question_answer

At\[{{90}^{o}}C,\]pure water has\[[{{H}_{3}}{{O}^{+}}]={{10}^{-6}}mol/L.\] The value of\[{{K}_{w}}\]at\[{{90}^{o}}C\]is:

A)  \[{{10}^{-6}}\]                                 

B)  \[{{10}^{-8}}\]

C)  \[{{10}^{-12}}\]                               

D)  \[{{10}^{-14}}\]                 In pure water: \[[{{H}^{+}}]=[O{{H}^{-}}]\] \[{{K}_{w}}=[{{H}^{+}}][O{{H}^{-}}]\] \[{{K}_{w}}={{10}^{-6}}\times {{10}^{-6}}={{10}^{-12}}\]