A) 66%
B) 42%
C) 34%
D) 58%
Correct Answer: C
Solution :
\[\underset{0.001-x}{\mathop{HCOOH}}\,\,{{\underset{x}{\mathop{HCOO}}\,}^{-}}+{{\underset{x}{\mathop{H}}\,}^{+}}\] As \[\alpha =\sqrt{\frac{{{K}_{a}}}{C}}=\sqrt{\frac{1.8\times {{10}^{-4}}}{0.001}}=0.42\] As \[\alpha >0.1\] so can?t be neglected, \[1.8\times {{10}^{-4}}=\frac{{{x}^{2}}}{0.001-x}\] and \[x=3.4\times {{10}^{-4}}\] Now, % ionization \[=\frac{\text{Ionised}\,\text{HCOOH}}{\text{Total}\,\text{HCOOH}}=\frac{3.4\times {{10}^{-4}}}{0.001}\times 100=34%\]
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