A) \[3\times {{10}^{-10}}\]
B) \[5\times {{10}^{-10}}\]
C) \[1\times {{10}^{-10}}\]
D) \[5\times {{10}^{-2}}\]
Correct Answer: B
Solution :
\[pH=-{{\log }_{10}}[{{H}^{+}}]\] or \[{{\log }_{10}}[{{H}^{+}}]=-4.7\] \[[{{H}^{+}}]={{10}^{-4.7}}\] \[={{10}^{-5}}\times {{10}^{+0.3}}\] \[={{10}^{-5}}\times anti\,\log \,0.3\] \[=2\times {{10}^{-5}}\] \[{{K}_{\omega }}=[{{H}^{+}}]\,[O{{H}^{-}}]\] \[\,[O{{H}^{-}}]=\frac{{{K}_{\omega }}}{[{{H}^{+}}]}=\frac{{{10}^{-14}}}{2\times {{10}^{-5}}}\] \[=0.5\times {{10}^{-9}}=5\times {{10}^{-10}}\]
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