A) 99.0%
B) 1.00%
C) 99.9%
D) 0.100%
Correct Answer: B
Solution :
\[HA\rightleftharpoons {{H}^{+}}+{{A}^{-}}\] \[{{K}_{a}}=\frac{[{{H}^{+}}]\,[{{A}^{-}}]}{[HA]}=\frac{{{[{{H}^{+}}]}^{2}}}{[HA]}\] \[[{{H}^{+}}]=\sqrt{{{K}_{a}}\,[HA]}=\sqrt{1\times {{10}^{-5}}\times 0.1}\] \[=\sqrt{1\times {{10}^{-6}}}=1\times {{10}^{-3}}\] \[\alpha =\frac{actual\,ionization}{molar\,concentration}=\frac{{{10}^{-3}}}{0.1}={{10}^{2}}\] % of acid dissociated = \[{{10}^{-2}}\times 100\] = 1%You need to login to perform this action.
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