Nernst Distribution Law and factors affecting distribution constant

Nernst Distribution Law and factors affecting distribution constant

• 1.
  Distribution Law Dr. MithilFal Desai Shree Mallikarjun and Shri Chetan Manju Desai College Canacona Goa
• 2.
  Nernst Distribution Law-statement. When a solute is distributed between two immiscible solvents in contact, at equilibrium, the ratio of its concentrations in the two solvents is constant at a given temperature, provided the solute exists in the same molecular state in both solvents. 𝐾𝑑 = 𝐶1 𝐶2 𝐾𝑑= Distribution constant/Partition coefficient/ Distribution coefficient C1 and C2= Concentration of substance in two immiscible solvents Association No Change Solvent A Solvent B (1891)
• 3.
  Factors affecting thedistribution constant 1. Temperature – Higher temperatures may change solubility 2. Nature of solvents – Polarity, hydrogen bonding, dielectric constant. 3. Nature of solute – Hydrophobic, hydrophilic, tendency to ionize. 4. Association/Dissociation of solute – If solute aggregates/ ionizes/dissociates in one solvent 5. Presence of common ions/impurities – Alters equilibrium distribution.
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  Validity of DistributionLaw 1. Two solvents are immiscible. 2. Solute exists in the same molecular form in both solvents. 3. Equilibrium is attained. 4. Temperature remains constant.
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  Modification of thedistribution law when molecules undergo a) association b) dissociation.
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  Association No Change HOW ISDISTRIBUTION LAW MODIFIED WHEN SOLUTE UNDERGOES ASSOCIATION Solvent A Solvent B
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  X nX Xn HOW ISDISTRIBUTION LAW MODIFIED WHEN SOLUTE UNDERGOES ASSOCIATION Let C1 be concentration of X in solvent A. C3 be concentration of X in solvent B C2 be concentration of Xn in solvent B. Solvent A Solvent B Applying distribution law to the equilibrium KD=C1 /C3 ----1 Kc= Xn /Xn = C3 n/ C2 ----2 Applying Mass Law to the chemical equilibrium Taking nth root on both sides in equation 𝑛 𝐾𝑐= C3 / 𝑛 C2 ----3 Dividing equation (1) by (3) KD/ 𝑛 𝐾𝑐 =(C1 / C3 )/ C3 / 𝑛 C2 ----4 KD/ 𝑛 𝐾𝑐 =(C1 / 𝑛 C2) ------5 K=(C1 / 𝑛 C2) Association No Change
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  X X A +B HOWIS DISTRIBUTION LAW MODIFIED When Solute undergoes Dissociation Let C1 be concentration of X in solvent A. C2 be total concentration of X in solvent B. Solvent A Solvent B If the degree of dissociation in solvent B is x, KD=C1 /C2(1-x) ----1 Applying distribution law to normal molecules in the two solvents K= C1 C2(1-x) Association No Change X A + B x x 1-x Hence, the concentration of the undissociated molecules in solvent B is C2(1–x). A case of this type arises in the distribution of a weak acid (e.g., succinic acid or oxalic acid) between ether and water.