A TLC plate showed two spots of Rf 0.25 and 0.26. The plate was dried carefully,...

A TLC plate showed two spots of Rf 0.25 and 0.26. The plate was dried carefully, and returned to the developing chamber. What would you expect to see after the second development was complete?

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Expert Solution

Assuming the distance between the baseline and solvent front be equal to 10 cm (a standard one).

Given, Rf value of spot 1 = 0.25 cm

Position of spot 1 from baseline = 2.5 cm

[using, Rf= d_spot / d_{solvent front,} where d= distance travelled by]

Rf value of spot = 0.26 cm

Position of spot 2 from baseline = 2.6 cm

Resolution = Rf2- Rf2 = 0.26 cm – 0.25 cm = 0.01 cm

Now, if the plate is dried and again subjected to developing chamber under identical conditions, the spots move along the solvent front like before.

However, with different distance of solvent front.

Total solvent front available for spot 1 = 10 cm – 2.5 cm = 7.5 cm

(because spot 1 already moved 2.5 cm during first incubation, it can only move along another 7.5 cm at maximum).

Total solvent front available for spot 2 = 10 cm – 2.6 cm = 7.4 cm

New position of spot 1 from baseline =

Initial distance from baseline + [Rf x solvent front (available distance to travel)]

=2.5 cm + (0.25 x 7.5) cm = (2.5 +1.875) cm = 4.375 cm

And, new Rf1 = 4.375 cm / 10 cm = 0.4375

Similarly, New position of spot 2 from baseline =

=2.6 cm + (0.26 x 7.4) cm = (2.6 +1.875) cm = 4.524 cm

And, new Rf2 = 4.524 cm / 10 cm = 0.4524

New resolution = 0.4524 – 0.4375 = 0.0149

Thus, subjecting to another round chromatography increases the resolution of spots.

queen_honey_blossom answered 11 months ago

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