Question

A 100.0-mL sample of spring water was treated to convert any iron present to Fe^2+. Addition of 25.00-mL of 0.002517 M K2Cr2O7. The excess K2Cr2O7 was back-titrated with 8.53 mL of 0.00949 M Fe^2+ solution. Calculate the concentration of iron in the sample in parts per million. a) 3.0 ppm b) 4.0 ppm c) 5.0 ppm d) 6.0 ppm

Answer 1

The concentration of iron in the sample in parts per million is
`\(166.096 \text{ ppm}\)`

How to determine the concentration of iron in the sample in parts?

Given data:

Volume of sample = 100 mL

Volume of
`\(K_2Cr_2O_7\)` = 25 mL

Molarity of
`\(K_2Cr_2O_7\)` = 0.002517 M

Volume of
`\(Fe^(2+)\)` solution = 8.53 mL

Molarity of
`\(Fe^(2+)\)` solution = 0.00949 M

Equation used:

Concentration in ppm =
`(Amount \; of \; solute)/(Amount \; of \; sample) * \(10^6\)`

Solution:

Six equivalents of
`\(Fe^(2+)\)` react with one equivalent of
`\(Cr_2O_7^(2-)\)`.

Milliequivalents of
`\(K_2Cr_2O_7\)` taken =
`\(25 * 0.002517 * 6\)`

Milliequivalents of excess
`\(K_2Cr_2O_7\) = \(8.53 * 0.00949 * 1\)`

Milliequivalents of
`\(Fe^(2+)\)` in water =
`\((25 * 0.002517 * 6) - (8.53 * 0.00949 * 1)\)`

`\(= 0.37755 - 0.0809497 = 0.2966 \text{ mmol}\)`

Mass of
`\(Fe^(2+)\)` in water sample =
`\(0.2966 * 10^(-3) \text{ mol} * 56 \text{ g/mol}\)`

`\(= 16.6096 * 10^(-3) \text{ g}\)`

Concentration in parts per million:

`\frac{16.6096 * 10^(-3) \text{ g}}{100 \text{ g}} * 10^6\) \(= 166.096 \text{ ppm}\)`

Therefore, the concentration of iron in the sample in parts per million is
`\(166.096 \text{ ppm}\)`.

Complete question:

A 100.0-mL sample of spring water was treated to convert any iron present to Fe21. Addition of 25.00-mL of 0.002517 M K2Cr2O7 resulted in the reaction 6Fe2+ + Cr2O7 2- + 14H+ → 6Fe3+ + 2Cr3+ + 7H2O

The excess K2Cr2O7 was back-titrated with 8.53 mL of 0.00949 M Fe2+ solution. Calculate the concentration of iron in the sample in parts per million.