A) \[-{{\log }_{e}}({{H}^{+}})\]
B) \[-{{\log }_{10}}({{H}^{+}})\]
C) \[{{\log }_{e}}({{H}^{+}})\]
D) \[{{\log }_{10}}({{H}^{+}})\]
Correct Answer: B
Solution :
\[pH=-\log \,\,[{{H}^{+}}]\].
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