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An acidic aqueous solution contains $0.02 M - {FeCl}_{2}$ and $0.05 M - {FeCl}_{3}$ , at 25° C. Now NaOH solution is added is this solution, drop by drop. The pH range in which only the precipitation of Fe(OH) is occur without precipitation of $Fe {(OH)}_{2}$ (s) differs by 'x' unit. The value of 'x' is given:
$K_{sp}$ of $Fe (OH)_{2} = 8 \times 10^{- 16}$
$K_{sp}$ of $Fe (OH)_{3} = 4 \times 10^{- 28}$

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answer is 2.

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Detailed Solution

For pptn. of $Fe (OH)_{3}, [{OH}^{-}] = \sqrt[3]{\frac{4 \times 10^{- 28}}{0.05}} = 2 \times 10^{- 9}$

For pptn. of $Fe (OH)_{2}, [{OH}^{-}] = \sqrt{\frac{8 \times 10^{- 16}}{0.02}} = 2 \times 10^{- 7}$

$Δ (pH)$ = 9-7 = 2

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