SOLUBILITY PRODUCT:If to a given amount of solvent at a particular temperature, a solute is added gradually in increasing amounts, a stage is reached when some of the solute remains undissolved, no matter how long we wait or how vigorously we stir. The solution is then said to be saturated. A

solution which remains in contact with undissolved solute is said to be saturated. At saturated stage, the quantity of the solute dissolved is always constant for the given amount of a particular solvent at a definite temperature.In case the solute is an electrolyte, its ionisation occurs in solution and degree of dissociation depends on the concentration of dissolved electrolyte at a particular temperature. Thus, in a saturated solution of an electrolyte two equilibria exist and can be represented as:

AB ↔ AB ↔

Solid unionized ions

(dissolved)

Applying the law of action to the

ionic equilibrium,[A

^{+}][B^{-}]/[AB]Since the solution is saturated, the concentration of unionised molecules of the electrolyte is constant at a particular temperature, i.e., [AB] = K'= constant.

Hence, [A

^{+}] [B^{-}] = K[AB] = KK = K_{s}(constant)K

_{s}is termed as thesolubility product. It is defined as the product of the concentration of ions in a saturated solution of an electrolyte at a given temperature.Consider, in general, the electrolyte of the type A

_{x}B_{y}which is dissociated as:A

_{x}B_{y}↔ xA^{y+}+ yB^{x-}Applying law of mass action,

[Ay+]x[Bx-]y/[A

_{x}B_{y}] = KWhen the solution is saturated,

[A

_{x}B_{y}] = K' (constant)or [A

^{y+}]^{x}[B^{x-}]^{y}= K [A_{x}B_{y}] = KK' = K_{s}(constant)Thus,

solubility productis defined as the product of concentrations of the ions raised to a power equal to the number of times the ions occur in the equation representing the dissociation of the electrolyte at a given temperature when the solution is saturated.Note:

Solubility productis not theionic productunder all conditions but only when the solution is saturated.

Different Expressions for Solubility Products(i) Electrolyte of the type AB:

Its ionisation is represented as:

AB ↔ A

^{+}+ B^{-}Thus, K

_{s}= [A^{+}][B^{-}]AgCl ↔ Ag

^{+}+ Cl^{-}; K_{s}= [Ag^{+}][Cl^{-}]BaSO

_{4}↔ Ba^{2+}+ SO^{-2}_{4}; K_{s}= [Ba^{2+}][SO^{-2}_{4}]

(ii) Electrolyte of the type AB

_{2}:Its ionisation is represented as:

AB

_{2}↔ A^{2+}+ 2B^{-}Thus, K

_{s}= [A^{2+}][B^{-}]^{2}PbCl

_{2}↔ Pb^{2+}+ 2Cl^{-}; K_{s}= [Pb^{2}^{+}][Cl^{-}]^{2}CaF

_{2}↔ Ca^{2+}+ 2F^{-}; K_{s}= [Ca^{2+}][F^{-}]^{2}

(iii) Electrolyte of the type A

_{2}B:Its ionisation is represented as:

A

_{2}B ↔ 2A^{2+}+ B^{2-}Thus, K

_{s}= [A^{+}]^{2}[B^{2-}]Ag

_{2}CrO_{4}↔ 2Ag^{+}+ CrO^{-2}_{4}; K_{s}= [Ag^{+}]^{2}[CrO^{-2}_{4}]H

_{2}S ↔ 2H^{+}+ S2^{-}; K_{s}= [H^{+}]^{2}[S^{2-}]

(iv) Electrolyte of the type A

_{2}B_{3}:Its ionisation is represented as:

A

_{2}B_{3}↔ 2A^{3+}+ 3B^{2-}Thus, K

_{s}= [A^{3+}]^{2}[B^{2-}]^{3}As

_{2}S_{3}↔ 2As^{3+}+ 3S^{2-}; K_{s}= [As^{3+}]^{2}[S^{2-}]^{3}Sb

_{2}S_{3}↔ 2Sb^{3+}+ 3S2^{-}; K_{s}= [Sb^{3+}]^{2}[S^{2-}]^{3}

(v) Electrolyte of the type AB

_{3}:Its ionisation is represented as:

AB

_{3}↔ A^{3+}+ 3B^{-}Thus, K

_{s}= [A^{3+}][B^{2-}]^{3}Fe(OH)

_{3}↔ Fe^{3+}+ 3OH^{-}; K_{s}= [Fe^{3+}][OH^{-}]^{3}AH

_{3}↔ Al^{3+}+ 3l^{-}; K_{s}= [Al^{3+}][I^{-}]^{3}

Solubility product of a weak electrolyte:Let degree of ionization of weak electrolyte A

_{m}B_{n}be 'α'.A

_{m}B_{n}↔ mA^{n+}+ nB^{m-}t = 0 s 0 0

t

_{eq}s-sα msα nsαK

_{sp}= [A^{n+}]^{m}[B^{m-}]^{n}= [msα ]

^{m}[nsα]^{n}K

_{sp}= m^{m}n^{n}(sα)^{m+n}

Criteria of precipitation of an electrolyte:A very useful conclusion is derived from the

solubility productconcept. No precipitation of the electrolyte occurs if theionic productis less than thesolubility product, i.e., the solution has not reached the saturation stage.

Case I:When K_{ip}<K_{sp}, then solution is unsaturated in which more solute can be dissolved.

Case II:When K_{ip}= K_{sp}, then solution is saturated in which no more solute can be dissolved.

Case III:When K_{ip}> K_{sp}, then solution is supersaturated and precipitation takes place.

When the

ionic productexceeds thesolubility product, the equilibrium shifts towards left hand side, i.e., increasing the concentration of undissociated molecules of the electrolyte. As the solvent can hold a fixed amount of electrolyte at a definite temperature, the excess of the electrolyte is thrown out from the solution as precipitate.Thus, for the precipitation of an electrolyte, it is necessary that the

ionic productmust exceed itssolubility productFor example, if equal volumes of 0.02 M AgN0_{3}solution and 0.02 M K_{2}Cr0_{4}solution are mixed, the precipitation of Ag_{2}Cr0_{4}occurs as theionic productexceeds thesolubility productof Ag_{2}Cr0_{4}which is 2 × 10^{-12}In the resulting solution,

[Ag

^{+}] = 0.02/2 = 0.01 = 1*10^{-2}Mand [CrO

^{2-}_{4}]= 0.02/2 = 0.01 = 1*10^{-2}M

Ionic productof Ag_{2}CrO_{4}= [Ag^{+}]^{2 }[CrO^{2-}_{4}]^{}= (1×10

^{-2})^{2}(1×10^{-2})= 1 × 10

^{-6}1 × 10

^{-6}is higher than 2 × 10^{-12}and thus precipitation of Ag_{2}CrO_{4}occurs.