A) 12
B) 9
C) 13
D) 10
Correct Answer: C
Solution :
\[\frac{\text{N}}{\text{10}}\] \[\text{NaOH}\] means, \[[O{{H}^{-}}]=\frac{1}{10}={{10}^{-1}}\] \[pOH=\log \frac{1}{[O{{H}^{-}}]}\] \[=\log \frac{1}{{{10}^{-1}}}=\log 10=1\] \[pH=14-pOH\] \[=14-1=13\]
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