Question

A 250.0 mL sample of spring water was treated to convert any iron present to Fe2+ . Addition of 24.00 mL of 0.002543 M K2Cr2O7 resulted in the reaction

6Fe2+ + Cr2O7 2- + 14H+  6Fe3+ + 2Cr3+ + 7H2O

The excess K2Cr2O7 was back titrated with 8.02 mL of 0.00971 M Fe2+ solution. Calculate the concentration of iron in the sample in ppm.

Please be specific and include details.

Answer 1

First we calculate the moles of Cr2O7-2 aggregates and in the excess titration:

n Cr2O7-2 add = M * V = 0.002543 M * 0.024 L = 6.1032x10 ^ -5 mol

n Cr2O7-2 excess = 0.00971 * 0.00802 * (1/6) = 1.2979x10 ^ -5 mol

We calculated the moles of Cr2O7-2 that reacted with Fe:

n Cr2O7-2 reacted = add - excess = (6.1032x10 ^ -5) - (1.2979x10 ^ -5) = 4.8053x10 ^ -5

We calculate the moles of Fe present:

n Fe = n Cr2O7-2 reacted * 6 = 4.8053x10 ^ -5 * 6 = 2.8832x10 ^ -4 mol

We calculate the ppm of Fe:

ppm Fe = (2.8832x10 ^ -4 mol / 0.25 L) * (55.85 g / 1 mol) * (1000 mg / 1 g) = 64.411 ppm

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