A) 2
B) 10
C) 11
D) 12
Correct Answer: D
Solution :
\[[NaOH]=\frac{0.4}{40}=0.01M;\,\,\,\,\,[O{{H}^{-}}]={{10}^{-2}}M\] \[[{{H}^{+}}]={{10}^{-12}},\,\,pH=-\log [{{H}^{+}}]=12\]You need to login to perform this action.
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