Solubility product
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] = Ksp (constant)

Ksp = [A+][B]

Ksp 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 AxB dissociate as follows-

AxBy   ⇌  x Ay+  + y Bx-

According to law of mass action,

K = [Ay+]x [Bx-]y / [AxBy]

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

K. K’= [Ay+]x [Bx-]y = Ksp (constant)

Ksp = [Ay+]x [Bx-]y

Ksp 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

Ksp = [A+ ] [B+ ]  = S x S = S2

S = √ Ksp

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

Ksp = [A2+ ] [B– ]2  = S x (2S)2

Ksp = 4S3

S = 3√ Ksp /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

Ksp = [A3+ ] [B– ]3  = S x (3S)3

Ksp = 27 S 4

S = 4√ Ksp /27