A 100mL sample of water was treated to convert any iron present to ${Fe}^{2 +}$ 25mL of $0.002 {M K}_{2} {Cr}_{2} O_{7}$
resulted in the reaction.
$6 {Fe}^{2 +} + {Cr}_{2} O_{7}^{2 -} + 14 H^{+} ⟶ 6 {Fe}^{3 +} + 2 {Cr}^{3 +} + 7 H_{2} O$
The excess $K_{2} {Cr}_{2} O_{7}$ was back-titrated with $7.5 mL$ of $0.01 M {Fe}^{2}$ solution. Calculate the parts per million (ppm) of iron in the water sample.

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answer is 126.

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Detailed Solution

milli equivalent of excess $K_{2} {Cr}_{2} O_{7} = 7.5 \times 0.01 \times 1$

milli equivalent of $K_{2} {Cr}_{2} O_{7}$ taken $= 25 \times 0.002 \times 6$

milli equivalent of ${Fe}^{2 +}$ in water sample

$\begin{array}{l} = 25 \times 0.002 \times 6 - 7.5 \times 0.01 \times 1 \\ = 0.225 \end{array}$

$∴$ Mass of ${Fe}^{2 +} = 0.225 \times 10^{- 3} \times 56 = 12.6 \times 10^{- 3} g$

$100 g ⟶ 12.6 \times 10^{- 3}$

$10^{6} g ⟶ \frac{12.6 \times 10^{- 8}}{100} \times 10^{6} = 126$

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