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The following reaction is usually carried out to test for the carbonate anion in an inorganic salt;

$$\ce{CO3^2- +BaCl2/CaCl2 -> BaCO3/CaCO3(s) + 2 Cl-}$$ The resulting carbonate is stated to be an insoluble precipitate, and also to be dissolvable in mineral acids like $\ce{HCl}$.

How does that happen? If the initial test is to actually work, the dissolution equilibrium should lie to the right; in the favour of the carbonate being formed. The dissolution requires the metal ions in the carbonate's solid lattice to break up and get solvated by the water molecules, ending up just like the strong electrolyte that we started with($\ce{BaCl2}$ or $\ce{CaCl2}$).

If you're going to just take some hydrochloric acid and treat the precipitate with it, and the reaction we just saw works out in the reverse, what does that mean about the equilibrium constant? It's supposed to lie to one side no matter what you start with, right?

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The ion $\ce{CO3^{2-}}$ introduced in the first reaction must be produced by dissolving a soluble carbonate like $\ce{Na2CO3}$. On the other hand $\ce{BaCl2}$ or $\ce{CaCl2}$ have to be dissolved in water to react. If they are introduced as solid these substances will not react, unless they slowly get dissolved. Nobody really knows why these chlorides are soluble in water, and the corresponding carbonates are insoluble. So the precipitation reaction should be rewritten as $$\ce{Ba^{2+} + CO3^{2-} <=> BaCO3(s) \ (1)}$$ $$\ce{Ca^{2+} + CO3^{2-} <=> CaCO3(s) \ (2)}$$ Usually the equation is displaced to the right side, and a mixture of some carbonate plus a solution containing Ba or Ca produces a dense precipitate of carbonate. But by adding an acidic solution, containing the ions $\ce{H+}$ (or $\ce{H3O^+}$), the following reaction happens that consumes the acidic ions : $$\ce{2 H+ + CO3^{2-} -> CO2(g) + H2O}$$ Here the ions $\ce{CO3^{2-}}$ are all destroyed and produce a gas which leaves the solution. This displaces the first equilibriums ($1$) and ($2$) to the left-hand side, and the precipitates disappear.

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  • $\begingroup$ Oh, so this is how dissolution actually works; it's not the dissolution of the solid substance, but the reaction with its solvated ions that causes the precipitates to disappear using Le Chatelier's principle. Can this be generally stated for all dissolutions? $\endgroup$
    – harry
    Commented May 10, 2021 at 22:59
  • $\begingroup$ That said, I guess what I did wrong in my answer was taking the dissolution to be just "precipitate+acid". Is that correct? $\endgroup$
    – harry
    Commented May 10, 2021 at 23:02
  • $\begingroup$ @Harry Holmes. Honestly my explanation explains the observed phenomena. But this does not prevent your explanation to be as valid as mine. Who knows ? $\endgroup$
    – Maurice
    Commented May 11, 2021 at 10:10
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On writing the equations for the initial precipitation and the dissolution reactions, it can be seen that their equilibrium constants are not the same. Taking $K_1,K_2$, and $K_3$ to be the constants for each of these and the double dissociation of $\ce{H2CO3}$ respectively, you get $$K_2=\frac{1}{K_1K_3} \neq K_1$$ This is why you can't reason out that the equilibrium in both reactions should be to one side; they're not the same reaction.

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