Question

Una muestra de 100 mL de agua de manantial se trató para convertir el hierro presente a la forma de Fe+2 . Al añadir 25.00 mL de K2Cr2O7 0.002107 M se produjo la siguiente reacción: 6 Fe+2 + Cr2O72- + 14H+ → 6 Fe+3 + 2Cr3+ + 7H2O

El exceso de K2Cr2O7 se tituló por retroceso con 7.47 mL de una solución 0.00979 en Fe+2. Exprese la concentración de hierro en el agua, en mg Fe+2/L .

A sample of 100 mL of spring water was treated to convert the iron present to the Fe + 2 form. When adding 25.00 mL of K2Cr2O7 0.002107 M the following reaction occurred: 6 Fe + 2 + Cr2O72- + 14H + → 6 Fe + 3 + 2Cr3 + + 7H2O

The excess of K2Cr2O7 was titrated by recoil with 7.47 mL of a 0.00979 solution in Fe + 2. Express the concentration of iron in water, in mg Fe + 2 / L.

Answer 1

From balanced reaction

Mole ratio of Fe²⁺ and Cr₂O₇^2- is 6:1

Hence, mmoles of Fe²⁺ = (mmoles of K₂Cr₂O₇)*6

mmoles of K₂Cr₂O₇ added to the iron sample

= molarity *volume = 25.00*0.002107 = 0.0527.

mmoles of Fe²⁺ added to titrate excess K₂Cr₂O₇

= 7.47*0.00979 = 0.0731 .

Hence, mmoles of excess K₂Cr₂O₇ = (0.0731/6) = 0.0122

Now, mmoles of K₂Cr₂O₇ reacted with the iron in water

= (0.0527 - 0.0122)

= 0.0405

So, mmoles Fe²⁺ in water sample = 0.0405*6 = 0.243

mass of Fe²⁺ = mmoles * molar mass

= 0.243*56 = 13.6 mg.

Now, volume of iron sample = 100 mL = 0.1 L.

So, concentration of iron in water

= (13.6/0.1)

= 136 mg/L