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Dissociation constant can be calculated from the Gibb's energy released by the dissociation, but as best I can tell, there's no way to determine solubility from dissociation constant, as things can dissolve without dissociating. Is this right?

A specific example that is perplexing me is disassociation of calcium chloride. I calculate from CRC Handbook values for aqueous $\ce{CaCl2}$ that at 300K, the pKd is 4.75, so the pCl is 1.48 and pCa is 1.78, if the pCaCl2 is 0. $\ce{CaCl2}$ is very soluble in water, so does this just mean that 1.7% of the $\ce{CaCl2}$ that is dissolved in water dissociates? I thought that salts generally dissociate almost fully, is calcium chloride just an exception? Is it really the case that salts can only partially dissociate like this even when highly soluble?

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  • $\begingroup$ How did you get pKd ? And how it comes pCa is so much bigger than pCl? That 1% should turn on a red light in your train of thoughts. $\endgroup$
    – Poutnik
    Commented Nov 21, 2022 at 22:19
  • $\begingroup$ It does turn a red light, that's why I'm here asking this. And the pCa is bigger because there's half as much Ca as Cl. $\endgroup$
    – Liam Clink
    Commented Nov 22, 2022 at 2:11
  • $\begingroup$ But what about pKd? $\endgroup$
    – Poutnik
    Commented Nov 22, 2022 at 5:33
  • $\begingroup$ Please do not confuse first time readers by striking things and including edit statements. If you change your question because you have made a mistake, simply do it. Then flag all comments relating to that change for deletion. $\endgroup$ Commented Nov 26, 2022 at 11:52
  • $\begingroup$ @Martin-マーチン I apologize, I wasn't aware that was possible, and I'm not sure how to do so. $\endgroup$
    – Liam Clink
    Commented Nov 28, 2022 at 16:34

2 Answers 2

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Suppose we have a partially soluble ionic salt in water of the form:

$$\pu{\ce{A_aB_b(s)}}$$

With the condition: $(a≠b)$ or $(a=1$ and $b=1)$

When this salt is added to water, the following equilibrium takes place first:

$$\pu{\ce{A_aB_b(s)<=>A_aB_b(aq)}}$$

Where the equilibrium constant associated with this process would be the following, since we can approximate negligible contribution from the solid phase:

$$\pu{K_s=\ce{[A_aB_b(aq)]}}$$

The next equilibrium system that takes place is:

$$\pu{\ce{A_aB_b(aq)<=>aA^b+(aq) +bB^a-(aq)}}$$

With the equilibrium expression:

$$\pu{K_d=\ce{\frac{[A^{b+}]^a\;[B^{a-}]^b}{[A_aB_b(aq)]}}}$$

When we multiply both equilibrium constants we get:

$\require{cancel}$

$$\pu{K_{sp}=K_s\;K_d=\ce{\cancel{\ce{[A_aB_b(aq)]}}}\times \ce{\frac{\ce{[A^{b+}]^a\;[B^{a-}]^b}}{\cancel{\ce{[A_aB_b(aq)]}}}}=\ce{[A^{b+}]^a\;[B^{a-}]^b}}$$

Which corresponds to the overall process:

$$ \require{cancel} \begin{align} \ce{A_aB_b(s)&<=>\cancel{\ce{A_aB_b(aq)}}} \tag{$K_s$} \\ \ce{\cancel{\ce{A_aB_b(aq)}}&<=>aA^b+(aq) +bB^a-(aq)} \tag{$K_d$} \\ \hline \ce{\ce{A_aB_b(s)}&<=>aA^b+(aq) +bB^a-(aq)} \tag{$K_{sp}$} \\ \end{align} $$

If $K_s$ is significantly higher than $K_d$, $K_{sp}$ would also be significantly higher, even though $K_d$ is small:

$$K_s>>K_d\implies K_{sp}>>K_{d}$$

The solubility $S$ of this salt could be calculated from the following expression, assuming no initial amounts of ions are present:

$$K_{sp}=(aS)^a\;(bS)^b$$

In conclusion, both $K_s$ and $K_d$ along with their respective equilibrium systems contribute to the solubility of the salt, and not just $K_d$ independently as one could wrongly assume.

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  • $\begingroup$ Ok, this is what I was thinking. So do highly soluble salts actually sometimes have low dissociation like I found for CaCl2? Also, there are four red "\cancel" in your equations $\endgroup$
    – Liam Clink
    Commented Nov 22, 2022 at 2:09
  • $\begingroup$ I don't know if your $K_d$ values for $\ce{CaCl2}$ are correct or incorrect, but as an example, at 25°C, the $K_s$ of $\ce{AgCl}$ is $3.6\times 10^{-7}$, while its $K_d$ is $5\times 10^{-4}$, giving a $K_{sp}$ of approximately $1.8\times 10^{-10}$. Also, I'm not sure why you're seeing those red terms, since both my computer and cellphone open the page normally. In short, the concentrations and text for the aqueous salt are cancelled out when multiplying $K_s$ and $K_d$ and adding both equilibrium systems, respectively. $\endgroup$
    – Sam202
    Commented Nov 22, 2022 at 2:21
  • $\begingroup$ It's showing correctly for me now. Also I checked my math and fixed it. I'm rusty with my chemistry, so I think I mixed up the solvation and dissociation because most of the experiments I did in high school were with strong acids, which by definition dissociate near fully. $\endgroup$
    – Liam Clink
    Commented Nov 22, 2022 at 2:44
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    $\begingroup$ Note that concept of Ksp is not applicable on ion molar concentrations of highly soluble salts, as ion TD activities are very different to their concentrations. $\endgroup$
    – Poutnik
    Commented Nov 22, 2022 at 9:34
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    $\begingroup$ Furthermore \pu{1 unit} is only for physical units and \ce{formula/equation} is for chemical equations. They don't escape within each other and they cannot be chained, and they should only be used for what they are intended. $\endgroup$ Commented Nov 26, 2022 at 12:02
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Dissociation constant can be calculated from the Gibb's energy released by the dissociation, but as best I can tell, there's no way to determine solubility from dissociation constant, as things can dissolve without dissociating. Is this right?

That is right. A more straightforward example would be acetylsalicylic acid (Aspirin), a weak acid. It has a certain solubility (quite low), and then dissociates at appropriate pH. More aspirin goes into solution because some of it is in the (charged) deprotonated state. The solubility of the neutral species and its acid dissociation constant are not linked directly - there are weak acids with a range of solubilities in water, and a range of acid strengths, and the two are not correlated.

I thought that salts generally dissociate almost fully, is calcium chloride just an exception?

At low concentrations, you would expect a salt to dissociate fully in water. At higher concentrations of one or both ions, ion pairs may form. They come in different flavors, contact ion pairs and solvent separated ion pairs. For calcium chloride, you could write them like this:

$$\ce{[Cl- Ca^2+ Cl-], [Cl- H2O Ca^2+ H2O Cl-] and [Cl- (H2O)_x Ca^2+ (H2O)_x Cl-]}$$

Also, you might have ion pairs with different stoichiometry, such as positively charged $\ce{[Ca^2+ Cl-]}$ or negatively charged $\ce{[Ca^2+ . 3Cl-]}$. The neutral ion pairs will not contribute to the conductivity. This is one fairly simple way to detect them.

How an ion speciates can be studied with advanced techniques such as solution state X-ray methods (EXAFS, e.g. this paper) or measuring the activity of the ions (e.g. through precipitation equilibria with a third ion). For calcium chloride, you have to have highly concentrated solutions at high temperature to see the effect of ion pair formation. An example of an ionic compound forming appreciable amounts of ion pairs is magnesium sulfate in the ocean.

There are other cases where different species co-existing in solution complicate things. Two common examples are carbon dioxide in water (either as carbon dioxide or as carbonic acid) and protons in water (either as hydronium ion or as higher-order complex with multiple waters). The pragmatic way of treating these complications is to talk about the total concentration, and use this in discussions of concentrations, reaction quotients and equilibrium constant expressions.

Is it really the case that salts can only partially dissociate like this even when highly soluble?

Most salts dissociate completely at low concentrations (behave like strong electrolytes). If they are soluble to high concentration (or you can add one or the other ion to high concentration), you increase the tendency to form coordination complexes (for transition metals) or ion pairs (for any cation).

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  • $\begingroup$ Is this tendency to form complexes related to activity? $\endgroup$
    – Liam Clink
    Commented Nov 28, 2022 at 16:38

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