Example:

i) In a paper chromatographic separation of cations- Ag⁺, Pb⁺ and Hg⁺, solvent front rises to 18.4 cm when cationic spots were observed at 15.8, 12.1 and 5.9 cm, correspondingly. Compute R_f values of the metal ions.

ii) Emission gases of a two wheeler were tested for pollutant metals by paper chromatography.  The spot equivalent to R_f value of 0.65 was observed. What is the possible pollutant metal ion in the emission gases?

Solution

Here,  R_f value for Ag⁺=      15.8/18.4      = 0.86

            R_f value for Pb⁺ =     12.1/18.4        = 0.66

            R_f value for Hg⁺ =     5.9 /18.4         = 0.32

As R_f value of 0.65 is comparable to 0.66 consequent to Pb⁺, it could be concluded in which the pollutant in the emission gases is likely to be lead.