A) \[3\]
B) \[11\]
C) \[4\]
D) \[-11\]
Correct Answer: B
Solution :
\[[NaOH]={{10}^{-3}}M\] \[\therefore \] \[[O{{H}^{-}}]={{10}^{-3}}M\]or \[[{{H}^{+}}]=\frac{{{10}^{-14}}}{{{10}^{-3}}}={{10}^{-11}}M\] \[pH=-\log [{{H}^{+}}]=-\log {{10}^{-11}}=11\]
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