a possible basic salt formation (as in calcium oxychloride), or with air/CO2 exposure, $\ce{CaCO3}$. So, if ones leaves anhydrous Calcium chloride open in even a low humidity atmosphere, expect some deterioration over time to the exposed product (with possible further implications for concrete, for example).

A suggested experiment, to newly purchased $\ce{CaCl2}$ take a small amount and completely dissolve in water. Note, the heating of the solution. Take a similar amount of $\ce{CaCl2}$ and place it on a plate in a low humidity environment. In a week, try again to dissolve the salt completely.