Starting with equation A bFeCN63 derive equation A d kFgl

Starting with equation (A = b[Fe(CN)₆]^3-) derive equation [A = d + kF_glucose]. Hint: write [Fe(CN)₆]^3- in terms of the formal concentration of glucose and the initial concentration of [Fe(CN)₆]^3- and recognize that ‘d’ and ‘k’ are arbitrary constants.

Solution

Beer’s law relates absorbance to both transmittance and to the concentration of the absorbing species

A = –logT = bC

The amount of [Fe(CN)6]3- is directly proportional to the amount of glucose
present.

here C=C0 + F[glucose] here C0= initial concentation of [Fe(CN)6]3-

substituting A=-logT =b(C0 +[glucose]) =bC0+bF[glucose] =d+kF[glucose]

here ‘d’ = bC0 and ‘k’ =b are arbitrary constants.