In a solution of 0.04 M FeCl2, 0.02 M FeCl3 and 0.01 M-HCl, how large may be its pH without there being precipitation of either Fe(OH)2 or Fe(OH)3 ? Ksp are 8 x 10-16 for Fe(OH) 2 and 4 x 10-28 for Fe(OH)3• (Give your answer excluding the decimal places).

Here, OH ion can be given as,OH−Fe(OH)2=8×10−160.04=2×10−7OH−Fe(OH)3=2.71×10−9Thus, the pH value,pH=5.43

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In a solution of 0.04 M ${FeCl}_{2}, 0.02 M {FeCl}_{3}$ and 0.01 M-HCl, how large may be its pH without there being precipitation of either Fe(OH)₂or Fe(OH)₃? K_sp are 8 x 10^-16 for Fe(OH) 2 and 4 x 10^-28 for Fe(OH)₃• (Give your answer excluding the decimal places).

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answer is 5.

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Detailed Solution

Here, OH ion can be given as,

$\begin{array}{l} {[{OH}^{-}]}_{Fe (OH)_{2}} = \sqrt{\frac{8 \times 10^{- 16}}{0.04}} = \sqrt{2} \times 10^{- 7} \\ {[{OH}^{-}]}_{Fe (OH)_{3}} = 2.71 \times 10^{- 9} \end{array}$

Thus, the pH value,

$pH = 5.43$

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