Solubility product

source : Chemistry Libre Texts

Solubility product-

A solution which remains in contact with undissolved solute is said to be saturated . In a saturated solution of an electrolyte two equilibria exists and can be written as,

AB ⇌ AB ⇌ A⁺ + B ^–

solid unionised ions

(dissolved)

According to law of mass action,

K = [A⁺][B^–] / [AB]

[AB] unionised = constant ( because solution is saturated) = K’

K. K’= [A⁺][B^–] = K_sp (constant)

K_sp = [A⁺][B^–]

K_sp is solubility product

” The product of concentration of ions in a saturated solution of an electrolyte at a given temperature is equal to the solubility product.”

Suppose , electrolyte of type A_xB dissociate as follows-

A_xB_y ⇌ x A^y+ + y B^x-

According to law of mass action,

K = [A^y+]^x [B^x-]^y / [A_xB_y]

[A_xB_y] = constant ( because solution is saturated) = K’

K. K’= [A^y+]^x [B^x-]^y = K_sp (constant)

K_sp = [A^y+]^x [B^x-]^y

K_sp is solubility product

” In a saturated solution of electrolyte at given temperature , the product of concentrations of the ions raised to the power equal to the number of times the ions occur in the equation.”

Relation between solubility and solubility product-

i) For 1: 1 type salt-

Ex – AgCl , BaSO₄, PbSO₄, AgI etc.

AB ⇌ A⁺ + B^–

[A⁺] = S mole / litre

K_sp = [A⁺] [B⁺] = S x S = S²

K_sp = S²

S = √ K_sp

ii) For 1: 2 or 2 : 1 type salt-

Ex – Ag₂ CO₃ , Ag₂CrO₄, PbCl₂, CaF₂ etc.

AB₂ ⇌ A²⁺ + 2 B^–

S 2S

[A²⁺] = S mole / litre, [B^–] = 2S

K_sp = [A²⁺] [B^–]2 = S x (2S)²

K_sp =4S³

S = 3√ K_sp /4

iii) For 1: 3 type salt-

Ex – Fe(OH)₃ , Al(OH)₃ , Cr(OH)₃ etc.

AB₃ ⇌ A³⁺ + 3 B^–

S 3S

[A³⁺] = S mole / litre, [B^–] = 3S

K_sp = [A³⁺] [B^–]³ = S x (3S)³

K_sp = 27 S ⁴

Solubility product