A) \[8\]
B) \[6\]
C) \[7.04\]
D) cannot be determined
Correct Answer: C
Solution :
At \[{{90}^{o}}C,\]\[[{{H}_{3}}{{O}^{+}}]={{10}^{-6}}\,mol\,litr{{e}^{-1}}\] \[\therefore \] \[[O{{H}^{-}}]={{10}^{-6}}\,mol\,\,litr{{e}^{-1}}\] hence, \[pOH=-\log [O{{H}^{-}}]=6\]You need to login to perform this action.
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