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Nernst Equation

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$100$ mL of a mixture of $F e S O_{4}$ and $F e_{2} (S O_{4})_{3}$ requires $50$ mL of $0.1$ N $K M n O_{4}$ solution, then the concentration of $F e^{2 +}$ is _________ $g L^{- 1}$ (in nearest integer).

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Solution

$1$ equivalent of $K M n O_{4}$ oxidizes $1$ equivalent of ferrous sulfate to ferric sulphate.
$50$ ml of $0.1$ N $K M n O_{4}$ corresponds to $\frac{50}{1000} \times 0.1 = 0.005$ moles of $K M n O_{4}$ .
It is also equal to the number of moles of ferrous sulphate.
The molar concentration of $F e^{2 +}$ ions is $\frac{0.005}{100} \times 1000 = 0.05$ M.
The concentration of $F e^{2 +}$ is $55.8 \times 0.05 = 2.8 g L^{- 1}$ $= 3 g L^{- 1}$ .

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