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^{–}]

_{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 _{x}B dissociate as follows-*

*A _{x}B_{y} ⇌ x A^{y+} + y B^{x-}*

*According to law of mass action,*

*K = [A ^{y+}]^{x} [B^{x-}]^{y} / [A_{x}B_{y}]*

*[A _{x}B_{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}

_{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 , BaSO _{4}, PbSO_{4}, AgI etc.*

*AB ⇌ A ^{+ } + B^{– }*

*[A ^{+ }] = S mole / litre*

*K _{sp} = [A^{+ }] [B^{+ }] = S x S = S^{2 }*

*K*_{sp} = S^{2 }

_{sp}= S

^{2 }

*S = ***√ **K_{sp}

**√**K

_{sp}

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

*Ex – Ag _{2} CO_{3} , Ag_{2}CrO_{4}, PbCl_{2}, CaF_{2} etc.*

*AB _{2} ⇌ A^{2+ } + 2 B^{–}*

* S 2S*

*[A ^{2+ }] = S mole / litre, [B^{–}] = 2S*

*K _{sp} = [A^{2+ }] [B^{– }]2 = S x (2S)^{2}*

*K _{sp} =4S^{3}*

*S = 3***√ **K_{sp} /4

**√**K

_{sp}/4

*iii) For 1: 3 type salt-*

*Ex – Fe(OH) _{3} , Al(OH)_{3} , Cr(OH)_{3} etc.*

*AB _{3} ⇌ A^{3+ } + 3 B^{–}*

* S 3S*

*[A ^{3+ }] = S mole / litre, [B^{–}] = 3S*

*K _{sp} = [A^{3+ }] [B^{– }]^{3} = S x (3S)^{3}*

*K _{sp} = 27 S ^{4}*

*S = 4***√ **K_{sp} /27

**√**K

_{sp}/27