A) \[0.8%\]
B) \[0.6%\]
C) \[0.4%\]
D) \[0.2%\]
Correct Answer: D
Solution :
We know that \[[{{H}^{+}}]={{10}^{-pH}}={{10}^{-5}}\] \[\alpha =\frac{actual\,\,concentration}{molar\,\,concentration}\] \[=\frac{{{10}^{-5}}}{0.005}=0.2\times {{10}^{-2}}\] \[\therefore \]percentage ionization\[=0.2\times {{10}^{-2}}\times 100\] \[=0.2%\]
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