A) 6
B) 6.9586
C) 7.9586
D) 8
Correct Answer: B
Solution :
\[[{{H}^{+}}]\]in\[HN{{O}_{3}}=1\times {{10}^{-8}}M\] \[[{{H}^{+}}]\]in water\[=1\times {{10}^{-7}}M\] Total\[[{{H}^{+}}]\]in solution \[=(1\times {{10}^{-8}}+1\times {{10}^{-7}})M\] \[=11\times {{10}^{-8}}\] \[pH=-log[{{H}^{+}}\left] =-log \right[11\times {{10}^{-8}}]\] \[=-log11+8\text{ }log\text{ }10\] \[=-1.0414+8\] \[=6.9586\]
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