Solubility and Solubility product - Ionic Equilibrium | Physical Chemistry PDF Download

Solubility (s) and Solubility Product (Ksp)

This is generally used for sparingly soluble salts.

Solubility product (ksp) is an equilibrium constant so like all the equilibrium constants, it depends only on the temperature.

Solubility product: in a saturated solution of a salt, there exists a dynamic equilibrium between the excess of solute and the ions furnished by the part of the solute which has gone in solution.

Solution

AgCl(s)     [Ag⁺][Cl^–]

AgCl         Ag⁺ + Cl^–

a-s                    s           s

K_sp = S²

Simple Solubility         Ax By  xA^y+ + yB^–x

Ksp = (xs)^x (ys)^y = x^x.y^y (s)^x+y

Factors Affecting Solubility:

• Effect of common ions on solubility

Because of the presence of common ion the solubility of the salt decreases.

AgCl       Ag⁺      +          Cl^–

-                       s                       s

Added NaCl, which is having common ion as that of sparingly soluble salt, AgCl.

            NaCl    →        Na⁺      +       Cl^–

            C                     0                    0

            -                       C         C

So, added electrolyte, will decrease the solubility of AgCl

AgCl  Ag⁺ + Cl^–

            S'            S'

Ksp = S' (S' + C)

Assumption: C >>> S^'

Hence S' + C »C

Ksp = s¹ x C

S¹ = ksp / C

• Simultaneous Solubility:

When sparingly soluble salts are added in water simultaneously, there will be two simultaneous equilibria in the solution

AgCl        Ag⁺      +     Cl^–

                        S₁+S₂          S₁                                 Ksp₁

AgCl        Ag⁺      +     Br^–

                        S₁+S₂          S₂                                  Ksp₂

Ksp₁ = (s₁ + s₂)s₁

Ksp₂ = (s₁ + s₂)s₂

Adding equation (1) and (2)

K₁P₁+K₁P₂ = (S₁+s₂)

From (1)

• Solubility in presence of hydrolysis:

When the ion from sparingly soluble salt, undergo hydrolysis Hence, we have to follow different procedure to calculate solubility of these salts. 

AgCN(s)  Ag⁺ + CN^–

                        s₁         s-x

CN^– + H₂O   HCN + OH^–

s-x                      x       x

k_sp=[Ag⁺] [CN^–]                                              … 1

                                      … 2

Multiply (1) and (2)

K_sp  x K_h = [Ag⁺] [HCN]

Since [HCN] = [OH^–]

Effect of complex formation: 

AgCl(s)   Ag⁺ (ag) + Cl^– (ag)

Ag⁺ + 2NH₃ [Ag(NH₃)₂]⁺ K_f = stability constant

S          C

-           C-2S                S

Reverse the Reaction,

[Ag(NH₃)₂]⁺ Ag⁺ + 2 NH₃

S                           -      C-2S

s-x                   x          C-2s+2x

Since, K_d is very very low,

Hence, c – 2s + 2x ≈ C – 2S

K_sp = x × s                                                                   … 1

Put value of K_d in eq. 1

S =

Concept of Precipitation: Less the value of S, less the solubility of the salt..Hence, more the extent of precipitation of salt.

AgCl  Ag⁺ + Cl^–

With the knowledge of solubility product of a sparingly soluble substance, we can predict whether under certain given conditions that substance would be precipitated or not. It may be remembered that a substance gets precipitated when the ionic product i.e. the product of concentrations of its ions present in a solution exceeds the solubility product of the substance,

Ki = [Ag⁺] [Cl^–]

If         K_ip>Ksp          Precipitation Occurs

If         K_ip<Ksp          Unsaturated Solution

If         k_ip = Ksp         Saturated Solution

Example : 10^–4 Molar Mg(NO₃)₂ with pH = 9 and if Ksp (Mg(OH)₂) = 8.9 × 10^–12

1. Will precipitation takes place or not.

2. At what minimum value of pH precipitation will occur.

a­)  Mg(NO₃)₂     →    Mg²⁺ + 2 NO₃^–

            10^–4                -             -

              -                  10^–4          10^–4

            Mg(OH₂)   Mg²⁺ + 20H^–

 K_ip = (10^–4) (10^–5)²

K_i = 10^–14

Ksp of Mg(OH)₂ = 8.9 × 10^–12

Ksp>K_ip

Hence solution is unsaturated. No precipitation will occur.

For Mg (OH)₂

            Ksp = [Mg²⁺][OH^–]²

            8.9 × 10^–12 = 10^–4[OH^–]²

            [OH^–] = 2.9 × 10^–4

pOH = 3.83

pH = 10.47

Hence pH should be greater than 10.47 for the precipitation to occur.

Preferential precipitation of an insoluble salt: 

Silver chloride is insoluble or sparingly soluble, to be more accurate. So is silver iodide. The question arises as to what would happen if potassium iodide solution is added to silver chloride.

AgCl + KI  KCl + AgI

Sparingly                     Sparingly
soluble                         soluble

Would the reaction take place towards the right, i.e. would the precipitated of siler chloride change into the precipitate of silver iodide.

For answer, it is necessary to look their respective solubility products.

Solubility product of silver chloride

Ksp (AgCl) = [Ag⁺] [Cl^–] = 1.56 × 10^–10

And that of silver iodide

Ksp(AgI) = (Ag⁺)(I^–) = 0.94 × 10^–16

Example :  0.1 Molar concentration of Cl^– ion and 10^–3 molar chromate CrO₄^2– ions are present in a solution. If AgNO₃ is added which will precipitate first

            If         Ksp (AgCl)=10^–12

                        Ksp (Ag₂CrO₄) 10^–10

2.  What will be the concentration of Cl^– ions when Ag₂CrO₄ begins to precipitate. Also find the percentage of Cl^– left with respect to original.

Solution           AgCl  Ag⁺ + Cl^–

                          -             S        S

                        K_sp = S²

                        S = 10^–6

For Ag₂CrO₄

                        Ag₂CrO₄ 2Ag⁺ + CrO₄^2–

                        Ksp = (2S)²S

                        10^–10 = 4S³

                        S = (0.25 × 10^–10) ^1/3

                        S = 2.9 × 10^–4

            Since, solubility of AgClis less than solubility of Ag₂CrO₄.

            Then, obviously AgCl will precipitate first

b)  When Ag₂CrO₄ begins to precipitate, concentration of CrO₄^2– would be

            10^–10 = [Ag⁺]²[CrO₄^2–]

            10^–10 = [Ag⁺] × 10^–3

            [Ag⁺] = 3.1 × 10^–4

For AgCl

            Ksp = [Ag⁺][Cl^–]

            10^–12 = 2.9 × 10^–4 [Cl^–]

            Cl^– = 3.4 × 10^–9

            Percentage of Cl^– left =

            Percentage of Cl^– left = 3.4 × 10^–6 %