A) \[2.63\times {{10}^{7}}\]
B) \[2.63\times {{10}^{-7}}\]
C) \[36.2\times {{10}^{7}}\]
D) \[36.2\times {{10}^{-7}}\]
Correct Answer: B
Solution :
\[pH=-\log [{{H}_{3}}{{O}^{+}}]\] \[\therefore \]\[\log [{{H}_{3}}{{O}^{+}}]=-pH\] \[=-6.58\] (given) \[\therefore \]\[[{{H}_{3}}{{O}^{+}}]=anti\log (-6.58)\] \[=antilog\overline{7}.42\] \[=2.63\times {{10}^{-7}}g\text{ }ion/L\]
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