Solubility equilibrium
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- Written by: Germán Fernández
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In this section we will study precipitation reactions. Many salts have low solubility in water and form precipitates. In the image we can see the formation of the lead iodide precipitate when mixing solutions of potassium iodide and lead nitrate.
$2I_{(aq)}^{-}+Pb_{(aq)}^{2+}\rightarrow PbI_{2(s)}$
Precipitation reactions are of great importance in the separation of ions using the fractional precipitation technique. This technique can be used both at the laboratory level to determine the amounts of an ionic compound present in a solution, and at the industrial level to recover certain metals from liquid effluents, either for economic or environmental interest.
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- Written by: Germán Fernández
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The solubility product constant, $K_{ps}$, is the equilibrium constant established between the solid and its ions in a saturated solution. It is an equilibrium constant as a function of concentrations.
Since the activity of pure solids is 1, only species in solution participate in the equilibrium constant. Let's see an example:
Silver chromate is a highly insoluble salt that is formed by mixing potassium chromate and silver nitrate. In the solution, the following equilibrium is established:
$Ag_2CrO_4(s)\rightleftharpoons 2Ag^+(aq)+CrO_4^{2-}(aq)$
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- Written by: Germán Fernández
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Write the solubility product constant for calcium fluoride.
Solution:
We write the equation for the equilibrium between salt and its ions:
$CaF_2(s)\rightleftharpoons Ca^{2+}(ac)+2F^-(ac)$
The solubility product constant is given by:
\begin{equation} K_{ps}=[Ca^{2+}][F^-]^2 \end{equation}
The solid does not participate in the constant and the concentrations must be raised to the stoichiometric coefficient.
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- Written by: Germán Fernández
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Solubility indicates the maximum molar concentration of ions in equilibrium with the solid in a saturated solution. For a salt it is possible to determine the solubility product constant from the solubility, and also the solubility from the solubility product constant. Let's see two examples:
The solubility of calcium sulfate in aqueous solution at 25ºC is 0.2 g of calcium sulfate in 100 mL. What is the value of the solubility product constant?
Read more: Relationship between solubility and solubility product constant
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- Written by: Germán Fernández
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The solubility of a salt decreases when a certain concentration of any of the ions that the salt generates upon dissolution (common ions) previously exists in the medium. The solubility of lead iodide in pure water at 25°C is $1.2x10 -3 $mol of lead iodide per liter of solution. If previously in the water we dissolve a certain amount of potassium iodide and then add lead iodide, we will observe a drastic decrease in its solubility, since iodide is a common ion. In the same way, if we dissolve lead iodide in water that contains certain amounts of lead, the solubility of the salt will be reduced by the presence of the common ion (in this case $Pb^{2+}$) We can also appreciate the effect of the common ion by adding to a saturated solution of lead iodide certain quantities of sodium iodide that will immediately precipitate the lead iodide. The explanation for this phenomenon lies in Le Châtelier's principle, according to which adding a product of a reaction in equilibrium produces a displacement towards the reactants. Adding iodide anions or lead cations shifts the solubility equilibrium towards the solid ($PbI_2$).
Calculate the molar solubility of lead iodide in a 0.1 M KI aqueous solution. $K_{ps}(PbI_2)=7.1x10^{-7}$
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In the previous point we have seen that the presence of a common ion decreases the solubility of salts. But, what happens if the ions present are different from those involved in the dissociation of the salt? In this case there is a slight increase in the solubility of the salt, called the saline effect. Thus, for example, the solubility of potassium nitrate in a sodium chloride solution is slightly higher than in pure water.
The increase in ionic concentration in the solution produces greater attractions between ions (increased ionic strength), which produces a decrease in activities (real ion concentrations) with respect to molar concentrations.
This decrease in activities forces a greater dissolution of the salt to reach the solubility product.
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- Written by: Germán Fernández
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The reaction quotient is defined identically to the equilibrium constant but input initial concentrations into it. Let's see an example with silver iodide:
$AgI(s)\rightleftharpoons Ag^+(aq)+I^-(aq)$
\begin{equation} K_{ps}=[Ag^+][I^-]=8.5x10^{-17} \end{equation}
- We mixed two solutions of silver nitrate and potassium iodide to prepare a solution of silver iodide with concentrations $[I^-]=0.01\;M$ and $[Ag^+]=0.02\;M$. To find out if the resulting solution is unsaturated, saturated or supersaturated, we calculate the reaction quotient:
$Q_{ps}=[Ag^+][I^-]=0.01x0.02=2x10^{-4}$
Comparing the reaction quotient with the equilibrium constant, we observe that: $Q_{ps}>K_{ps}$. This means that the concentrations of $I^-$ and $Ag^+$ in the solution are higher than those that a saturated solution would present, producing the precipitation of silver iodide, the solution is supersaturated.
Read more: The reaction quotient: Saturated, unsaturated or supersaturated solution