A) \[4\times {{10}^{-8}}\]
B) \[2\times {{10}^{-8}}\]
C) \[4\times {{10}^{-4}}\]
D) \[2\times {{10}^{-4}}\]
Correct Answer: A
Solution :
\[pH=-{{\log }_{10}}[{{H}^{+}}]\] \[7.4=-{{\log }_{10}}[{{H}^{+}}]\] \[{{\log }_{10}}[{{H}^{+}}]=-7.4\] \[[{{H}^{+}}]={{10}^{-7.4}}\] \[={{10}^{-8}}\times {{10}^{0.6}}\] \[={{10}^{-8}}\times anti\,\log {{10}^{0.6}}={{10}^{-8}}\times 4\] \[[{{H}^{+}}]=4\times {{10}^{-8}}\]
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