Ionic EquilibriumHard
Question
An aqueous solution contains 0.02 M-FeCl2 and 0.05 M-FeCl3. The solubility products are 8 × 10−16 for Fe(OH)2 and 4 × 10−28 for Fe(OH)3. Identify the correct option(s) among the following regarding the precipitation of metal hydroxides.
Options
A.At pH = 9.0, neither Fe(OH)2 nor Fe(OH)3 will precipitate.
B.At pH = 6.0, neither Fe(OH)2 nor Fe(OH)3 will precipitate.
C.If pH of the solution is in between 5.3 and 7.3, only Fe(OH)3 will precipitate but not Fe(OH)2.
D.If pOH of the solution is in between 6.7 and 8.7, only Fe(OH)2 will precipitate but not Fe(OH)3.
Solution
For precipitation of Fe(OH)2,
$\left\lbrack OH^{-} \right\rbrack_{\text{min}} = \sqrt{\frac{8 \times 10^{- 16}}{0.02}} = 2 \times 10^{- 7} \Rightarrow P_{\max}^{OH_{\min}^{H}}$
For precipitation of Fe(OH)3,
${\left\lbrack OH^{-} \right\rbrack{\left( \frac{4 \times 10^{- 28}}{0.05} \right)^{1/3}}^{- 9}}_{\min} $$$\Rightarrow P_{\max}^{OH_{\min}^{H}}$$
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