A) 1
B) 0
C) 14
D) \[{{10}^{-14}}\]
Correct Answer: C
Solution :
Given, \[[O{{H}^{-}}]=1\,mol\,1\text{ }{{L}^{-1}}\] So, we know that \[pOH=-log[O{{H}^{-}}]\] \[pOH=-log1=0\] \[\because \] \[pH+pOH=14\] \[\therefore \] \[pH=14-0=14\]You need to login to perform this action.
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