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A sample of 100.0 mL fresh water was treated to convert any iron present in Fe...

A sample of 100.0 mL fresh water was treated to convert any iron present in Fe When 25.00 mL of K2Cr2O7 0.002517 M was added the following reaction occurred: 6Fe + Cr2O + 14H6Fe + 2Cr + 7H20 The excess of Kr Cr2O7 was titrated by recoiling with 8.53 mL of a solution of Fe 0.00949 M. Calculate the concentration of iron in the sample in parts per million.

A sample of 100.0 mL fresh water was treated to convert any iron present in Fe2+ . When 25.00 mL of K2Cr2O7 0.002517 M was added the following reaction occurred: 6Fe^2++ Cr2O7^-2+ 14H—>6Fe^3+ + 2Cr^3+ + 7H20

The excess of Kr2 Cr2O7 was titrated by recoiling with 8.53 mL of a solution of Fe^2+ 0.00949 M. Calculate the concentration of iron in the sample in parts per million.

Expert Solution

Six equivalents of Fe⁺² react with one equivalent of Cr₂O₇^2- Two sources of Fe⁺² reacted with the Cr₂O₇^2- so the total molar amount of Cr₂O₇^2- must be equal to one-sixth the sum of the molar amount of Fe⁺² from the two sources. Each molar amount m can be determined as the product of concentration c and volume V.

6m_Cr2O7^2- = m_{Fe⁺²,water} + m_Fe⁺²

6 c_Cr2O7² V_Cr2o7^2- = m_{Fe⁺²,water} + C_Fe+2 V_Fe+2

6 (2.517 X 10^-3 M)(2.500 X 10^-2L) = m_{Fe⁺²,water} + (9.49 X 10^-3 M)(8.53 X 10^-3 L)

3.776 X 10^-4 mol = m_{Fe⁺²,water} +8.09 X 10^-5 mol

m_{Fe⁺²,water} = 2.967 X 10^-4 mol

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

? mg Fe = (2.967 X 10^-4 mol Fe) X (55.845 g Fe / 1 mol Fe) X (10³ mg / 1 g)

? mg Fe = 16.56 mg Fe

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

? mg Fe = 16.56 mg Fe / 0.1000 L

? mg Fe = 165.6 ppm Fe

the concentration of iron in the sample is 165.6 ppm

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queen_honey_blossom answered 2 years ago

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