A) 1
B) 6
C) 8
D) 10
Correct Answer: B
Solution :
The \[{{K}_{\omega }}\]at\[373K={{10}^{-12}}\] In case of pure water \[{{K}_{\omega }}=10{{~}^{12}}\] \[=[{{H}^{+}}][O{{H}^{-}}]\] \[\therefore \] \[[{{H}^{+}}]=[O{{H}^{-}}]={{10}^{-6}}\] We know that\[pH=-\log [{{H}^{+}}]\] \[=-\log [{{10}^{-6}}]=6\] The \[{{K}_{\omega }}\]at \[373\text{ }K={{10}^{-12}}\]
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