Soluble - Solubility > 0.1 M
Slightly soluble - 0.01 M < solubility < 0.1M
Sparingly soluble – solubility < 0.01M
We have now consider the equilibrium between the sparingly soluble ionic salt and its saturated aqueous solution. A solution which remains in contact with excess of the solute is said to be saturated. The amount of a solvent () in 100 ml or 1L) to form a saturated solution at a given temperature is termed as the solubility of the solute in the solvent at that temperature.
For a sparingly soluble salts like AgCl,PbI2, BaSO4 etc. Ionisation is very small and concentration of the salt may be considered as constant. Thus, for AgCl,
AgCl <========> Ag+ + Cl-
K = [Ag+][Cl-]/[AgCl]
For a pure solid substance the concentration remains constant and we can write
Ksp = K[AgCl]
=[Ag+][Cl-]
Where Ksp is the solubility product constant or solubility product.
If S represents solubility of AgCl (in mol L-1)
Then [Ag+]= [Cl-] = S molL-1
Ksp = [Ag+][cl-] = S2
S = √Ksp
A solubility product constant expression is the product of the chemical equation for a solubility equilibrium, with each term raised to the power given by the coefficient in the chemical equation.
Example:-
For BaSO4 (binary solute giving two ions)
BaSO4 <=======> Ba2++ SO42-
Ksp = [Ba2+][SO42-] = S2
For PbI2 (Ternary solute giving three ions)
PbI2 <=======> Pb2+ + 2I-
Ksp = [Pb2+][I-]2 = (s)(2s)2 = 4S3
For Al(OH)3 (Quaternary solute giving four ions)
Al(OH)3 <=======> Al3+ + 3OH-
Ksp = [Al3+][OH-]3 = (S)(3S)3 = 27Ss4
For a solute AxBy, (giving (x+y) ions)
Ksp = [Ay+]x [Bx-]y = (xS)x (yS)y = XxYy(S)(x+y)
Precipitation reactions
For the sparingly soluble solute AB,
AB <=======> A+ +B-
Q is the equilibrium constant with the given [A+] and [B-]
Q = [A+][B-]
When applied to solubility product, Q is generally called the ionic product. Precipitation occurs if Q>Ksp, then solution is said to be super saturated and precipitation cannot occur if Q
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