In a solution $0.04 M$ $F e {C l}_{2}$ , $0.02 M$ $F e {C l}_{3}$ and $0.01 M$ $H C l$ , how large may be its $p H$ (nearest integer) without there being precipitation of either $F e {(O H)}_{2}$ or $F e {(O H)}_{3}$ ? $K_{s p}$ of $F e {(O H)}_{2}$ and $F e {(O H)}_{3}$ are $8 \times 10^{- 16}$ and $4 \times 10^{- 38}$ respectively.

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Solution

Since the solubility product of ferric hydroxide has lower value, it will precipitate first, The expression for the solubility product is $K_{s p} = [F e^{3 +}] [O H^{-}]^{3}$ .
Substitute values in the above expression.
$4 \times 10^{- 38} = 0.02 \times [O H^{-}]^{3}$
Henc, $[O H^{-}] = 1.25 \times 10^{- 12}$
$p O H = - l o g [O H^{-}] = - l o g 1.25 \times 10^{- 12} = 11.90$ .
$p H = 14 - p O H = 14 - 11.90 = 2.10$

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