A) \[Ba{{\left( OH \right)}_{2}}\left( {{K}_{sp}}=1.1\times {{10}^{-4}} \right)\]
B) \[Mg{{\left( OH \right)}_{2}}\left( {{K}_{sp}}=3.5\times {{10}^{-4}} \right)\]
C) \[Fe{{(OH)}_{2}}\left( {{K}_{sp}}=8.1\times {{10}^{-16}} \right)\]
D) \[Ca{{\left( OH \right)}_{2}}\left( {{K}_{sp}}=2.1\times {{10}^{-5}} \right).\]
Correct Answer: C
Solution :
[c] \[pH=8,pOH=6;[O{{H}^{-}}]={{10}^{-6}}M\] Ionic product of \[Fe{{(OH)}_{2}}=0.2\times {{(1\times {{10}^{-6}})}^{2}}\] \[=2\times {{10}^{-13}}>{{K}_{sp}}\left( =8.1\times {{10}^{-16}} \right)\]
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