I. \[Ba{{(OH)}_{2}}({{K}_{sp}}=5\times {{10}^{-3}})\] |
II. \[Ni{{(OH)}_{2}}({{K}_{sp}}=1.6\times {{10}^{-16}})\] |
III. \[Mn{{(OH)}_{2}}({{K}_{sp}}=2\times {{10}^{-13}})\] |
IV. \[Fe{{(OH)}_{2}}({{K}_{sp}}=8\times {{10}^{-16}})\] |
A) I and IV
B) III and IV
C) II and IV
D) II Ill and IV
Correct Answer: C
Solution :
When 1L each solution are mixed \[{{[OH]}^{-}}={{10}^{-6}}M\] (Buffer solution) \[{{M}^{n+}}=0.05\,M\] for \[Q=[0.5]{{[{{10}^{-6}}]}^{2}}=5\times {{10}^{-14}}\] \[Q>{{K}_{sp}}\]for \[F{{e}^{2+}}\]and \[N{{i}^{2+}}\] Thus,\[Fe{{(OH)}_{2}}\]and\[Ni{{(OH)}_{2}}\]are precipitate.You need to login to perform this action.
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