A) \[3.84\]
B) \[2.42\]
C) \[4.44\]
D) \[1.42\]
Correct Answer: B
Solution :
\[pH=-\log [{{H}^{+}}]\] \[=-\log (3.8\,\times \,{{10}^{-3}})\] \[=-\log 3.8-\log {{10}^{-3}}\] \[=-0.5798+3=2.4202\]
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