A) \[1.96\times {{10}^{-2}}mole\text{ }li{{t}^{-1}}\]
B) \[1.96\times {{10}^{-3}}mole\text{ }li{{t}^{-1}}\]
C) \[1.5\times {{10}^{-4}}mole\text{ }li{{t}^{-1}}\]
D) \[1.5\times {{10}^{-3}}mole\text{ }li{{t}^{-1}}\]
Correct Answer: C
Solution :
: \[pH=-\log [{{H}^{+}}]\] \[[{{H}^{+}}]=\]Antilog \[(-pH)\] \[\therefore \]\[[{{H}^{+}}]=1.5\times {{10}^{-4}}mole\,li{{t}^{-1}}\]
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