A) \[{{10}^{-8}}\]
B) \[{{10}^{-6}}\]
C) \[{{10}^{-2}}\]
D) \[{{10}^{-4}}\]
Correct Answer: C
Solution :
[c] \[Mg{{(OH)}_{2}}\,M{{g}^{2+}}\,+2O{{H}^{-}}\] |
\[{{K}_{sp}}=[M{{g}^{2+}}]{{[O{{H}^{-}}]}^{2}}\] |
\[1\times {{10}^{-14}}=[M{{g}^{2+}}]\,{{[{{10}^{-6}}]}^{2}}\] |
\[\because \] \[pH=8\] |
\[\therefore \] \[[{{H}^{+}}]={{10}^{-8}}\,mol\,{{L}^{-1}}\]. |
and \[[O{{H}^{-}}]={{10}^{-6}}\,\,mol\,\,{{L}^{-1}}]\] |
\[[M{{g}^{2+}}]=\frac{{{10}^{-14}}}{{{10}^{-12}}}={{10}^{-2}}mol\,{{L}^{-1}}\] |
Solubility \[[M{{g}^{2+}}]={{10}^{-2}}\,mol\,\,{{L}^{-1}}\] |
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