A) \[{{10}^{-14}}\]
B) \[{{10}^{-12}}\]
C) \[{{10}^{-8}}\]
D) \[{{10}^{-6}}\]
Correct Answer: D
Solution :
\[{{K}_{\omega }}=[{{H}^{+}}]\,[O{{H}^{-}}]=[{{H}_{3}}{{O}^{+}}][O{{H}^{-}}]\] \[={{10}^{-6}}\times {{10}^{-6}}\] \[={{10}^{-12}}\] \[\because \] \[[{{H}_{3}}{{O}^{+}}]={{10}^{-6}}mole/litre=[O{{H}^{-}}]\]You need to login to perform this action.
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