Question

A 240.0-mL sample of spring water was treated to convert any iron present to Fe²⁺. Addition of 27.00-mL of 0.002520 M K₂Cr₂O₇ resulted in the reaction

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

The excess K₂Cr₂O₇ was back-titrated with 8.80 mL of 0.00936 M Fe²⁺ solution. Calculate the concentration of iron in the sample in parts per million.

Concentration of iron = _______ ppm?

Answer 1

The balanced Redox reaction between Fe⁺² and dichromate in acidic medium is

6 Fe⁺² + Cr₂O₇^–2+ 14 H+ --------> 6 Fe⁺³ + 2 Cr⁺³+ 7 H₂O

Number of moles of Cr₂O₇^–2 initially added = M*V = 0.002520 M * 27.00 mL = 0.06804 mmol

Number of moles of Fe⁺² used for back titration = 0.00936 M * 8.80 mL = 0.082368 mmol

According to balanced equation,

6 mole of Fe⁺² reacts with 1 mole of Cr₂O₇^–2

So,

0.082368 mmol of Fe⁺² would react with 0.082368 *1/6 = 0.013728 mmol of Cr₂O₇^–2 .

Net moles of Cr₂O₇^–2 used for estimation = 0.06804 – 0.013728 = 0.054312 mmol

According to balanced equation,

1 mole of Cr₂O₇^–2 reacts with 6 moles of Fe⁺²

So,

0.054312 mmol of Cr₂O₇^–2 would react with 6* 0.054312 mmol = 0.325872 mmol of Fe⁺².

Number of moles of Fe⁺² = 0.325872 mmol = 0.325872 × 10^–3 mol

Mass of Fe⁺² = moles * mol.wt. = 0.325872 × 10^–3 mol * 55.845 g/mol = 0.018198 g

Volume of Fe⁺² solution = 240.0 mL

Concentration of Iron (in ppm) = 0.018198 g/240 mL * 10⁶ = 75.83 ppm