A) 9
B) 10
C) 12
D) 13
Correct Answer: D
Solution :
Normality of \[NaOH\] = its molarity \[=\frac{M}{10}\] [\[NaOH\] is monoacidic base and molarity = acidity \[\times N\]] \[[O{{H}^{-}}]=\frac{1}{10}=0.1M\] As \[[{{H}_{3}}{{O}^{+}}][O{{H}^{-}}]={{10}^{-14}}\] \[[{{H}_{3}}{{O}^{+}}]=\frac{{{10}^{-14}}}{0.1}={{10}^{-13}}\] \[pH=-\log [{{H}_{3}}{{O}^{+}}]=-\log {{10}^{-13}}\] \[pH=13\]
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