A) \[1\times {{10}^{-3}}\]
B) \[1\times {{10}^{-6}}\]
C) \[1\times {{10}^{-9}}\]
D) \[1\times {{10}^{-12}}\]
Correct Answer: D
Solution :
The \[[{{H}_{3}}{{O}^{+}}]\] in distilled water at \[{{80}^{o}}C={{10}^{-6}}M\] We know that \[{{H}_{2}}O\] is neutral \[\therefore \] \[[{{H}_{3}}{{O}^{+}}]=[O{{H}^{-}}]\] and \[{{K}_{w}}=[{{H}_{3}}{{O}^{+}}]\times [O{{H}^{-}}]\] \[\Rightarrow \] \[{{K}_{w}}={{10}^{-6}}\times {{10}^{-6}}={{10}^{-12}}\] (\[{{K}_{w}}\] increases with increase in temperature due to increased ionization of \[{{H}_{2}}O\])
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