Gel filtration / size-exclusion chromatography relation For a solute in size-exclusion chromatography, which general relationship correctly gives the elution volume V in terms of void volume V0, distribution constant kD, and internal water volume Vi?

Introduction / Context:
In size-exclusion (gel filtration) chromatography, molecules separate by their ability to access pore volume. The classic linear relationship between elution volume and distribution behavior is fundamental for estimating molecular size and calibrating columns.

Given Data / Assumptions:

• V0 is the void volume (outside the beads).
• Vi is the internal solvent volume within the beads accessible to solute.
• kD is the distribution constant ranging from 0 (totally excluded) to 1 (fully included).

Concept / Approach:
The total volume sampled by a solute equals the volume outside beads plus a fraction of the internal volume it can enter. Hence, elution volume is V = V0 + kD * Vi. This relationship allows plotting V versus log(MW) using standards.

Step-by-Step Solution:
Recognize limiting cases: a very large solute has kD = 0 → V ≈ V0.For a very small solute, kD = 1 → V ≈ V0 + Vi (total accessible volume).Interpolate for intermediate sizes: V = V0 + kDVi.Select the expression consistent with these boundary conditions.

Verification / Alternative check:
Empirical calibration curves using proteins of known molecular mass produce linear plots when V (or Kav) is related to log(MW), supporting the stated formula.

Why Other Options Are Wrong:
V = V0 / Vi (option b) has incorrect units and ignores kD.V = V0 - kDVi (option c) would predict decreasing volume with greater inclusion, which is opposite of reality.V / V0 = kDVi (option d) mixes dimensions and is not the standard relationship.

Common Pitfalls:
Confusing V0 (outside beads) with Vi (inside beads); forgetting that increased pore access increases elution volume (later elution), not decreases it.

Final Answer:
V = V0 + kDVi.