In liquid–liquid extraction, how is selectivity (S_ij) between solute i and competing solute j most appropriately expressed?

Introduction:
Selectivity quantifies how preferentially a solvent extracts one solute relative to another. High selectivity favors separation at lower solvent usage and fewer stages, making it a central metric in screening solvents and designing extraction cascades.

Given Data / Assumptions:

• K_i = (concentration of solute i in extract) / (concentration of solute i in raffinate).
• K_j defined analogously for solute j.
• Equilibrium is established between phases.

Concept / Approach:
For two solutes i and j, selectivity S_ij compares their distribution coefficients: S_ij = K_i / K_j. When S_ij > 1, the solvent prefers i over j. Values close to 1 imply poor discrimination, necessitating more stages or different solvents.

Step-by-Step Solution:
Step 1: Define K for each solute as extract/raffinate concentration ratio at equilibrium.Step 2: Form the ratio S_ij = K_i / K_j.Step 3: Interpret S_ij to assess separation effectiveness.

Verification / Alternative check:
McCabe–Thiele-style analyses and solvent screening tables use S_ij to rank solvent systems for selective recovery of target solutes.

Why Other Options Are Wrong:

• mass fraction terms: Describe compositions but not relative preference between two different solutes.
• ratio of extract to raffinate for one solute: That is K, not selectivity between two solutes.
• difference K_i − K_j: Lacks the right scaling; dimensionless ratio is standard.

Common Pitfalls:
Confusing distribution coefficient K (single solute) with selectivity S_ij (two-solute comparison) and neglecting activity-coefficient effects at high concentrations.

Final Answer:
ratio of the distribution coefficients of two solutes, S_ij = K_i / K_j