A) 7
B) 7.3010
C) 6.6990
D) none of these
Correct Answer: B
Solution :
Given-that, \[[O{{H}^{-}}]={{10}^{-7}}+{{10}^{-7}}\](from water) \[\therefore \] \[[O{{H}^{-}}]=2\times {{10}^{-7}}\] \[\therefore \] \[pOH=-\log [O{{H}^{-}}]\] \[=-\log 2\times {{10}^{7}}\] \[=6.6990\] \[\therefore \] \[pH=14-6.6990\] \[=7.3010\]
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