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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