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We have come across many precipitation reactions in inorganic qualitative analysis. For example, dil HCl is used to precipitate Pb2+ ions as PbCl2 which is sparingly soluble in water.
Kidney stones are developed over a period of time due to the precipitation of Ca2+ (as calcium oxalate etc.). To understand the precipitation, let us consider the solubility equilibria that exist between the undissociated sparingly soluble salt and its constituent ions in solution.
For a general salt XmYm
The equilibrium constant for the above is
In solubility equilibria, the equilibrium constant is referred as solubility product constant (or) Solubility product. In such heterogeneous equilibria, the concentration of the solid is a constant and is omitted in the above expression
Ksp = [Xn+]m[Ym]n
The solubility product of a compound is defined as the product of the molar concentration of the constituent ions, each raised to the power of its stoichiometric co – efficient in a balanced equilibrium equation.
Solubility product finds useful to decide whether an ionic compound gets precipitated when solution that contains the constituent ions are mixed.
When the product of molar concentration of the constituent ions i.e., ionic product, exceeds the solubility product then the compound gets precipitated.
The expression for the solubility product and the ionic product appears to be the same but in the solubility product expression, the molar concentration represents the equilibrium concentration and in ionic product, the initial concentration (or) concentration at a given time ‘t’ is used.
In general we can summarise as,
Ionic product > Ksp, precipitation will occur and the solution is super saturated. Ionic product < Ksp, no precipitation and the solution is unsaturated.
Ionic product = Ksp, equilibrium exist and the solution is saturated.
Determination of Solubility Product from Molar Solubility
Solubility product can be calculated from the molar solubility i.e., the maximum number of moles of solute that can be dissolved in one litre of the solution. For a solute XmYn,
XmYn(s) ⇄ mXn+(aq) + nYm-(aq)
From the above stoichiometrically balanced equation we have come to know that 1 mole of XmYn(s) dissociated to furnish ‘m’ moles of Xn+ and ‘n’ moles of Ym- is ‘s’ is molar solubility of XmYn, then
[Xn+]=ms and [Ym+]=ns
∴ Ksp = [Xn+]m[Ym+]n