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-
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.
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
Test Your Knowledge
Question 1: When a solute undergoes association in one of the solvents (forming An from nA), which of the following expressions represents the modified distribution law?
- A) CI / CII(1 - α) = K
- B) CI / n√CII = K
- C) CI / CA = KD
- D) CI × n√CII = K
View Answer
Correct Answer: B
Explanation: When association occurs (nA ⇌ An), the concentration of the normal species in the second solvent is proportional to the nth root of the total concentration CII, leading to the modified ratio CI / n√CII = K. Option A represents dissociation, and C is the original Nernst Distribution Law which does not account for association.
Question 2: In the case of solute dissociation in one of the solvents, if α represents the degree of dissociation, what does the term CII(1 - α) represent?
- A) The total concentration of the solute in the second solvent.
- B) The concentration of the dissociated ions in the second solvent.
- C) The concentration of the undissociated solute molecules in the second solvent.
- D) The equilibrium constant of the dissociation reaction.
View Answer
Correct Answer: C
Explanation: Since CII is the total analytical concentration, subtracting the dissociated portion (CIIα) leaves CII(1 - α), which represents the concentration of species A that remains in the undissociated form, to which the Distribution Law actually applies.
Question 3: Why must the Distribution Law be modified when a solute associates or dissociates in one of the phases?
- A) Because the temperature of the system changes significantly.
- B) Because the law only applies to ideal gases.
- C) Because the Distribution Law is valid only for the same molecular species present in both solvents.
- D) Because the volumes of the solvents change during the process.
View Answer
Correct Answer: C
Explanation: The Nernst Distribution Law assumes that the solute exists in the same molecular state in both solvents. When association or dissociation occurs, the molecular state changes in one phase, so the concentration of the common molecular species must be used to maintain a constant ratio.
Related Topics: