## Advanced Level Chemistry - Redox Volumetric Titration Question 1 - ANSWERS to the calculation

### Thoughts on style of question? * Worksheet of Redox Titration Questions * Updated Doc Brown's Chemistry April 29th 2010 *

(a) MnO4-(aq) + 8H+(aq) + 5Fe2+(aq) ==> Mn2+(aq) + 5Fe3+(aq) + 4H2O(l)

(b) (i) mol MnO4- = 0.02[2dp] x 24.3[1dp] / 1000[4sf] = 0.000486[6dp]

mol Fe2+ = 5 x 0.000486[6dp] (from 1 MnO4- : 5 Fe2+ in equation)
mol Fe2+ = 0.00243[5dp] in 20 cm3[2sf]

so scaling up to 1 dm3, molarity of Fe2+ = 0.00243[5dp] x 1000[4sf] / 20[2sf]
molarity Fe2+ = 0.1215 mol dm-3 [4dp]

(ii) The end point is the first faint permanent pink due to a trace of excess KMnO4

(c) mol MnO4- = 0.02[2dp] x 25.45[4sf] / 1000 = 0.000509 [6dp]

mol Fe = mol Fe2+ = 5 x 0.000509[6dp] = 0.002545 [6dp]

mass Fe = 0.002545 x 55.9[1dp] = 0.1423g [4dp] per titration.

total Fe in wire = 0.1423 x 10[2sf] = 1.423g (1/10th of the made up solution used in titration)

so % Fe = 1.423[3dp] x 100 / 1.51 = 94.2% [3sf]