A) 0.8%
B) 0.6%
C) 0.4%
D) 0.2%
Correct Answer: D
Solution :
We know that \[[{{H}^{+}}]={{10}^{-pH}}={{10}^{-5}}\] \[\alpha =\frac{\text{actual}\,\text{concentration}}{\text{molar}\,\text{concentration}}\] \[=\frac{{{10}^{-5}}}{0.005}=0.2\times {{10}^{-2}}\] \[\therefore \] percentage ionisation \[=0.2\times {{10}^{-2}}\times 100\] \[=0.2%\]You need to login to perform this action.
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