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+ + Cr2O72- + 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.

Answer 1

ans)

from above data that

we have to calculate the

The initial moles of K2Cr2O7 = 0.002543 M x 0.024 L = 6.10 x 10-5 mols

moles of Fe2+ used for back titration = 0.00971 M x 0.00802 L = 7.78 x 10^-5 mols

1 mole of K2Cr2O7 reacts with 6 moles of Fe2+

so moles of K2Cr2O7 consumed per Fe2+ in back titration = 7.78 x 10^-5 /6

moles of K2Cr2O7 consumed per Fe2+ in back titration= 1.296x 10^-5 mols

now we have to calculate the

moles of K2Cr2O7 consumed = 6.10 x 10^-5 - 1.296x 10^-5

moles of K2Cr2O7 consumed= 4.80 x 10^-5 mols

This would have reacted with = 6 x 4.80 x 10^-5 = 2.88 x 10^-4 mols of Fe2+ present in 250 ml solution

concentration of Fe2+ in the original spring water sample = 2.88 x 10^-4/0.25 = 0.001152 M

finally we have to calculate in ppm we get

molar mass= 55.845 g/mole

In ppm concentration of Fe2+ = 0.001152x 55.845 x 1000 = 64.33 ppm