Question

Determination of iron by redox titration

What error (+, - or 0) is introduced in the % Fe in the unknown and briefly explain why in each case.
(a) The permanganate solution becomes contaminated with organic matter before standardization.
(b) The original KMnO4 solution contained some MnO2 that was not filtered out.
(c) The sodium oxalate was not totally dry when it was weighed out.
(d) The unknown had some of its iron already oxidized to Fe3+.

Answer 1

Error % is calculated as {(Experimental value - Actual value)/Actual value}x100. If the experimental value is greater than actual, a positive error will arise and if it is lesser, the error will be negative.

a) If the permanganate solution was contaminated with organic matter, it will react with those and thus have a concentration lesser than expected. This is result in increased consumption of KMnO4, resulting in a higher than actual concentration of iron as it can be seen from the law of equivalence that N_Fe = (V_KMnO4 x N_KMnO4)/V_Fe implying that the volume of permanganate added is directly propotional to the concentration and thus the % of Fe in the sample. Therefore, this case will end in a positive error.

b) If the permanganate solution has some manganese dioxide that hadn't been filtered out, while weighing, the weight of permanganate and the oxide will be taken as that of the former only. This will result in a lower than calculated concentration of permanganate giving a higher volume of permanganate required than actual resulting in a greater than actual concentration of iron. Thus, it will give a positive error.

c) If the oxalate salt was not completely dry when weighed out, it will possess an unknown quantity of water along with the salt. This will result in a lower concentration of the salt. So, in the redox titration with permanganate, the analyte will consume less permanganate than what it should. This will result in a higher overall concentration of KMnO4 as N_KMnO4 = (V_oxalate x N_oxalate)/V_KMnO4 and a lower V_KMnO4 gives a higher normality. This will in turn result in a higher than actual normality of iron again giving a positive error.

d) If the iron present in the analyte had already been oxidised, the volume of permanganate required will be decreased. This will end in a lower percentage of iron found experimentally as some of the iron present does not answer to the titration. This scenario will give a negative error.