A) \[1\times {{10}^{1}}\text{ }mol\text{ }d{{m}^{-3}}\]
B) \[1\times {{10}^{-12}}\text{ }mol\text{ }d{{m}^{-3}}\]
C) \[1\times {{10}^{-1}}\text{ }mol\text{ }d{{m}^{-3}}\]
D) \[1\times {{10}^{-14}}\text{ }mol\text{ }d{{m}^{-3}}\]
Correct Answer: B
Solution :
\[[{{H}^{+}}][O{{H}^{-}}]={{10}^{-14}}\] \[[{{10}^{-2}}][O{{H}^{-}}]={{10}^{-14}}\] \[[O{{H}^{-}}]=\frac{{{10}^{-14}}}{{{10}^{-2}}}={{10}^{-12}}mol\,\,d{{m}^{-3}}\]
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