Question

A 220.0-mL sample of spring water was treated to convert any iron present to Fe2+. Addition of 27.00-mL of 0.002516 M K2Cr2O7 resulted in the reaction.

6Fe2+ +Cr2O7(2-) + 14H+ --> 6Fe3+ +2Cr(3+) + 7H2O

The excess was back-titrated with 8.50 mL of 0.00987 M Fe2+ solution. Calculate the concentration of iron in the sample in parts per million. Concentration of iron = ppm

Answer 1

Let us consider given reaction,

6Fe2+ +Cr2O7(2-) + 14H+ --> 6Fe3+ +2Cr(3+) + 7H2O

Each molar amount m can be determined as the product of concentration c and volume v

6 m_Cr2O7 = m_Fe2+_,water +m_Fe^2+

6C_Cr2O7 V_{Cr2O7 =} mFe^2_, water + C_Fe^2+V_Fe^2+

6(2.516 x 10^-3M )(2.700 x 10^-2L) =  m_Fe2+_,water +( 9.87 x 10^-3 M) (8.50 x 10^-3 L)

4.07 x 10^-4 mol = m_Fe2+_,water +8.38 x 10^-5 mol

m_Fe2+_,water = 3.23 x 10^-4

Now, use the molar mass of Fe to determine the milligram amount of Fe in the sample.

? mg Fe = (3.23 x 10^-4 mol Fe) (55.845 g Fe / 1 mol Fe) (10³ mg / 1g)

? mg Fe = 18.03 mg Fe

In dilute solutions, ppm is equivalent to milligrams per liter.Use this to determine the ppm concentration of Fe in the sample.

? ppm Fe = 18.03 mg Fe / 0.2200 L

= 81.95 ppm Fe

Therefore, the concentration of iron in the sample is 81.95 ppm Fe.