# Solubility Product

When a sparingly soluble substance, say AB is kept in contact with water for some time at a definite temperature the following equilibrium is established:

$\underset{\text{Solid}}{AB} \leftrightharpoons \underset{\text{Dissolved in solvent}}{AB} \leftrightharpoons \underset{Ions}{A^+ + B^+}$

Applying the law of mass action,

$\dfrac{[A^+][B^-]}{[AB]} = K$

But the concentration of the unionised [AB] is constant at a given temperature; if excess of AB is present.

Hence $[A^+][B^-] = K[AB] = \text{Constant} (K_{sp} or S)$

Hence in saturated solution, the product of the ionic concentrations of sparingly soluble substance is constant at a given temperature. This constant product is called the solubility products $(S or K_{sp})$.

Relation between solubility ‘s’ and solubility product $(K_{sp})$. Consider a reaction:

$AB \leftrightharpoons \underset{\underset{(s = \text{solubility})} s s }{A^+ + B^-}$

$\text{Now} K_{sp} = [A^][B^-] \\ K_{sp} = s^2\hspace{3mm} \text{At constant temperature}$

$s = \sqrt{K_{sp}} \\[3mm] AgX \leftrightharpoons Ag^+ + X^- \\[3mm] BaSO_4 \leftrightharpoons Ba^{2+} + SO^{2-}_4 \\[3mm] CaCO_3 \leftrightharpoons Ca^{+2} + CO^{-2}_3$

For $A_XB$ By type of Salt:

$A_XB_Y \leftrightharpoons \underset{XS}{XA^{+Y}} + \underset{YS}{YB^{-X}} \\[3mm] K_{sp} = (X.s)^x. (Y.s)^y \\[3mm] K_{sp} = X^x. Y^Y (s)^{x + y} \\[3mm] E.g., (i) AlCl_3 \leftrightharpoons Al^{+3} + 3Cl^- \\[3mm] K_{sp} = s^1(3s)^3 = 27 s^4 \\[3mm] E.g. (ii) Al_2(SO_4)_3 \leftrightharpoons 2Al^{+3} + 3SO^{-2}_4 \\[3mm] K_{sp} 2^23^3(s)^{2 + 3} = 108 s^5$

When the ionic product exceeds the solubility product, the solution is super saturated and precepitation occurs and if the ionic product is less than the solubility product, the solution will be unsaturated and the precipitation will not occur.

This is also called the theory of precipitation.

Ionic product < Solubility product, the salt dissolves.

Ionic product < $K_{sp}$, more salt can dissolve.

Ionic product > solubility product, precipitation takes place.

Ionic product > $K_{sp}$, the salt precipitates.

Since precipitation is governed by the principles of solubility product, the latter receives important applications in the field of analytical chemistry.

Precipitation as sulphides of radicals of Group II and IV: Precipitation of the sulphides can occur only when the ionic product $[M^{2+}][S^{2-}]$ exceeds the solubility product-of the sulphide (MS) at that temperature I.e. , $[M^{2+}][S^{2-}]$ > solubility product of MS.

In the presence of acid, the ionisation of $H_2S$  is low,

$H_2S \leftrightharpoons 2H^+ + S^{2-}$

I. e. , suppressed due to the increase of $H^+$ ions (produced by acid).

Due to low $[S^{2-}]$in the solution, the solubility product of the sulphides of Group IV radicals is not reached. It is, however, enough to cause the precipitation of the IInd group radicals e.g., CuS, CdS, PbS etc., which possess a low solubility product.

In the presence of $NH_4OH \text{the} OH^-$ ions obtained from it $[NH_4OH \leftrightharpoons NH^+_4 OH]$  combine with the $H^+$ions produced from $H_2S(H_2S \leftrightharpoons 2H^+ + S^{2-} )$ to give unionised water, so that more of $H_2S$  ionises and thus the concentration of $S^{2-}$  ions in solution increases. In this way it becomes so high that the solubility product of IV group radicals e. g. , ZnS, MnS etc. is exceeded and the precipitation takes place.

$[M^{2+} \underset{\text{Precipitation occurs}}{[S^{2-} ] > S[ \text{solubility}} \text{Product} ]$

Since the solubility product governs all the precipitation reactions in qualitative analysis, it is of great importance and is intensively used in qualitative analysis.

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