Modified form of Distribution Law

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

nA → A_n
Number of particles decreases after association.
If a solute A present in solvent-I where its concentration is C_I and in solvent-II, n molecules of solute A associates to form A_n and a few molecules of solute A are also present in solvent-II. If the concentration of A and A_n be C_A and C_II in solvent-II, respectively, then from distribution law-

C_I/C_A = K_D -----equation(1)
now equilibrium constant for the reaction nA ⇌ A_n is-
K_C = [A_n]/[A]ⁿ
or, K_C = C_II/Cⁿ_A
Taking nth root , we get-
n√C_II/C_A = n√K_C -----equation(2)
Now, dividing equation (1) by equation (2), we get-
(C_I/C_A) x (C_A/n√C_II) = (K_D/n√K_C)
C_I/n√C_II = K_D/n√K_C = 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 C_I. When it dissociates into x and y in solvent-II having total concentration C_II.

If α be the degree of dissociation of solute A in solvent-II, then-
      A       ⇌       x       +       y
C_II(1 − α)       C_IIα            C_IIα
so, the concentration of undissociated molecules of solute A in solvent-II will be C_II(1 − α)
Hence, the modified form of distribution law when dissociation of solute occurs in one of the solvent will be-
C_I/C_II(1 − α) = K

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