A) \[0.1\times {{10}^{-2}}\]
B) \[0.2\times {{10}^{-2}}\]
C) \[0.5\times {{10}^{-4}}\]
D) \[0.6\times {{10}^{-6}}\]
Correct Answer: B
Solution :
Acid, \[[{{H}^{+}}]={{10}^{-pH}}={{10}^{-5}}\] \[\alpha =\frac{\text{actual}\,\text{ionisation}}{\text{molar}\,\text{concentration}}=\frac{{{10}^{-5}}}{0.005}=0.2\times {{10}^{-2}}\]
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