Solubility Equilibria of Sparingly Soluble Salts: The solubility of a particular salt in a solvent is always a very significant fact to be taken into consideration when studying its properties and the properties of the solution it forms with a particular solvent. While salts like \(\mathrm{NaCl}, \mathrm{KCl}\) and others such as potassium dichromate are completely soluble in water, there are some salts that do not dissolve at all or dissolve only to a very minimal extent. These salts form a special case when defining the solubility equilibria and are called sparingly soluble salts. 

Such sparingly soluble salts present solubility equilibria which are measured using an entity called a solubility product. This article will discuss the solubility product and how it can be applied to the solubility equilibria of sparingly soluble salts.

What is Solubility Product?

A sparingly soluble salt is so-called because when it is stirred into water or mixed with water, only a very small amount of the salt goes into the solution, and most of it remains undissolved. The solution becomes saturated with that little amount of salt dissolved, and the salt immediately dissociates into its ions. For example, when a sparingly soluble salt like \(\mathrm{AgCl}\) goes into the solution, it dissociates as \(\mathrm{Ag}^{+}\) and \(\mathrm{Cl}^{-}\) ions immediately. So, a dynamic equilibrium exists between the two entities: the undissolved salt and the dissolved ions which are in the solution. 

At equilibrium, the dissolving of solid AgCl and precipitation of undissolved AgCl will take place at the same rate. Hence, the equilibrium between the two opposite processes can be given by a reversible equation:

\({\rm{AgCl(s)Dissolution/Precipitation}} \Leftrightarrow {\rm{AgCl(aq)}} \leftrightarrow {\rm{A}}{{\rm{g}}^{\rm{ + }}}{\rm{(aq) + C}}{{\rm{l}}^{\rm{ – }}}{\rm{(aq)}}\)

The equation can also be written as:

\({\rm{AgCl(aq)}} \leftrightarrow {\rm{A}}{{\rm{g}}^{\rm{ + }}}{\rm{(aq) + C}}{{\rm{l}}^{\rm{ – }}}{\rm{(aq)}}\)

When the law of chemical equilibrium is applied to the above equation, it can be written as:

\({\rm{K = }}\frac{{\left[ {{\rm{A}}{{\rm{g}}^{\rm{ + }}}} \right]\left[ {{\rm{C}}{{\rm{l}}^{\rm{ – }}}} \right]}}{{{\rm{[AgCl]}}}}\)

The concentration of the undissolved solid AgCl usually remains constant, and hence, the above equation can be written as:

\(\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right]=\mathrm{K}^{*}[\mathrm{AgCl}]=\mathrm{K}_{\mathrm{sp}}\)

The \(\mathrm{K}_{\mathrm{sp}}\) is the solubility product and is equal to the ionic product: \(\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right]\) in the saturated solution.

So, for any sparingly soluble salt solution, \({{\rm{A}}_{\rm{x}}}{{\rm{B}}_{\rm{y}}}\) the equilibrium of the reaction can be represented as:

\({{\rm{A}}_{\rm{x}}}{{\rm{B}}_{\rm{y}}} \leftrightarrow {\rm{x}}{{\rm{A}}^{{\rm{y}} + }} + {\rm{y}}{{\rm{B}}^{{\rm{x}} – }}\)

The solubility product for \({{\rm{A}}_{\rm{x}}}{{\rm{B}}_{\rm{y}}}\) can be written as:

\({{\rm{K}}_{{\rm{sp}}}} = {\left[ {{{\rm{A}}^{{\rm{y}} + }}} \right]^{{\rm{x}}*}}{\left[ {\;{{\rm{B}}^{{\rm{x}} – }}} \right]^{\rm{y}}}\)

The \({\rm{x}}\) and \({\rm{y}}\) represent the number of ions present in the formula of the sparingly soluble salt.

Hence, the solubility product at a specified temperature for an electrolyte can be defined as the product of the molar concentrations of the ions present in its saturated solution, with each concentration raised to the power equal to the number of ions produced on the dissociation of one molecule of the electrolyte. 

Solubility Product Examples

The below table gives the solubility product of some of the common sparingly soluble salts at \({\rm{298K}}{\rm{.}}\) 

Sparingly Soluble Salt	Solubility product, \(\mathrm{K}_{\mathrm{sp}}\)
\(\mathrm{AgCl}\)	\(1.8 * 10^{-10}\)
\(\mathrm{AgBr}\)	\(5.0 * 10^{-13}\)
\(\mathrm{CaF}_{2}\)	\(5.3^{*} 10^{-9}\)
\(\mathrm{CaCO}_{3}\)	\(1.3^{*} 10^{-33}\)
\(\mathrm{Mg}(\mathrm{OH})_{2}\)	\(1.8 * 10^{-11}\)
\({\rm{PbS}}\)	\(8.0 * 10^{-28}\)
\({\rm{CuS}}\)	\(6.3^{*} 10^{-36}\)
\({\rm{H}}{{\rm{g}}_2}{\rm{C}}{{\rm{l}}_2}\)	\(1.3^{*} 10^{-18}\)

Difference between Solubility Product and Ionic Product

As we know, the solubility product and the ionic product of an electrolyte is represented by the product of their concentrations, raised to a power that is equal to the number of ions present on its dissociation. So, how do we differentiate between the two of them? Here is how:

a. Ionic product is used for all types of solutions, both saturated and unsaturated. However, the solubility product is only used for saturated solutions, where a dynamic equilibrium exists between the undissolved salt and the dissolved ions present in the solution. So, one can call the solubility product the ionic product in saturated solutions.

b. For any salt, its solubility product is constant at a constant temperature. The ionic product, however, will depend upon the number of ions present at that particular time.

Solubility products of any sparingly soluble salt can be determined using the value of its solubility product in water at that temperature. 

Significance of Solubility Product

The solubility product value can be applied in various aspects and, therefore, is very useful. Below are some of the applications of the solubility product of salts.

a. To Calculate the Solubility of Sparingly Soluble Salt: The solubility of a sparingly soluble salt is very difficult to measure. However, the knowledge of its solubility product can help in determining its solubility at any temperature.

The relationship between solubility \({\rm{(in\, mol/L)}}\) and solubility product \(\left( {{{\rm{K}}_{{\rm{sp}}}}} \right)\) depends upon the nature of the salt. 

i. Salts of type AB: (Example: \(\mathrm{PbSO}_{4}, \mathrm{AgCl}, \mathrm{AgBr}\) etc.

\({\rm{AB}} \leftrightarrow {{\rm{A}}^{\rm{ + }}}{\rm{ + }}{{\rm{B}}^{\rm{ – }}}\)

\({{\rm{K}}_{{\rm{sp}}}}{\rm{ = }}\left[ {{{\rm{A}}^{\rm{ + }}}} \right]\left[ {{{\rm{B}}^{\rm{ – }}}} \right]{\rm{ = }}{{\rm{s}}^{\rm{*}}}{\rm{s = }}{{\rm{s}}^{\rm{2}}}\)

ii. Salts of type \({\rm{A}}{{\rm{B}}_2}:\left( {{\rm{PbC}}{{\rm{l}}_2},{\rm{Sr}}{{\rm{F}}_2}} \right)\)

\({\rm{A}}{{\rm{B}}_{\rm{2}}} \leftrightarrow {{\rm{A}}^{\rm{ + }}}{\rm{ + 2}}{{\rm{B}}^{\rm{ – }}}\)

\(\mathrm{K}_{\mathrm{sp}}=\left[\mathrm{A}^{+}\right]\left[\mathrm{B}^{-}\right]=\mathrm{s}^{*}(2 \mathrm{~s})^{2}=\mathrm{s} * 4 \mathrm{~s}^{2}=4 \mathrm{~s}^{3}\)

Similarly, it can be calculated for all other salts with different types to determine the relationship between solubility and solubility product.

b. To Predict Precipitation Reactions: We know that the solubility product of salt is its maximum saturation point at a particular temperature. So, it represents the upper limit in the ionic product of a sparingly soluble salt.

What that means is that if the ionic product is equal to \(\mathrm{K}_{\mathrm{sp}}\) then no more solute can go into that solution at that particular temperature. This can help in predicting the precipitation.

When the ionic product or the product of the concentration of ions in any solution is more than that of the value of its solubility product of that sparingly soluble salt, then the excess ions will result as a precipitate or precipitation will occur.

So, when two solutions containing known concentrations of ions are mixed together, the knowledge of their solubility product can help predict whether the precipitation will occur or not.

c. In Qualitative Analysis: The separation and identification of basic radicals are made based upon the principles of common ion effect and solubility product principle in qualitative analysis. 

d. Precipitation of Soluble Salts: The use of solubility product and common ion effect explains the phenomenon of precipitation when a saturated solution is mixed with a suitable electrolyte containing a common ion. The increase in the ionic product over the solubility product results in precipitation. This phenomenon is used in the purification of common salt, salting out of soap and also in the manufacture of baking soda or sodium bicarbonate.

Summary

The solubility of a sparingly soluble salt can be calculated using the principle of solubility product, which is the ionic product of the salt at saturation. At this point, the sparing solubility salt-containing some ions in the solution and some precipitate of the undissolved salt. A solubility product is the same as an ionic product when it is taken at saturation, wherein the undissolved salt and the dissolved ions are in dynamic equilibrium with each other. The values solubility product of the sparingly soluble salts are used in various applications, including in finding its solubility, which is otherwise difficult to analyse, in qualitative analysis, in predicting precipitation reactions and also in precipitation of soluble salts.

FAQs

Q.1. How do you find the solubility of a sparingly soluble salt?
Ans: The solubility of the sparingly soluble salt can be found using its solubility product value in water. Depending upon the type of salt, there are different ways in which the solubility product can be found. For example, for salt type \(\mathrm{AB}\), the solubility of the salt can be found using:
\({\rm{AB}} \leftrightarrow {{\rm{A}}^{\rm{ + }}}{\rm{ + }}{{\rm{B}}^{\rm{ – }}}\)
\({{\rm{K}}_{{\rm{sp}}}} = \left[ {{{\rm{A}}^ + }} \right]\left[ {{{\rm{B}}^ – }} \right] = {{\rm{s}}^*}\;{\rm{s}} = {{\rm{s}}^2}\)

Q.2. What is the symbol for the equilibrium constant of sparingly soluble salt?
Ans: The symbol for the equilibrium constant of a sparingly soluble salt or solubility product is \(\mathrm{K}_{\mathrm{sp}}\).

Q.3. What is the solubility product of a sparingly soluble salt?
Ans: The solubility product at a specified temperature for an electrolyte can be defined as the product of the molar concentrations of the ions present in its saturated solution, with each concentration raised to the power equal to the number of ions produced on the dissociation of one molecule of the electrolyte.

Q.4. Are the equilibria of slightly soluble salts product or reactant favoured?
Ans: The sparingly soluble salts, at saturation, exists in a dynamic equilibrium. At this point, the ions in the solution and undissolved salt are in equilibrium. So, both forward and reverse reactions take place at the same time or are at a standstill when the solution is in equilibrium. 

Q.5. How do you find a solubility product?
Ans: The solubility product of a sparingly soluble salt, such as \({{\rm{A}}_{\rm{x}}}{{\rm{B}}_{\rm{y}}}\) can be determined using the formula:
\({{\rm{K}}_{{\rm{sp}}}} = {\left[ {{{\rm{A}}^{{\rm{y}} + }}} \right]^{{\rm{x}}*}}{\left[ {\;{{\rm{B}}^{{\rm{x}} – }}} \right]^{\rm{y}}}\)

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