A) 1
B) 2
C) 3
D) 4
Correct Answer: B
Solution :
\[\text{HN}{{\text{O}}_{\text{3}}}\] completely ionizes as \[HN{{O}_{3}}+{{H}_{2}}O\xrightarrow{{}}{{H}_{3}}{{O}^{+}}+NO_{3}^{-}\] \[\therefore \] \[[{{H}_{3}}{{O}^{+}}]=HN{{O}_{3}}={{10}^{-2}}M\](given) \[pH=-=\log [{{H}_{3}}{{O}^{+}}]\] \[=-\log [{{10}^{-2}}]\] \[=-(-2\,log10)\] \[=2\]You need to login to perform this action.
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