A) \[{{10}^{-6}}\]
B) \[{{10}^{-12}}\]
C) \[{{10}^{-14}}\]
D) \[{{10}^{-8}}\]
Correct Answer: B
Solution :
Key Idea: \[{{K}_{w}}=[{{H}^{+}}]\,[O{{H}^{-}}]\] \[{{H}_{2}}O[{{H}^{+}}]+[O{{H}^{-}}]\] \[\therefore \] \[[{{H}^{+}}]=[O{{H}^{-}}]\,([{{H}^{+}}])=[{{H}_{3}}{{O}^{+}}]\] \[\therefore \] \[{{K}_{w}}={{[{{H}_{3}}{{O}^{+}}]}^{2}}\] Given, \[[{{H}_{3}}{{O}^{+}}]={{10}^{-6}}mol/L\] \[\therefore \] \[{{K}_{w}}={{({{10}^{-6}})}^{2}}\] \[={{10}^{-12}}\]
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