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, \[[{{H}^{+}}]={{10}^{-pH}}={{10}^{-5}}\] \[\alpha =\frac{actual\text{ }ionisation}{molar\text{ }concentration}=\frac{{{10}^{-5}}}{0.005}=0.2\times {{10}^{-2}}\]
You need to login to perform this action.
You will be redirected in
3 sec
