CBSE Class 11 Chemistry Notes For Solubility And Solubility Product

Solubility And Solubility Product

The terms ‘solubility’ and ‘solubility product’ are often used in the context of the dissolution of a solute in a liquid. It is important to note that these two terms do not have the same meaning and therefore cannot be used interchangeably.

Comparison between solubility and solubility product:

• The term solubility applies to all kinds of solutes (ionic, neutral, sparingly soluble, or highly soluble) whereas the term solubility product is mainly used for sparingly soluble compounds.
• The solubility of a solute in a solution may change due to common ion effects or formation of complex salt but its solubility product remains constant at a fixed temperature.
• Both solubility and solubility products of a solute in a liquid change with the temperature variation.

Solubility products of sparingly soluble salts

Salts having solubilities less than 0.01 mol-L^-1 at ordinary temperature are commonly known as sparingly soluble salts. Some examples of such salts are AgCl, BaSO₄, CaSO₄, etc. A salt of this type in its saturated aqueous solution remains virtually insoluble and only a small part of it gets dissolved but the dissolved portion of the salt, however small it may be, gets completely dissociated.

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Solubility Product:

The Solubility Product of A sparingly soluble salt at a given temperature is defined as the product of the molar concentrations of the constituent ions in its saturated solution, each concentration term raised to a power representing the number of ions of that type produced by dissociation of one formula unit of the salt.

Characteristics of solubility product:

• At a particular temperature, the solubility product of a sparingly soluble salt has a fixed value. Its value changes with temperature variation.
• At a particular temperature, the solubility product of a sparingly soluble salt in its saturated solution remains unaltered if the concentrations of the constituent ions are changed. It remains fixed even in the presence of other ions in the solution.

Equation of solubility product:

Let’s consider the equilibrium that exists in a saturated solution of the sparingly soluble salt, AxBy. In this solution, undissolved AxBy remains in equilibrium with the ions (A+ and B- ) produced from the dissociation of dissolved A_xB_y.

⇒ \(\mathrm{A}_x \mathrm{~B}_y(s \text {, undissolved }) \rightleftharpoons x \mathrm{~A}^{y+}(a q)+y \mathrm{~B}^{x-}(a q)\)

Applying the law of mass action we have, an equilibrium constant,

⇒ \(K=\frac{\left[\mathrm{A}^{y+}\right]^x \times\left[\mathrm{B}^{x-}\right]^y}{\left[\mathrm{~A}_x \mathrm{~B}_y(s)\right]}\)

Where, [A^y+], [B^x-], and [A_x B_y(s)] are the molar concentrations of A^y+,B^+-– and A_xB_y(s) respectively, at equilibrium.

Now, the molar concentration of pure solid at a particular temperature is constant. Again, at a given temperature, the equilibrium constant, K is also constant.

So, at a given temperature, K  ×  [A_x × B_y(s)] = constant.

∴ K ×  [A_x × B_y(s)]=[A^y+]^x ×  [B^x-]^y

Or K_sp = [A^y+] × [B^x-]^y [ sp = solubility product ]

Where, K_sp = K ×  [A_x B_y(s)] = constant.

K_sp is called the solubility product or solubility product constant of the sparingly soluble salt, A_xB_y.

Relation between solubility and solubility product

In case of a sparingly soluble salt of the type AB: Equilibrium formed by a sparingly soluble salt of the type AB in its saturated aqueous solution:

AB(s) ⇌ A⁺ (aq) + B^–(aq)

∴ The solubility product of, Ksp = [A⁺] × [B⁺]

Let the solubility of the salt, AB, in its saturated aqueous solution at a given temperature be S mol.L^-1

Hence, in the solution, the molar concentration of A⁺ ions [A⁺] = S mol.L^-1 and that of B^– ions, [B-] = S mol.L^-1 [v one formula unit of AB on dissociation gives one A+ and one B^– ion].

∴ K_sp = [A⁺][B^–] = S ×  S = S²

∴ K_sp  = s² ……………………….. (1)

and S = \(\sqrt{K_{s p}}\)……………………… (2)

Equation (1) is the relation between the solubility the and solubility product of a sparingly soluble salt of the type AB.

Equation (2) indicates that the solubility of a sparingly soluble salt, AB is equal to the square root of its solubilityproduct.

Example: AgCl is a sparingly soluble salt. In a saturated solution of AgCl, the following equilibrium is formed:

⇒ \(\mathrm{AgCl}(s) \rightleftharpoons \mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q)\)

So, the solubility product of AgCl, K_sp = [Ag²⁺] × [Cl^–]

Suppose, at a certain temperature, the solubility of AgCl in its saturated aqueous solution = S mol.L^-1. Hence, in the solution,

[Ag⁺] tS mol.L^-1 and [Cl^–] = S mol.L^-1

∴ K_sp  = [Ag+] × [Cl^–] = S × S = S²

Similarly, in the case of other sparingly soluble salts ofthe type AB (CaSO₄, BaSO₄, AgBr, etc.), K_sp = S²

In the case of a sparingly soluble salt of type AB₂:

The following equilibrium exists in a saturated aqueous solution of a sparingly soluble salt of the type AB₂:

⇒ \(\mathrm{AB}_2(s) \rightleftharpoons \mathrm{A}^{2+}(a q)+2 \mathrm{~B}^{-}(a q)\)

∴ The solubility product of AB₂, K_sp = [A²⁺] × [B^–]²

Let the solubility of AB₂ in its saturated solution at a given temperature be S mol.L-1. Therefore, in the solution, [A²⁺] = S mol-L^-1, [B^–] = 2S mol.L^-1 formula unit of AB₂ dissociates into one A²⁺ ion and two B_ ions].

∴ K_sp = [A²⁺] × [B^–]² = S ×  (2S)² = 4S³

∴ K_sp = 4S³ …………………………..(1)

Equation (1) represents the relation between the solubility and the solubility product of the sparingly soluble salt AB₂.

Example:

In a saturated aqueous solution of CaF₂, the following equilibrium exist:

CaF₂(s) ⇌  Ca²⁺(aq) + 2F^–(aq)

The solubility product of CaF₂, K_sp = [Ca²⁺] [F^–]² Let the solubility of CaF2 in its saturated aqueous solution at a certain temperature = S mol.L-1. So, in the saturated solution,

[Ca²⁺] = S mol.L^-1 and [F^–] =2S mol.L^-1

K_sp = [Ca²⁺ ] ×  [F^–]² = S ×  (2S)² = 4S³

Similarly, in the case of other sparingly soluble salts ofthe type

AB₂ (BaF2, PbCl₂, etc.), K_sp= 4S³

In case sparingly soluble salt of the type A_xB_y :

In a saturated aqueous solution of a salt of the type A_xB_y, the following equilibrium is established:

A_xB_y (s) ⇌ xA^y+(aq) + y B^x-(aq)

∴ The solubility product of A_xB_y,

K_sp = [A^y+]^x×  [B^x-]y

Suppose, the solubility of its saturated solution is = S mol.L^-1

Hence, in the solution, [A^y+] = xS mol-L^-1 and [B^x- ] = yS mol.L^-1

Since one formula unit of AxBy dissociates into x number of A^y+ and y number of B^x- ions]

∴ K_sp = [A^y+]^x×  [B^x-]y

= (xS)^x × (yS)^y=(x^x y^y) S^x+y

∴ K_sp = x^x y^y(S)^x+y

Equation [1] represents the relation between the solubility and the solubility product of sparingly soluble salt A_xB_y

Relation between solubility (s) and solubility product (k) of some sparingly soluble salts:

Values of solubility  products (Ksp) of some sparingly soluble salt at 25 °C:

Common Ion Effect On The Solubility Of A Sparingly Soluble Salt 

In a saturated solution of a sparingly soluble salt, if a strong electrolyte having a common ion is added, then the solubility of the sparingly soluble salt decreases.

Hence, the solubility of a sparingly soluble salt in an aqueous solution of a salt having a common ion is less than its solubility m pure water.

Example:

The solubility of AgCl in an aqueous 0.1(M) KCl solution is less than its solubility in pure water. Here, the communion is Cl^– ion

Explanation:

 The following equilibrium is established in the saturated aqueous solution of, AgCl

The solubility product of AgCl, K_sp = [Ag⁺] × [Cl^–]. At a certain temperature, the value of K_sp is constant. Hence, the product of molar concentrations [Ag⁺] × [Cl^–] ) of

Ag⁺ and Cl are always constant at a particular temperature.

Therefore, if the molar concentration of one of the ions between Ag⁺ and Cl^– is increased, then the molar concentration of the other will be decreased to maintain a constant value of K_sp.

Let strong electrolyte KCl (having common ion Cl^– ) be added to the saturated solution AgCl. This will increase the molar concentration of the common ion (Cl^–) in the solution.

According to Le Chatelier’s principle, to maintain the constancy of K_sp some of the Cl^– ions combine with an equal number of Ag⁺ ions to form solid AgCl. This causes the equilibrium to shift to the left. As a result, the solubility of AgCl in the solution decreases.

At the new equilibrium, [Cl^–] << [Ag⁺], and the solubility of AgCl becomes equal to [Ag⁺]. If an aqueous AgNO₃ solution instead of KCl is added to the saturated solution of AgCl, then the solubility of AgCl will also decrease because of the communion (Ag⁺) effect.

Increase in solubility due to the presence of common ion:

AgCN is a sparingly soluble salt and its solubility increases when KCN is added to its aqueous solution. However, this increase in solubility does not contradict the principle of solubility product.

KCN reacts with sparingly soluble AgCN, and forms a water-soluble complex, K [Ag(CN)₂]. Thus, the solubility of AgCN increases.

KCN(aq) + AgCN(aq) K [Ag(CN)₂] (aq) A similar phenomenon is observed in the case of mercuric iodide (Hgl₂).

It shows greater solubility in an aqueous KI solution due to the formation of a water-soluble complex, 2KI[HgI₄].

2KI(aq) + Hgl₂ (aq) ⇌  K₂ [HgI₄](aq)

Effect of pH on solubility

The solubility of salts of weak acids like phosphate increases at lower pH. This is because, at a lower pH, the concentration of the anion decreases due to its protonation. This in mm increases the solubility of that salt.

Example: Consider the solubility equilibrium:

CaF₂(s) ⇌ Ca²⁺+(aq) + 2F^–(aq)……………….(1)

If the solution is made more acidic, it results in the protonation of some ofthe fluoride ions.

⇒ \(\mathrm{H}_3 \mathrm{O}^{+}(a q)+\mathrm{F}^{-}(a q) \rightleftharpoons \mathrm{HF}(a q)+\mathrm{H}_2 \mathrm{O}(l)\)

As acid is added, the concentration of F^– ion decreases due to the reaction(2).

This causes the equilibrium [1] to shift to the right, thereby increasing the concentration of Ca²⁺ ions. Thus, more CaF₂ dissolves in a more acidic solution.

Principle of solubility product and precipitation of sparingly soluble salts

Principle of solubility product:

At a particular temperature, the precipitation of salt from its solution or its dissolution in the solution depends on the value ofthe solubility product of the salt at that temperature.

If the product of the molar concentrations of the constituent ions, with suitable powers, of the salt in the solution exceeds the value of the solubility product of the salt at that temperature, then the salt will be precipitated from the solution.

If this value does not exceed the solubility product of that salt, then the salt will not be precipitated. This principle is known as the principle of solubility product. This principle is extensively used in different chemical analysis and industrial processes.

Explanation:

At 25°C, the solubility product of AgCl is 1.8 × 10^-10. Hence, in the saturated solution of AgCl at 25°C, the highest value of the product of molar concentrations of Ag+ and Cl- ions is 1.8× 10^-10

If the product of molar concentrations of Ag+ and Cl^– ions is less than, then the solution will be unsaturated. Under this condition, AgCl will be dissolved in the solution to increase the concentration of Ag+ and Cl^– ions and this will continue until the product of ofmolar concentrations of Ag⁺ and Cl^– becomes equal to.

If the product of molar concentrations of Ag⁺ and Cl^– is greater than K_sp(AgCl), then AgCl will be precipitated and this will continue unless the product ofmolar concentrations of Ag⁺ and Cl^– becomes equal to K_sp (AgCl).

Applications of solubility product

Application-1:

At a certain temperature, if the solubility product of a sparingly soluble salt is known, then the solubility of the salt, as well as the molar concentrations of the constituent ions of the salt in its saturated solution, can be determined.

Conditions for precipitation:

In the solution, if the product of the molar concentrations (with appropriate powers) of the constituent cation and anion of the salt at a particular temperature.

Is greater than the solubility product of the salt at that temperature, then the salt will be precipitated, Is less than the solubility product of the salt at that temperature, then the salt will remain dissolved in the solution, Is equal to the solubility product of the salt at that temperature.

Then the solution will just be saturated. In this case, the dissolved portion of the salt remains in equilibrium with the undissolved part.

Applications of solubility product principle

Preparation of pure NaCl from impure NaCl:

Pure sodium chloride can be prepared from impure sodium chloride by applying this principle. When dry HCl gas is introduced into the saturated aqueous solution of impure sodium chloride, the concentration of Cl- ions increases to a great extent and the ionic product of Na⁺ and Cl^–, i.e., [Na⁺] × [Cl^–] exceeds the solubility product of NaCl. Consequently, sodium chloride gets separated as pure crystals.

During this precipitation, impurities like MgCl₂, CaCl₂, etc., do not precipitate out as the values of the ionic product of the constituent ions of these impurities do not exceed their respective solubility products.

Separation of NaHCO₃ by Solvay process:

In this process, a saturated solution of sodium chloride is added to a solution of NH₄HCO₃.

The reaction concerned is:

⇒  \(\mathrm{NH}_4 \mathrm{HCO}_3(a q)+\mathrm{NaCl}(a q) \rightarrow \mathrm{NaHCO}_3(a q)+\mathrm{NH}_4 \mathrm{Cl}(a q)\)

Among the compounds participating in the reaction, NaHCO₃ has the least value of solubility product. The product of the molar concentrations of Na+ and HCO₃^– ions present in the solution can easily exceed the solubility product of NaHCO₃, and hence NaHCO₃. gets precipitated.

[KHCO₃ cannot be prepared by the Solvay process because the solubility product of KHCO₃ Is very high.]

Manufacture of soaps:

Soap consists of sodium or potassium salts of organic fatty acids of high molecular mass, such as stearic acid, oleic acid, etc. During the manufacturing of soaps, some amount of soap remains in a colloidal state in the mother liquor. If some NaCl is added to this solution, then the concentration of Na⁺ ions increases, thereby decreasing the solubility of soap. As a result, soap gets precipitated.

Application in analytical chemistry:

The principle of solubility product is extremely helpful in the qualitative analysis of basic radicals. Based on the values of the solubility product, the basic radicals of inorganic salts are divided into different groups.

Group 1:

Basic radicals \(\mathbf{A g}^{+}, \mathbf{P b}^{2+}, \mathbf{H g}_2^{2+}(\text { ous })\)

Group reagent:  6(N) HCl

As the values of the solubility product of chlorides of group-1 metals are sufficiently low [At 25°C, K_sp(AgCl)

= 1.8 × 10^-10, K_sp(Hg₂Cl₂)

= 1.2 × 10 K_sp(PbCl₂)

= 1.7 × 10^-18],

In addition to (N)HCI, the solubility products of these chlorides are exceeded by the products of the molar concentration of the corresponding ions. So, these chlorides get precipitated.

On the other hand, the chlorides of other groups have higher values of solubility product and thus are not precipitated.

[The solubility product of PbCl₂ is much higher than that of both AgCl and Hg₂Cl₂. So, Pb²⁺. ions are not completely precipitated as PbCl₂ in group 1. Pb²⁺. ions left in the solution are separated as precipitate in group 2.

Group 2:

Basic radicals: Cu²⁺+, Pb²⁺, Cd²⁺, Hg²⁺ (i.e ), Bi³⁺, As³⁺, Sb³⁺, Sn²⁺.

Group reagent:

H₂S gas in the presence of dilute HCl In an aqueous solution, H₂S ionizes partially to produce S^2- ions, which precipitate the basic radicals of this group as their corresponding sulfides (CuS, PbS, CdS, etc.).

The solubility product ofthe metallic sulphides of group-2

[K_sp(CuS) = 8.0 × 10^-37, K_sp (PbS) = 3.0 × 10^-28,

K_sp(CdS) = 1.0 × 10^-27 ,K_sp(HgS) = 2 × 10^-53

K_sp (Bi₂S₃) = 1.6 × 10^-72]  are much smaller as compared to those of the sulfides of the subsequent groups.

So, when H₂S gas is passed through the solution consisting of different basic radicals in the presence of HCl, the product of molar concentrations of the corresponding ions of each of the sulfides of this group exceeds its corresponding solubility product. Hence, only the sulfides of this group are precipitated.

It is important to note that the degree of ionization of weak acid H₂S in the presence of HCl decreases due to the common ion (H⁺) effect. As a result, except for group- 2, the solubility products of sulfides of the subsequent groups are not exceeded. Therefore, except for the sulfide salts of group 2, sulfides of all other groups are not precipitated.

Another notable point is that among the sulfides of group-2, the solubility products of CdS and PbS are sufficiently high. So, as a result of the common ion effect, if the concentration of S^2-ions decreases to a large extent, CdS and PbS are not precipitated. It has been experimentally proved that if H₂S gas is passed through the solution using 0.3(N) HCl solution, then all the sulfides of this group are completely precipitated, but the precipitation of the sulfides of other groups does not occur.

Group 3A Basic radicals: Fe³⁺ , Al³⁺ , Cr³⁺ Group reagent:

NH4OH solutions presence of NH₄Cl The basic radicals of this group are precipitated as their corresponding hydroxides [Fe(OH)₃, Al(OH)₃, Cr(OH₃) ]. The solubility products of the hydroxides of the basic radicals of group 3A are much smaller than those of the hydroxides belonging to the subsequent groups.

So, when NH₄OH is added to the solution of different basic radicals in the presence of NH₄Cl, only the solubility products of the hydroxides of group-3A are exceeded, resulting in the preferential precipitation of the hydroxides of group-3A basic radicals. The hydroxides of the basic radicals of subsequent groups are not removed by this way of precipitation

Here, it is to be noted that the degree of ionization of weak base NH₃ in the presence of NH₄Cl is decreased due to the common ion (NH₄⁺) effect. As a result, the concentration of OH ions decreases to such an extent that except for group-3A, the solubility product of metallic hydroxides of the subsequent groups is not exceeded. Therefore, except for group-3A, metallic hydroxides of other groups are not precipitated.

Group 3 Basic radicals: Co²⁺, Ni²⁺, Mn²⁺, Zn²⁺

Group reagent:

H₂S in the presence of NH3 For precipitation of the basic radicals of this group as their corresponding sulfides [CoS, ZnS, MnS, NiS], H₂S gas is passed through the test solution in the presence of NH3.

In the presence of OH^– ions produced by the dissociation of NH₄OH, the equilibrium involving the dissociation of the weak acid H2S gets shifted to the right, forming sulfide ions (S^2-) to a greater extent

[H₂S(aq) + 2H₂O ⇌  2H₃O⁺(aq) + S^2-(aq)].

Due to a sufficient increase in the concentration of S^2- ions, the solubility products of the sulfides of group-3B are exceeded. Hence, the basic radicals of this group are precipitated as sulfides.