Precipitation

Precipitation is when a solid solute spontaneously forms from a solution. This might occur if we mixed two solutions together that contained a mixture of ions that form an insoluble compound. For example, imagine we mixed a solution of BaCl₂ and solution of NaNO₃. This would result in a solution with Na⁺ ions, Ba²⁺, Cl^-1, and NO₃^- ions. As a result of mixing these solutions we could form Ba(NO₃)₂ a compound with a low solubility. We could write this reaction as

\[ \rm{BaCl_2(aq) + 2NaNO_3(aq)} \rightleftharpoons \rm{Ba(NO_3)_2(s) + 2NaCl(aq)}\]

We would also write this as a total ionic equation as

\[ \rm{Ba^{2+}(aq) + 2Cl^-(aq) + 2Na^+(aq) + 2NO_3^-(aq)}\] \[\rightleftharpoons \rm{Ba(NO_3)_2(s) + 2Na^+(aq) + 2Cl^-(aq)}\]

Or ignoring the spectator ions (sodium and chloride) we would simply write the net ionic equation

\[ \rm{Ba^{2+}(aq) + 2NO_3^-(aq)} \rightleftharpoons \rm{Ba(NO_3)_2(s)}\]

You can see the net ionic reaction in this case is essentially simply the dissolution of barium nitrate but written in reverse. Therefore this "reaction" will only happen at certain concentrations. There are concentrations of Ba²⁺ and NO₃^- at which we would expect to form some solid precipitate, while at other concentrations we would not (since the solid would be completely dissolved under these conditions). We can predict these concentration from the solubility product.

The ion product is not only useful for predicting equilibrium conditions; it is also useful to determine if a system is in or out of equilibrium. We therefore need to distinguish between the ion product itself and its value at equilibrium. At equilibrium the ion product is the solubility product K_sp. However, we can calculate this product for any solution (or any imagined solution). For any given set of conditions we can calculate the ion product and denote it with the letter Q. To note we are looking at solubility it would be given the symbol Q_sp. For example above imagine we used a 0.1M BaCl₂ solution and a 1M Na(NO₃) solution. This would yield a mixture with a concentration of barium ions, [Ba²⁺] = 0.1 M, and the concentration of nitrate ions, [NO₃^-] = 1 M. At these concentrations Ba(NO₃)₂ would we form a precipitate? To answer this question we need to compare the ion product for the given conditions, Q_sp, to that at the equilibrium conditions, K_sp.

If we find

\[ Q_{sp} > K_{sp} \hskip24pt {\rm then\;precipitate\;will\;form}\]

This is because the concentrations in the given solution are higher than we would find at equilibrium. When the concentrations are higher than the equilibrium concentrations, then solid will precipitate until the concentrations are reduced to the equilibrium concentrations. However, if we find

\[ Q_{sp} < K_{sp} \hskip24pt {\rm then\;no\;precipitate\;will\;form}\]

This is because the concentrations are below the equilibrium concentrations. Any solid would dissolve spontaneously at these concentrations. Therefore no solid will form. If we find that

\[ Q_{sp} = K_{sp} \hskip24pt {\rm then\;the\;system\;is\;at\;equilibrium}\]

No solid is precipitate. No more solid can dissolve into this solution. We have formed a saturated solution.

Returning to the Ba(NO₃)₂ solution, does a precipitate form when [Ba²⁺] = 0.1 M and [NO₃^-] = 1 M? We need to compare Q_sp and K_sp.

\[ Q_{sp} =\rm{[Ba^{2+}][NO_3^-]^2} = (0.1)(1)^2 = 0.1.\]

A table of K_sp values shows that K_sp for Ba(NO₃)₂ is 4.6 x 10^-3. Since Q_sp > K_sp a precipitate will form. Note: nitrate salts are general soluble. However, even with soluble salts it is possible to get the concentrations so high that they will precipitate. It all depends on what you have now (Q) vs what you would have at equilibrium (K).

A problem to determine if a precipitate forms

Super-Saturated Solutions

It is possible to increase the concentration beyond this equilibrium point. A solution with a concentration higher than the saturation concentration is called a super-saturated solution. In such a condition the system is no longer at equilibrium and will spontaneously form a solid precipitate since Q_sp > K_sp . However, as the formation of the solid crystals takes time the solution can stay trapped in this super-saturated condition for a long time. Typically it is possible to make a super-saturated solution by first making a saturated solution at higher temperature where the solubility is higher. The solution can then be slowly cooled to a lower temperature where the solubility is lower.