Modified form of Distribution Law

Modified form of Distribution Law | Modified Distribution Law

Modified form of Distribution Law

Modified form of distribution law when association of solute occurs in one of the solvent

nA → An
Number of particles decreases after association.
If a solute A present in solvent-I where its concentration is CI and in solvent-II, n molecules of solute A associates to form An and a few molecules of solute A are also present in solvent-II. If the concentration of A and An be CA and CII in solvent-II, respectively, then from distribution law-
Modified form of distribution law when association of solute occurs in one of the solvent
CI/CA = KD -----equation(1)
now equilibrium constant for the reaction nA ⇌ An is-
KC = [An]/[A]n
or, KC = CII/CnA
Taking nth root , we get-
n√CII/CA = n√KC -----equation(2)
Now, dividing equation (1) by equation (2), we get-
(CI/CA) x (CA/n√CII) = (KD/n√KC)
CI/n√CII = KD/n√KC = K -----equation (3)
Equation (3) is a modified form of distribution law when association of solute occurs in one of the solvent.

Modified form of distribution law when dissociation of solute occurs in one of the solvent

Let a solute molecule A which does not dissociate in solvent-I has concentration CI. When it dissociates into x and y in solvent-II having total concentration CII.
Modified form of distribution law when dissociation of solute occurs in one of the solvent
If α be the degree of dissociation of solute A in solvent-II, then-
      A       ⇌       x       +       y
CII(1 − α)       CIIα            CIIα
so, the concentration of undissociated molecules of solute A in solvent-II will be CII(1 − α)
Hence, the modified form of distribution law when dissociation of solute occurs in one of the solvent will be-
CI/CII(1 − α) = K

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